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Electric Charges and Fields Class 12 Notes CBSE Physics Chapter 1 (Free PDF Download)

Last updated date: 15th Jul 2024
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Electric Charge and Field Notes Class 12 For CBSE - Free PDF Downloa

Physics Class 12 chapter 1, Electric Charges and Fields comes under Unit -1 Electrostatics. As per the latest CBSE syllabus for the Class 12 board exam, unit- 1 has quite a good weightage of 8 marks. This chapter emphasizes conceptual understanding and mathematical calculations. You must also know that the Class 12 Physics board exam asks 3-4 numerical questions with a weightage of 10-15 marks. For this, you need to have all the concepts of Electric Charges and Fields at your fingertips. For that Vedantu provides you Electric Charges and Fields Class 12 Notes. 

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Also, check CBSE Class 12 Physics revision notes for all chapters:

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Class12 Physics Ch1 Notes - Basic Subjective Questions

Section-A (1 Marks Questions)

1. Which orientation of an electric dipole in a uniform electric field would correspond to stable equilibrium?

Ans. When dipole moment vector is parallel to electric field $vector \vec{P}\parallel \vec{E}$ .

2. If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the Gaussian surface change?

Ans. Electric flux $\phi _{E}$ is given by

$\phi _{E}=\oint \vec{E}\cdot \vec{ds}=\dfrac{Q}{\varepsilon _{0}}$

where [Q is total charge inside the closed surface ∴ on changing the radius of sphere, the electric flux through the Gaussian surface remains same

3. Figure shows three point charges, +2q, -q and + 3q. Two charges +2q and -q are enclosed within a surface ‘S’. What is the electric flux due to this configuration through the surface ‘S’ 

seo images


Ans.Electric flux $=\oint_{s}^{}\vec{E}\cdot \vec{dS}$

According to Gauss’s’ law, $\phi =\oint_{s}^{}\vec{E}\cdot \vec{dS}=\dfrac{q_{1}}{\varepsilon _{0}}$ … where [q1 is the total charge enclosed by the surface S $\phi =\dfrac{2q-q}{\varepsilon _{0}}=\dfrac{q}{\varepsilon _{0}}\therefore$ electric flux, $\phi =\dfrac{q}{\varepsilon _{0}}$

4. What is the electric flux through a cube of side 1cm which encloses an electric dipole?

Ans. Zero because the net charge of an electric dipole (+ q and – q) is zero.

5. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?

Ans. The electric flux due to a point charge enclosed by a spherical gaussian surface remains ‘unaffected’ when its radius is increased.

Section – B (2 Marks Questions)

6. Differentiate between uniform electric field and non-uniform electric field. 

Ans. Uniform electric field: In case of uniform electric field, the field lines are straight and equally spaced. Non- uniform electric field: in case of non uniform electric field, the field lines are not straight.

Uniform electric field and Non- uniform electric field

7. Is the electric field due to a charge configuration with total charge zero, necessarily zero? Justify. 

Ans. No, it is not necessarily zero. If the electric field due to a charge configuration with total charge is zero because the electric field due to an electric dipole is non-zero.

8. Define the term electric dipole moment of a dipole. State its S.I. unit

Ans. τ = OE sin θ

If E = 1 unit, θ = 90°, then τ = P

Dipole moment may be defined as the torque acting on an electric dipole, placed perpendicular to a uniform electric dipole, placed perpendicular to a uniform electric field of unit strength. or Strength of the electric dipole is called the dipole moment.

$\left | \vec{P} \right | =q\left | 2a \right |$

∴ SI unit is Cm.

9. In which orientation, a dipole placed in a uniform electric field is in

(a) stable,

(b) unstable equilibrium? 

Ans. For stable equilibrium, a dipole is placed parallel to the electric field.

For unstable equilibrium, a dipole is placed antiparallel to the electric field.

10. Define one coulomb. What is the SI unit of a charge?

Ans. One coulomb is the charge transferred in one second across a section of wire carrying current of one ampere.

SI unit of a charge is coulomb (C).

PDF Summary - Class 12 Physics Electric Charges and Fields Notes (Chapter 1)

1. Electric Charge

1.1 Definition- Charge is that property that is associated with the matter due to which it produces and experiences electrical and magnetic effects.

1.2 Type

There exist two types of charges in nature. They are:

i. Positive charge

ii. Negative charge

Charges with the same electrical sign repel each other while charges with opposite electrical signs attract each other.

1.3 Unit and Dimensional Formula

S.I. unit of charge is coulomb (C),

$\left( {1{\text{mC}} = {{10}^{ - 3}}{\text{C}},1\mu {\text{C}} = {{10}^{ - 6}}{\text{C}},\ln {\text{C}} = {{10}^{ - 9}}{\text{C}}} \right)$

C.G.S. the unit of charge is e.s.u. $1{\text{C}} = 3 \times {10^9}$ esu

The Dimensional formula is given by $[{\text{Q}}] = [{\text{AT}}]$.

1.4 Point Charge

Whose spatial size is negligible as compared to other distances.

1.5 Properties of Charge

(i) Charge is a Scalar Quantity: Charges can be added or subtracted algebraically.

(ii) Charge is transferable: When a charged body is put in contact with an uncharged body, the uncharged body becomes charged due to transfer of electrons from the charged body to the uncharged body.

(iii) Charge is always associated with mass: Charge cannot exist without mass though mass can exist without charge.

(iv) Charge is conserved: Charge can neither be created nor be destroyed.

(v) Invariance of charge: The numerical value of an elementary charge is independent of velocity.

(vi) Charge produces an electric field and magnetic field: When a charged particle is at rest it only produces an electric field in the space surrounding it. However, if the charged particle is in unaccelerated motion it produces both electric and magnetic fields. And if the motion of the charged particle is accelerated it not only produces electric and magnetic fields but also radiates energy in the space surrounding the charge in the form of electromagnetic waves.

(vii) Charge resides on the surface of conductor: Charge resides on the outer surface of a conductor because like charges repel and try to get as far away as possible from one another and stay at the farthest distance from each other which is the outer surface of the conductor. Therefore a solid and hollow conducting sphere of the same outer radius will hold a maximum equal charge and a soap bubble expands on charging.

(viii) Quantization of charge: When a physical quantity can have only discrete values rather than any value, the quantity is said to be quantised. The smallest charge that can exist in nature is the charge of an electron. If the charge of an electron $\left( { - 1.6 \times {{10}^{ - 19}}{\text{C}}} \right)$ is taken as elementary unit i.e. quanta of charge the charge on anybody will be some integral multiple of e i.e., ${\text{Q}} =  \pm ne$ with ${\text{n}} = 0$, $1,2,3 \ldots  \ldots $

Charge on a body can never be $0.5\,e,\, \pm 17.2e$ or $ \pm {10^{ - 5}}e$ etc.

1.6 Comparison of Charge and Mass

We are familiar with the role of mass in gravitation and we have just studied some features of electric charge. The comparison between the two are as follows:

Sl. No




Electric charge can be positive, negative or zero.

Mass of a body is a positive quantity.


Charge carried by a body does not depend upon the velocity of the body.

Mass of a body increases with its velocity as 

$m=\dfrac {m_o}{\sqrt{1-{\dfrac{v^2}{c^2}}}}$ where $c$ id velocity of light in vacuum, $m$ is the mass of the velocity $v$ and ${m_0}$ is rest mass of the body.


The charge is quantized.

The quantization of mass is yet to be established.


Electric charge is always conserved.

Mass is not conserved as it can be changed into energy and vice-versa.


Force between charges can be attractive(unlike charges) or repulsive(like charges) in nature.

The gravitational force between two masses is always attractive.

1.7 Methods of Charging

A body can be charged by following methods:

i. By friction: 

In friction when two bodies are rubbed together, electrons are transferred from one body to the other. As a result of this one body becomes positively charged while the other is negatively charged, e.g., when a glass rod is rubbed with silk, the rod becomes positively charged while the silk becomes negatively charged. However, ebonite on rubbing with wool becomes negatively charged making the wool positively charged. Clouds also become charged by friction. In charging by friction in accordance with conservation of charge, both positive and negative charges in equal amounts appear simultaneously due to the transfer of electrons from one body to the other.

ii. By electrostatic induction: 

If a charged body is brought near an uncharged body, the charged body will attract the opposite charge and repel a similar charge present in the uncharged body. As a result of this one side of the neutral body (closer to charged body) becomes oppositely charged while the other is similarly charged. This process is called electrostatic induction.

(Image will be uploaded soon)

Note: Inducting body neither gains nor loses charge. 

(iii) Charging by conduction: 

Take two conductors, one charged and the other uncharged. Bring the conductors in contact with each other. The charge (whether negative or positive) under its own repulsion will spread over both the conductors. Thus, the conductors will be charged with the same sign. This is called as charging by conduction (through contact).

Note: A truck carrying explosives has a metal chain touching the ground, to conduct away the charge produced by friction.

1.8 Electroscope

(Image will be uploaded soon)

It is a simple apparatus with which the presence of an electric charge on a body is detected (see figure). When a metal knob is touched with a charged body, some charge is transferred to the gold leaves, which then diverges due to repulsion. The separation gives a rough idea of the amount of charge on the body. If a charged body is brought near a charged electroscope the leaves will also diverge. If the charge on the body is like that on an electroscope and will usually converge if opposite. If the induction effect is strong enough leaves after converging may again diverge.

(1) Uncharged electroscope

(2) Charged electroscope

(Image will be uploaded soon)

2. Coulomb's Law

If two stationary and point charges ${Q_1}$ and ${Q_2}$ are kept at a distance $r$, then it is found that force of attraction or repulsion between them is Mathematically, Coulomb's law can be written as

${\text{F}} = {\text{k}}\dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}$

where ${\text{k}}$ is a proportionality constant.

In SI units ${\text{k}}$ has the value,

${\text{k}} = 8.988 \times {10^9}{\text{N}}{{\text{m}}^2}/{{\text{C}}^2}$ 

${\text{k}} = 9.0 \times {10^9}{\text{N}}{{\text{m}}^2}/{{\text{C}}^2}$

(a) The direction of force is always along the line joining the two charges.

(b) The force is repulsive if the charges have the same sign and attractive if their signs are opposite.

(c) This force is conservative in nature.

(d) This is also called inverse square law.

2.1 Variation of $k$

Constant $k$ depends upon a system of units and medium between the two charges.

  1. In C.GS. for air 

$k=1,\;{\text{F}}=\dfrac{{{{\text{Q}}_1}{{\text{Q}}_2}}} {{{{\text{r}}^2}}} {\text{Dyne}}$

  1. In S.I. for air $k = \dfrac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\dfrac{{{\text{N}} - {{\text{m}}^2}}}{{{{\text{C}}^2}}}$,

${\text{F}} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{{{\text{Q}}_1}{{\text{Q}}_2}}}{{{{\text{r}}^2}}}{\text{ Newton }}\left( {1{\text{ Newton }} = {{10}^5}} \right.{\text{ Dyne) }}$


  • ${\varepsilon _0} = $ Absolute permittivity of air or free space

$ = 8.85 \times {10^{ - 12}}\dfrac{{{C^2}}}{{N - {m^2}}}\left( { = \dfrac{{{\text{ Farad }}}}{m}} \right)$

Dimension is $\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]$

  • ${\varepsilon _0}$ Relates with absolute magnetic permeability $\left( {{\mu _0}} \right)$ and velocity of light $(c)$ according to the following relation $c = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$

2.1.2 Effect of Medium

(a) When a dielectric medium is completely filled in between charges rearrangement of the charges inside the dielectric medium takes place and the force between the same two charges decreases by a factor of $K$ known as dielectric constant, $K$ is also called relative permittivity ${\varepsilon _r}$ of the medium (relative means with respect to free space).

Hence in the presence of medium

${F_m} = \dfrac{{{F_{{\text{air }}}}}}{K} = \dfrac{1}{{4\pi {\varepsilon _0}K}} \cdot \dfrac{{{Q_1}{Q_2}}}{{{r^2}}}$

Here ${\varepsilon _0}\;{\text{K}} = {\varepsilon _0}{\varepsilon _{\text{r}}} = \varepsilon $ (permittivity of medium)












$\infty $

2.2 Vector Form of Coulomb's Law

It is helpful to adopt a convention for subscript notation.

${{\text{F}}_{12}} = $ force on 1 due to $2\quad \;{{\text{F}}_{21}} = $ force on 2 due to 1

(Image will be uploaded soon)

Suppose the position vectors of two charges ${{\text{q}}_1}$ and ${{\text{q}}_2}$ are ${\overrightarrow {\text{r}} _1}$ and ${\overrightarrow {\text{r}} _2}$, then, electric force on charge ${{\text{q}}_1}$ due to charge ${{\text{q}}_2}$ is,

${\overrightarrow {\text{F}} _{12}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{{\left| {{{\overrightarrow {\text{r}} }_1} - {{\overrightarrow {\text{r}} }_2}} \right|}^3}}}\left( {{{\overline {\text{r}} }_1} - {{\overrightarrow {\text{r}} }_2}} \right)$

Similarly, electric force on ${{\text{q}}_2}$ due to charge ${{\text{q}}_1}$ is

${{\text{\vec F}}_{{\text{21}}}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{{\left| {{{\vec r}_2} - {{\vec r}_1}} \right|}^3}}}\left( {{{\vec r}_2} - {{\vec r}_1}} \right)$

Force is a vector, so in vector form the Coulomb's law is written as

${\overrightarrow {\text{F}} _{12}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}{\widehat {\text{r}}_{12}}$

Where ${\widehat {\text{r}}_{12}}$ is a unit vector, directed toward ${{\text{q}}_1}$ from ${{\text{q}}_2}$.


\[{\widehat r_{12}} =  - {\widehat r_{21}}\]

\[{\overrightarrow {\text{F}} _{12}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{{{\text{r}}^2}}}{\widehat {\text{r}}_{12}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{{\text{q}}_2}{{\text{q}}_1}}}{{{{\text{r}}^2}}}\left( { - {{\widehat {\text{r}}}_{21}}} \right)\]

\[{\overrightarrow {\text{F}} _{12}} =  - \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{{\text{q}}_2}{{\text{q}}_1}}}{{{{\text{r}}^2}}}{\widehat {\text{r}}_{21}} =  - {\overrightarrow {\text{F}} _{21}} \]

Remember convention for $\widehat {\text{r}}$.

Here ${q_1}$ and ${q_2}$ are to be substituted with sign. Position vector of charges ${q_1}$ and ${q_2}$ are ${\vec r_1} = {x_1}\hat i + {y_1}\hat j + {z_1}\hat k$ and ${\vec r_2} = {x_2}\hat i + {y_2}\hat j + {z_2}\hat k$ respectively. Where $\left( {{x_1},{y_1},{z_1}} \right)$ and  $\left( {{x_2},{y_2},\,\,{z_2}} \right)$ are the co-ordinates of charges ${{\text{q}}_1}$ and ${{\text{q}}_2}$.

2.3 Principle of Superposition

According to the principle of superposition, the total force acting on a given charge due to a number of charges is the vector sum of the individual forces acting on that charge due to all the charges.

Consider number of charges Q1, Q2, Q3…..are applying force on a charge $Q$

Net force on $Q$ will be

${\overrightarrow {\text{F}} _{{\text{net }}}} = {\overrightarrow {\text{F}} _1} + {\overrightarrow {\text{F}} _2} +  \ldots  \ldots  \ldots  + {\overrightarrow {\text{F}} _{{\text{n}} - 1}} + {\overrightarrow {\text{F}} _{\text{n}}}$

The magnitude of the resultant of two electric force is given by ${\text{F}} = \sqrt {{\text{F}}_1^2 + {\text{F}}_2^2 + 2\;{{\text{F}}_1}\;{{\text{F}}_2}\cos \theta } $ and the force direction is given by,

 $\tan \alpha  = \dfrac{{{F_2}\sin \theta }}{{{F_1} + {F_2}\cos \theta }}$

3. Electric Field

A positive charge or a negative charge is said to create its field around itself. Thus space around a charge in which another charged particle experiences a force is said to have an electrical field in it.

3.1 Electric Field Intensity $\left( {\overrightarrow {\text{E}} } \right)$

The electric field intensity at any point is defined as the force experienced by a unit positive charge placed at that point. 

$\vec E = \dfrac{{\vec F}}{{{q_0}}}$

Where ${q_0} \to 0$ so that presence of this charge may not affect the source charge $Q$ and its electric field is not changed, therefore expression for electric field intensity can be better written as: 

$\vec E = \mathop {\lim }\limits_{{q_0} \to 0} \dfrac{{\vec F}}{{{q_0}}}$

(a) Unit and Dimensional formula: 

It's S.I. unit is,

$\dfrac{{{\text{ Newton }}}}{{{\text{ coulomb }}}} = \dfrac{{{\text{ volt }}}}{{{\text{ meter }}}} = \dfrac{{{\text{ Joule }}}}{{{\text{ coulomb }} \times {\text{ meter }}}}$ and

C.G.S. unit =${\text{Dyne/stat}}\,\,{\text{coulomb}}.$

Dimension: $\left[ {\text{E}} \right] = \left[ {{\text{ML}}{{\text{T}}^{ - 3}}{{\text{A}}^{ - 1}}} \right]$

(b) Direction of electric field: Electric field (intensity) $\overrightarrow {\text{E}} $ is a vector quantity. Electric field due to a positive charge is always away from the charge and that due to a negative charge is always towards the charge.

3.2 Relation Between Electric Force and Electric Field

In an electric field ${\text{\vec E}}$ a charge $(Q)$ experiences a force $F = QE$. If the charge is positive then force is directed in the direction of the field while if the charge is negative force acts on it in the opposite direction of field.

3.3 Superposition of Electric Field

The resultant electric field at any point is equal to the vector sum of electric fields at that point due to various charges.

$\overrightarrow {\text{E}}  = {\overrightarrow {\text{E}} _1} + {\overrightarrow {\text{E}} _2} + {\overrightarrow {\text{E}} _3} +  \ldots $

The magnitude of the resultant of two electric fields are given by

${\text{E}} = \sqrt {{\text{E}}_1^2 + {\text{E}}_2^2 + 2{{\text{E}}_1}{{\text{E}}_2}\cos \theta } $ and the direction is given by,

$\tan \alpha  = \dfrac{{{E_2}\sin \theta }}{{{E_1} + {E_2}\cos \theta }}$

3.4 Point Charge

Point charge produces its electric field at a point ${\text{P}}$ which is distance $r$ from it given by,

${{\text{E}}_{\text{P}}} = \dfrac{{\text{Q}}}{{4\pi {\varepsilon _0}{{\text{r}}^2}}}$ (Magnitude)

For positive point charge, ${\text{E}}$ is directed away from it.

For negative point charge, ${\text{E}}$ is directed towards it.

3.5 Continuous Charge Distributions

There is an infinite number of ways in which we can spread a continuous charge distribution over a region of space. Mainly three types of charge distributions will be used. We define three different charge densities.



SI units

(lambda)$\lambda $

Charge per unit length


(sigma) $\sigma $

Charge per unit area


(rho)$\rho $

Charge per unit volume


If a total charge $q$ is distributed along a line of length $\ell $, over a surface area A or throughout a volume V, we can calculate charge densities from.

$\lambda  = \dfrac{{\text{q}}}{\ell },\sigma  = \dfrac{{\text{q}}}{{\text{A}}},\rho  = \dfrac{{\text{q}}}{{\text{V}}}$

3.6 Properties of Electric Field Lines

1. Electric field lines originate from a positive charge & terminate on a negative charge.

2. The number of field lines originating/terminating on a charge is proportional to the magnitude of the charge.

3. The number of Field Lines passing through the perpendicular unit area will be proportional to the magnitude of the Electric Field there.

4. Tangent to a Field line at any point gives the direction of the Electric Field at that point. This will be the instantaneous path charge will take if kept there.

5. Two or more field lines can never intersect each other.

(they cannot have multiple directions)

6. Uniform field lines are straight, parallel & uniformly placed.

7. Field lines cannot form a loop.

8. Electric field lines originate and terminate perpendicular to the surface of the conductor. Electric field lines do not exist inside a conductor.

9. Field lines always flow from higher potential to lower potential.

10. If in a region electric field is absent, there will be no field lines.

3.7 Motion of Charged Particle in an Electric Field

(a) When charged particle initially at rest is placed in the uniform field:

Let a charge particle of mass $m$ and charge $Q$ be initially at rest in an electric field of strength $E$.

(Image will be uploaded soon)

(i) Force and acceleration

The force experienced by the charged particle is $F = QE$. Positive charge experiences force in the direction of electric field while negative charge experiences force in the direction opposite to the field. [Fig. (A)]

Acceleration produced by this force is ${\text{a}} = \dfrac{{\text{F}}}{{\text{m}}} = \dfrac{{{\text{QE}}}}{{\text{m}}}$

Since the field ${\text{E}}$ in constant the acceleration is constant, thus motion of the particle is uniformly accelerated.

(ii) Velocity: 

Suppose at point $A$ particle is at rest and in time $t$, it reaches the point $B$ 

(Image will be uploaded soon)

$V = $ Potential difference between $A$ and $B$

$S = $ Separation between $A$ and $B$

(a) By using

$v = {\text{u}} + {\text{at}},\quad v = 0 + {\text{Q}}\dfrac{{\text{E}}}{{\text{m}}}$

${\text{t}} \Rightarrow v = \dfrac{{{\text{QEt}}}}{{\text{m}}}$

(b) By using,

${{\text{v}}^{\text{2}}}{\text{ = }}{{\text{u}}^{\text{2}}}{\text{ + 2as,}}{{\text{v}}^{\text{2}}}{\text{ = 0 + 2 \times }}\dfrac{{{\text{QE}}}}{{\text{m}}}{\text{ \times s = }}\sqrt {\dfrac{{{\text{2QEs}}}}{{\text{m}}}} $

(iii) Momentum: 

Momentum $p = mv$,

${\text{p}} = {\text{m}} \times \dfrac{{{\text{QEt}}}}{{\text{m}}} = {\text{QEt}}$

(iv) Kinetic energy: 

Kinetic energy gained by the particle in time $t$ is

${\text{K}} = \dfrac{1}{2}{\text{m}}{{\text{v}}^2} = \dfrac{1}{2}\;{\text{m}}\dfrac{{{{({\text{QEt}})}^2}}}{{\;{\text{m}}}} = \dfrac{{{{\text{Q}}^2}{{\text{E}}^2}{{\text{t}}^2}}}{{2\;{\text{m}}}}$

(b) When a charged particle enters with an initial velocity at a right angle to the uniform field.

When a charged particle enters perpendicularly in an electric field, it describes a parabolic path as shown.

(i) Equation of trajectory: 

Throughout the motion, a particle has uniform velocity along$x$-axis and horizontal displacement$(x)$ is given by the equation $x = ut$

Since the motion of the particle is accelerated along $y$-axis, we will use the equation of motion for uniform acceleration to determine displacement y.

From ${\text{S}} = {\text{ut}} + \dfrac{1}{2}{\text{a}}{{\text{t}}^2}$

We have $u = 0$ (along $y - $axis) so $y = \dfrac{1}{2}a{t^2}$

i.e., displacement along $y$-axis will increase rapidly with time (since y $\left. { \propto {{\text{t}}^2}} \right)$

From displacement along ${\text{x}}$-axis, ${\text{t}} = {\text{x}}/{\text{u}}$

So $y = \dfrac{1}{2}\left( {\dfrac{{QE}}{m}} \right){\left( {\dfrac{x}{u}} \right)^2};$ this is the equation of parabola which shows ${\text{y}} \propto {{\text{x}}^2}$.

(ii) Velocity at any instant: 

At any instant $t,{v_x} = u$ and ${v_{\text{y}}} = \dfrac{{{\text{QEt}}}}{{\text{m}}}$.


$v = |\vec v| = \sqrt {v_{\text{x}}^2 + v_{\text{y}}^2}  = \sqrt {{{\text{u}}^2} + \dfrac{{{{\text{Q}}^2}{{\text{E}}^2}{{\text{t}}^2}}}{{\;{{\text{m}}^2}}}} $

If $\beta $ is the angle made by $v$ with $x$-axis than

$\tan \beta  = \dfrac{{{v_y}}}{{{v_x}}} = \dfrac{{{\text{QEt}}}}{{{\text{mu}}}}{\text{.}}$

4. Electric Potential Energy

4.1 Potential Energy of 2 Charges System

It is always change in potential energy that is designed as $\Delta U =  - {W_{{\text{corcersatire fore }}}} =  - {W_{{\text{Coulamb forre }}}}$

Potential energy is defined of a system of charges in a particular configuration.

Consider a system of two charges ${q_1}$ and ${q_2}.$ Suppose, the charge ${{\text{q}}_1}$ is fixed and the charge ${q_2}$ is taken from a point A to B.

The electric force on the charge ${q_2}$ is $\vec F = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}\hat r$

The total work done as the charge ${q_2}$ moves from $B$ to $C$ is ${\text{W}} = \int_{{{\text{r}}_1}}^{{{\text{r}}_2}} {\dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{4\pi {\varepsilon _0}{{\text{r}}^2}}}} {\text{dr}} = \dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{{\text{r}}_1}}} - \dfrac{1}{{{{\text{r}}_2}}}} \right)$

The change in potential energy ${\text{U}}\left( {{{\text{r}}_2}} \right) - {\text{U}}\left( {{{\text{r}}_1}} \right)$ is, therefore,

${\text{U}}\left( {{{\text{r}}_2}} \right) - {\text{U}}\left( {{{\text{r}}_1}} \right) =  - {\text{W}} = \dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{{{\text{r}}_2}}} - \dfrac{1}{{{{\text{r}}_1}}}} \right)$

The potential energy of the two-charge system is assumed to be zero when they have infinite separation.

The potential energy when the separation is $r$ is,

${\text{U}}({\text{r}}) = {\text{U}}({\text{r}}) - {\text{U}}(\infty ) = \dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{4\pi {\varepsilon _0}}}\left( {\dfrac{1}{{\text{r}}} - \dfrac{1}{\infty }} \right) = \dfrac{{{{\text{q}}_1}{{\text{q}}_2}}}{{4\pi {\varepsilon _0}{\text{r}}}}$

The potential energy depends essentially on the separation between the charges and is independent of the spatial location of the charged particles.

The equation gives the electric potential energy of a pair of charges.


Electric potential energy is a scalar quantity so in the above formula take sign of ${Q_1}$ and ${Q_2}$.

4.2 Electron Volt (eV)

It is the smallest practical unit of energy used in atomic and nuclear physics. As electron volt is defined as "the energy acquired by a particle having one quantum of charge 1 e when accelerated by 1 volt i.e.,

$1{\text{eV}} = 1.6 \times {10^{ - 19}}{\text{C}} \times \dfrac{{1\;{\text{J}}}}{{\text{C}}} = 1.6 \times {10^{ - 19}}\;{\text{J}} = 1.6 \times {10^{ - 12}}{\text{erg}}$

4.3 Potential Energy of a System of n Charges

In a system of $n$ charges electric potential energy is calculated for each pair and then all energies so obtained are added algebraically. i.e., 

${\text{U}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{{{\text{Q}}_1}{{\text{Q}}_2}}}{{{{\text{r}}_{12}}}} + \dfrac{{{{\text{Q}}_2}{{\text{Q}}_3}}}{{{{\text{r}}_{23}}}} +  \ldots  \ldots  \ldots ..} \right]$  and in case of continuous distribution of charge.

As ${\text{dU}} = {\text{dQ}} \cdot {\text{V}} \Rightarrow {\text{U}} = \int {\text{V}} {\text{dQ}}$

e.g., Electric potential energy for a system of three charges:

${\text{ Potential energy }} = \dfrac{1}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{{Q_1}{Q_2}}}{{{r_{[2}}}} + \dfrac{{{Q_2}{Q_3}}}{{{r_{23}}}} + \dfrac{{{Q_3}{Q_1}}}{{{r_{31}}}}} \right]$

5. Electric Potential

Suppose a test charge $q$ is moved in an electric field from a point ${\text{A}}$ to a point ${\text{B}}$ while all the other charges in question remain fixed. If the electric potential energy changes by ${{\text{U}}_{\text{B}}} - {{\text{U}}_{\text{A}}}$  due to this displacement, we define the potential difference between the point ${\text{A}}$ and the point ${\text{B}}$ as

${\text{\Delta V = }}\dfrac{{{\text{\Delta U}}}}{{\text{q}}}\,\,{\text{i}}{\text{.e}}{\text{.,}}{{\text{V}}_{\text{B}}} - {{\text{V}}_{\text{A}}} = \dfrac{{{{\text{U}}_{\text{B}}} - {{\text{U}}_{\text{A}}}}}{{\text{q}}}{\text{ = }}\dfrac{{{{\text{W}}_{{\text{ext}}}}}}{{\text{q}}}\,\,\,\left[ {{\text{\Delta KE}} = 0} \right]$

Conversely, if a charge $q$ is taken through a potential difference 

${{\text{V}}_{\text{B}}} - {{\text{V}}_{\text{A}}}$, the electric potential energy is increased by  ${{\text{U}}_{\text{B}}} - {{\text{U}}_{\text{A}}} = {\text{q}}\left( {{{\text{V}}_{\text{B}}} - {{\text{V}}_{\text{A}}}} \right)$

Also ${W_{{\text{ext }}}} = q\left( {{V_B} - {V_\beta }} \right)[\Delta KE = 0]$

Potential difference between two points give us an idea about work which has to be done in moving a charge between those points.

5.1 Electric Potential Due to a Point Charge

Consider a point charge Q placed at a point A. The potential at P is,

${{\text{V}}_{\text{P}}} - {{\text{V}}_\infty } = \dfrac{{{{\text{U}}_{\text{P}}} - {{\text{U}}_\infty }}}{{\text{q}}} = \dfrac{{\text{Q}}}{{4\pi {\varepsilon _0}{\text{r}}}} \Rightarrow {{\text{V}}_{\text{P}}} = \dfrac{{\text{Q}}}{{4\pi {\varepsilon _0}{\text{I}}}}$

$\left( {\because {{\text{V}}_\infty }} \right.$ is taken as 0 )

The electric potential due to a system of charges may be obtained by finding potentials due to the individual charges using equation and then adding them. Thus,

${\text{V}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\sum {\dfrac{{{{\text{Q}}_{\text{i}}}}}{{{{\text{r}}_{\text{i}}}}}} $

Electric potential is a scalar quantity, hence sign of charges is to taken in expression it is denoted by $V$

5.2 Unit and Dimensional Formula

\[{\text{S}}{\text{. I}}{\text{. unit}} - \dfrac{{{\text{ Joule }}}}{{{\text{ Coulomb }}}} = {\text{volt}} \]

\[[{\text{V}}] = \left[ {{\text{M}}{{\text{L}}^2}\;{{\text{T}}^{ - 3}}\;{{\text{A}}^{ - 1}}} \right] \] 

5.3 Types of Electric Potential

According to the nature of charge potential is of two types

(i) Positive potential: Due to positive charge.

(ii) Negative potential: Due to negative charge.


  • At the centre of two equal and opposite charge ${\text{V}} = 0$ but ${\text{E}} \ne 0$.

  • At the centre of the line joining two equal and similar charge ${\text{V}} \ne 0,{\text{E}} = 0$.

  • If left free to move,

The positive charge will always move from higher to lower potential points.

A negative charge will always move from lower to higher potential points.

(Because this motion will decrease the potential energy of a system)

6. Relation Between Electric Field & Potential

In an electric field rate of change of potential with distance is known as a potential gradient. It is a vector quantity and it's direction is opposite to that of the electric field. Potential gradient relates with electric field according to the following relation ${\text{E}} =  - \dfrac{{{\text{dV}}}}{{{\text{dr}}}}$; This relation gives another unit of electric field is $\dfrac{{{\text{ volt }}}}{{{\text{ meter }}}}$. In the above relation negative sign indicates that in the direction of the electric field potential decreases.

In space around a charge distribution, we can also write,

$\overrightarrow {\text{E}}  = {{\text{E}}_{\text{Z}}}\widehat {\text{i}} + {{\text{E}}_{\text{y}}}\widehat {\text{j}} + {{\text{E}}_z}\widehat {\text{k}}$

where ${{\text{E}}_{\text{x}}} =  - \dfrac{{{\text{dV}}}}{{{\text{dx}}}},{{\text{E}}_{\text{y}}} =  - \dfrac{{{\text{dV}}}}{{{\text{dy}}}}$ and ${{\text{E}}_{\text{z}}} =  - \dfrac{{{\text{dV}}}}{{{\text{d}}z}}$

Suppose A, B and C are three points in an uniform electric field as shown in figure.

(Image will be uploaded soon)

  1. Potential difference between point $A$ and $B$ is,

${{\text{V}}_{\text{B}}} - {{\text{V}}_{\text{A}}} =  - \int_{\text{A}}^{\text{B}} {\overrightarrow {\text{E}} }  \cdot \overrightarrow {{\text{dr}}} $

Since displacement is in the direction of electric field, hence $\theta  = {{0}^{0}}$

So, ${V_B} - {V_A} =  - \int_a^B E .dr\cos 0 =  - \int_a^B E  \cdot dr =  - Ed$

Equipotential Surface or Lines

(1) If every point of a surface is at the same potential, then it is said to be an equipotential surface dark for a given charge distribution, locus of all points having the same potential is called "equipotential surface" regarding equipotential surface following points should keep in mind:

(2) The direction of the electric field is perpendicular to the equipotential surfaces or lines.

(3) The equipotential surfaces produced by a point charge or a spherically charge distribution are a family of concentric spheres.

(4) For a uniform electric field, the equipotential surfaces are a family of planes perpendicular to the field lines.

(5) A metallic surface e of any shape is an equipotential surface e.g. When a charge is given to a metallic surface, it distributes itself in a manner such that its every point comes at the same potential even if the object is of irregular shape and has sharp points on it.

(6) Equipotential surfaces can never cross each other. It is a common misconception that the path traced by a positive test charge is a field line but actually the path traced by a unit positive test charge represents a field full line only when it moves along a straight line.

7. Electric Dipole

7.1 General Information

A system of two equal and opposite charges separated by a small fixed distance is called a dipole.

(i) Dipole axis: Line joining negative charge to positive charge of a dipole is called its axis. It may also be termed as its longitudinal axis.

(ii) Equatorial axis: Perpendicular bisector of the dipole is called its equatorial or transverse axis as it is perpendicular to the length.

(iii). Dipole length: The distance between two charges is known as dipole length (d).

(iv). Dipole moment: It is a quantity that gives information about the strength of dipole. It is a vector quantity and is directed from negative charge to positive charge along the axis. It is denoted as p and is defined as the product of the magnitude of either of the charge and the dipole length.

i.e., $\vec p = q(\vec d)$

Its S.I. unit is coulomb-metre or Debye (1 Debye = $\left. {3.3 \times {{10}^{ - 30}}C \times m} \right)$ and its dimensions are ${M^0}{L^1}{T^1}{A^1}$.


  • A region surrounding a stationary electric dipole has electric field only.

  • When a dielectric is placed in an electric field, its atoms or molecules are considered as tiny dipoles.

(a) Electric Potential due to a dipole

${{\text{V}}_{\text{P}}} = \dfrac{{{\text{k}}( - q)}}{{{\text{AP}}}} + \dfrac{{{\text{k}}( + {\text{q}})}}{{{\text{BP}}}}$

$r >  > d$ (distance ' $r$ ' is large as compared to d)

${\text{AP}} \approx {{\text{O}}^{\prime \prime }}{\text{P}};\quad {\text{BP}} \approx {{\text{O}}^\prime }{\text{P}}$

${O^{\prime \prime }}{\text{P}} = {\text{r}} + {\text{d}}/2\cos \theta ,\quad O{\text{P}} = {\text{r}} - {\text{d}}/2\cos \theta $

${{\text{V}}_{\text{p}}} = \dfrac{{{\text{k}}( - {\text{q}})}}{{({\text{r}} + {\text{d}}/2\cos \theta )}} + \dfrac{{{\text{k}}( + {\text{q}})}}{{({\text{r}} - {\text{d}}/2\cos \theta )}}$

$ = {\text{k}}( + {\text{q}})\left[ {\dfrac{1}{{{\text{r}} - {\text{d}}/2\cos \theta }} - \dfrac{1}{{{\text{r}} + {\text{d}}/2\cos \theta }}} \right]$

$ = {\text{kq}}\left[ {\dfrac{{r + {\text{d}}/2\cos \theta  - {\text{r}} + {\text{d}}/2\cos \theta }}{{{{\text{r}}^2} - \dfrac{{{{\text{d}}^2}}}{4}{{\cos }^2}\theta }}} \right] = \dfrac{{{\text{kqd}}\cos \theta }}{{{{\text{r}}^2} - \dfrac{{{{\text{d}}^2}}}{4}{{\cos }^2}\theta }}$

${{\text{V}}_p} = \dfrac{{{\text{k}}({\text{qd}})\cos \theta }}{{{{\text{r}}^2} - \dfrac{{{{\text{d}}^2}}}{4}{{\cos }^2}\theta }} = \dfrac{{{\text{kp}}\cos \theta }}{{{{\text{r}}^2} - \dfrac{{{{\text{d}}^2}}}{4}{{\cos }^2}\theta }}[\because {\text{p}} = {\text{qd}}]$

since $r >  > d$

${{\text{V}}_{\text{p}}} = \dfrac{{{\text{kp}}\cos \theta }}{{{{\text{r}}^2}}} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{{\text{p}}\cos \theta }}{{{{\text{r}}^2}}}$

$\theta $ is angle with the axis of dipole; ${\text{r}}$ is distance from centre of dipole.

(b) Electric Field due to dipole

(i) For points on the axis

Let the point ${\text{Pbe}}$ at distance ${\text{r}}$ from the centre of the dipole on the side of the charge $q$,as shown in the below figure.

(Image will be uploaded soon)


${E_{ - q}} =  - \dfrac{q}{{4\pi {z_0}{{(r + a)}^2}}}\hat p$

where $\widehat {\text{p}}$ is the unit vector along the dipole axis (from \[ - q\] to \[q\] ). Also, ${{\text{E}}_{ + {\text{q}}}} = \dfrac{{\text{q}}}{{4\pi {\varepsilon _0}{{({\text{r}} - {\text{a}})}^2}}}\widehat {\text{p}}$

The total field at P is,

\[E = {E_{ + q}} + {E_{ - q}} = \dfrac{q}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{{{{(r - a)}^2}}} - \dfrac{1}{{{{(r + a)}^2}}}} \right]\hat p \]

\[= \dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{{4ar}}{{{{\left( {{r^2} - {a^2}} \right)}^2}}}\hat p \]

For $r >> a$

$E = \dfrac{{4qa}}{{4\pi {\varepsilon _0}{r^3}}}\hat p\quad (r >  > a)$…(i)

(ii) For points on the equatorial plane

The magnitudes of the electric fields due to the two charges $ + {\text{q}}$ and $ - {\text{q}}$ are given by,

\[{E_{ + q}} = \dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{1}{{{r^2} + {a^2}}} \]

\[{E_{ - q}} = \dfrac{q}{{4\pi {\varepsilon _0}}}\dfrac{1}{{{r^2} + {a^2}}} \]

and are equal.

The directions of ${E_{ + q}}$ and ${E_{ - q}}$ are as shown in fig. (b). Clearly, the components normal to the dipole axis cancel away. The components along the dipole axis add up. The total electric field is opposite to $\widehat {\text{p}}$. We have,

\[{\text{E}} =  - \left( {{{\text{E}}_{ + {\text{q}}}} + {{\text{E}}_{ - \alpha }}} \right)\cos \theta \widehat {\text{p}} \]

 \[=  - \dfrac{{2{\text{qa}}}}{{4\pi {\varepsilon _0}{{\left( {{{\text{r}}^2} + {{\text{a}}^2}} \right)}^{3/2}}}}\widehat {\text{p}} \]

At large distances $(r >  > a)$, this reduces to

$E =  - \dfrac{{2qa}}{{4\pi {\varepsilon _0}{r^3}}}{\hat p_{(r >  > a)}}$…(ii)

From Eqs. (i) and (ii), it is clear that the dipole field at large distances does not involve $q$ and a separately; it depends on the product $qa$. This suggests the definition of dipole is defined by

${\text{P}} = {\text{q}} \times 2{\text{a}}\widehat {\text{p}}$

that is, it is a vector whose magnitude is charge $q$ times the separation $2a$ (between the pair of charges $q$, $ - q$) and the direction is along the line from $ - q$ to $q$. In tarms of p, the electric field of a dipole at large distances takes simple forms:

At a point on the dipole axis

$E = \dfrac{{2p}}{{4\pi {\varepsilon _0}{r^3}}} = \dfrac{{2kp}}{{{r^3}}}(r >  > a)$

At a point on the equatorial plane.

${\text{E}} = \dfrac{{\text{p}}}{{4\pi {\varepsilon _0}{{\text{r}}^3}}} = \dfrac{{ - {\text{kp}}}}{{{{\text{r}}^3}}}({\text{r}} >  > {\text{a}})$

7.3 Electric Dipole in a Uniform Electric Field

(i) Force and Torque: If a dipole is placed in a uniform field such that dipole (i.e. $\vec p$ ) makes an angle $\theta $ with the direction of field then two equal and opposite forces acting on dipole constitute a couple whose tendency is to rotate the dipole hence torque is developed in it and dipole tries to align itself in the direction of the field. Consider an electric dipole in placed in a uniform electric field such that dipole (i.e., $\vec p$) makes an angle $\theta $ with the direction of the electric field as shown.

(a) Net force on electric dipole ${{\text{F}}_{{\text{net}}}} = 0$

(b) $\quad  \ldots  \ldots \quad \tau  = {\text{pE}}\sin \theta (\vec \tau  = \overrightarrow {\text{p}}  \times \overrightarrow {\text{E}} )$

(ii) Work: From the above discussion it is clear that in an uniform electric field dipole tries to align itself in the direction of electric field (i.c. equilibrium position). To change it's angular position some work has to be done.

Suppose an electric dipole is kept in an uniform electric field by making an angle ${\theta _1}$ with the field, if it is again turn so that it makes an angle ${\theta _2}$ with the field, work done in this process is given by the formula

${\text{W  =  qE}}\left( {{\text{cos}}{{\text{\theta }}_{\text{1}}} - {\text{cos}}{{\text{\theta }}_{\text{2}}}} \right)$

(iii) Potential energy: In case of a dipole (in a uniform field), potential energy of dipole is defined as work done in rotating a dipole from a direction perpendicular to the field to the given direction i.e. if ${\theta _1} = {90^o}$ and ${\theta _2} = \theta $ then

$W=\delta U={U_{\theta}}-{U_{90^o}}=-PE \cos \theta $

$ \Rightarrow {U_{\theta}}= -PE \cos \theta \left ( \because U_{90}=0\:or\:U=-PE \right )$

8.2 Neutral Point

A neutral point is a point where resultant electrical field is zero. Thus neutral points can be obtained only at those points where the resultant field is subtractive.

(a) At an internal point along the line joining two like charges (Due to a system of two like point charge): Suppose two like charges. ${Q_1}$ and ${Q_2}$ are separated by a distance ${\text{x}}$ from each other along a line as shown in following figure.

(Image will be uploaded soon)

If ${\text{N}}$ is the neutral point at a distance ${{\text{x}}_1}$ from ${{\text{Q}}_1}$ and at a distance ${x_2}\left( { = x - {x_1}} \right)$ from ${Q_2}$ then for natural pt. at $N$,

| E.F. due to ${Q_1}| = |E.F$ due to ${Q_2}\mid $ i.e.,

$\dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{\left| {{Q_1}} \right|}}{{x_1^2}} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{\left| {{Q_2}} \right|}}{{x_2^2}} \Rightarrow \dfrac{{\left| {{Q_1}} \right|}}{{\left| {{Q_2}} \right|}} = {\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right)^2}$

Short trick:

${x_1} = \dfrac{x}{{1 + \sqrt {\left| {{Q_2}} \right|/\left| {{Q_1}} \right|} }}{\text{ and }}{x_2} = \dfrac{x}{{1 + \sqrt {{Q_1}|/|{Q_2}\mid } }}$


In the above formula if ${Q_1} = {Q_2}$, neutral point lies at the centre so remember that resultant field at the midpoint of two equal and like charges is zero.

(b) At an external point along the line joining two unlike charges (Due to a system of two unlike point charge): 

Suppose two unlike charge ${Q_1}$ and ${Q_2}$ separated by a distance $x$ from each other.

Here neutral point lies outside the line joining two unlike charges and also it lies nearer to charge which is smaller in magnitude.

If $\left| {{Q_1}} \right| < \mid {Q_2}$ then neutral point will be obtained on the side of ${Q_1}$, suppose it is at a distance $l$ from ${Q_1}$ Hence at neutral point;

$\dfrac{{{\text{k}}\left| {{Q_1}} \right|}}{{{\ell ^2}}} = \dfrac{{{\text{k}}\left| {{{\text{Q}}_2}} \right|}}{{{{(x + \ell )}^2}}} \Rightarrow \dfrac{{\left| {{Q_1}} \right|}}{{\left| {{Q_2}} \right|}} = {\left( {\dfrac{\varepsilon }{{x + 2}}} \right)^2}$

Short trick: $\ell  = \dfrac{x}{{\left( {\sqrt {{Q_2}\left| {/{Q_1}} \right|}  - 1} \right)}}$


In the above discussion if $\left| {{Q_1}} \right| = \left| {{Q_2}} \right|$ neutral point will be at infinity.

8.3 Equilibrium of Charge

(a) Definition: A charge is said to be in equilibrium, if net force acting on it is zero. A systern of charges is said to be in equilibrium if each charge is in equilibrium.

(b) Type of equilibrium: Equilibrium can be divided in following type:

(i) Stable equilibrium: After displacing a charged particle from it's equilibrium position, if it returns back then it is said to be in stable equilibrium. If ${\text{U}}$ is the potential energy then in case of stable equilibrium ${\mathbf{U}}$ is minimum.

(ii) Unstable equilibrium: After displacing a charged particle from it's equilibrium position, if it never returns back then it is said to be in unstable equilibrium and in unstable equilibrium, U is maximum.

(iii) Neutral equilibrium: After displacing a charged particle from it's equilibrium position if it neither comes back, nor moves away but remains in the position in which it was kept it is said to be in neutral equilibrium and in neutral equilibrium, U is constant.

(c) Different cases of equilibrium of charge

Suppose three similar charges ${Q_1},q$ and ${Q_2}$ are placed along a straight line as shown below.

Case -1:

Charge $q$ will be in equilibrium if $\left| {{{\text{F}}_1}} \right| = \left| {{{\text{F}}_2}} \right|$ ie.,,$\dfrac{{{{\text{Q}}_1}}}{{{{\text{Q}}_2}}} = {\left( {\dfrac{{{{\text{x}}_1}}}{{{{\text{x}}_2}}}} \right)^2}$;

This is the condition of equilibrium of charge $q$. After following the guidelines, we can say that charge $q$ is in stable equilibrium and this system is not in equilibrium.

${x_1} = \dfrac{x}{{1 + \sqrt {{Q_2}/{Q_1}} }}{\text{ and }}{x_2} = \dfrac{x}{{1 + \sqrt {{Q_1}/{Q_2}} }}$

e.g. if two charges $ + 4\mu C$ and $ + 16\mu C$ are separated by a distance of $30\;{\text{cm}}$ from each other then for equilibrium a third charge should be placed between them at a distance

${{\text{x}}_1} = \dfrac{{30}}{{1 + \sqrt {16/4} }} = 10\;{\text{cm or }}{{\text{x}}_2} = 20\;{\text{cm}}$


Two similar charge ${{\text{Q}}_1}$ and ${{\text{Q}}_2}$ are placed along a straight line at a distance $x$ from each other and a third dissimilar charge $q$ is placed in between them as shown below

Charge $g$ will be in equilibrium if $\left| {{{\text{F}}_1}} \right| = \left| {{{\text{F}}_2}} \right|$

i.e., $\dfrac{{{Q_1}}}{{{Q_2}}} = {\left( {\dfrac{{{x_1}}}{{{x_2}}}} \right)^2}$.


Same short trick can be used here to find the position of charge $q$ as we discussed in Case-1 i.e.,

${x_1} = \dfrac{x}{{1 + \sqrt {{Q_2}/{Q_1}} }}$ and ${x_2} = \dfrac{x}{{1 + \sqrt {{Q_1}/{Q_2}} }}$

It is very important to know that magnitude of charge $q$ can be determined if one of the extreme charge (either ${Q_1}$. or $\left. {{Q_2}} \right)$ is in equilibrium ie. if ${Q_2}$ is in equilibrium then$\left| q \right| = {{\text{Q}}_1}{\left( {{{\text{x}}_2}/{\text{x}}} \right)^2}$ and if ${{\text{Q}}_1}$ is in equilibrium then $|{\text{q}}| = {{\text{Q}}_2}{\left( {{{\text{x}}_1}/{\text{x}}} \right)^2}({\text{It}}$ should be remember that sign of $q$ is opposite to that of $\left. {{Q_1}\left( {{\text{or }}{Q_2}} \right)} \right)$.

Case -3:

Two dissimilar charge ${Q_1}$ and ${Q_2}$ are placed along a straight line at a distance ${\text{x}}$ from each other, a third charge q should be placed outside the line joining ${Q_1}$ and ${Q_2}$ for it to experience zero net force.

(Let $\left. {\left| {{Q_2}} \right| < \mid {Q_1}} \right)$

Short Trick :

For it's equilibrium. Charge $q$ lies on the side of charge which is smallest in magnitude and

$d = \dfrac{x}{{\sqrt {{Q_1}/{Q_2} - 1} }}$

(d) Equilibrium of suspended charge in an electric field

(i) Freely suspended charged particle: To suspend a charged a particle freely in air under the influence of electric field it's downward weight should be balanced by upward electric force for example if a positive charge is suspended freely in an electric field as shown then,

In equilibrium ${\text{QE}} = {\text{mg}} \Rightarrow {\text{E}} = \dfrac{{{\text{mg}}}}{{\text{Q}}}$


In the above case if direction of electric field is suddenly reversed in any figure then acceleration of charge particle at that instant will be $a = 2g$.

(ii) Charged particle suspended by a mass less insulated string (like simple pendulum): Consider a charged particle (like Bob) of mass $m$, having charge $Q$ is suspended in an electric field as shown under the influence of electric field. It turned through an angle (say $\theta )$ and comes in equilibrium.

So, in the position of equilibrium $\left( {{O^r}} \right.$ position)

${\text{T}}\sin \theta  = {\text{QE}}$ …(i)

$T\cos \theta  = mg$ …(ii)

By squaring and adding equation (i) and (ii),

$T = \sqrt {{{(QE)}^2} + {{(mg)}^2}} $

Dividing equation (i) by (ii)

$\tan \theta  = \dfrac{{{\text{QE}}}}{{{\text{mg}}}}$

$ \Rightarrow \quad \theta  = {\tan ^{ - 1}}\dfrac{{{\text{QE}}}}{{{\text{mg}}}}$

(iii) Equilibrium of suspended point charge system: Suppose two small balls having charge $ + Q$ on each are suspended by two strings of equal length $l$. Then for equilibrium position as shown in figure.

(Image will be uploaded soon)

${\text{T}}\sin \theta  = {{\text{F}}_e}$ …(i)

${\text{T}}\cos \theta  = {\text{mg}}\quad  \ldots $..(ii)

${{\text{T}}^2} = {\left( {{{\text{F}}_{\text{e}}}} \right)^2} + {({\text{mg}})^2}$

and $\tan \theta  = \dfrac{{{{\text{F}}_{\text{c}}}}}{{{\text{mg}}}}$;

her e ${F_e} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{Q^2}}}{{{x^2}}}$ and $\dfrac{x}{2} = \ell \sin \theta $.

(iv) Equilibrium of suspended point charge system in a liquid: In the previous discussion if point charge system is taken into a liquid of density $\rho $ such that $\theta $ remain same then

In equilibrium

${\text{F}}{{\text{e}}^\prime } = {{\text{T}}^t}\sin \theta {\text{ and }}({\text{mg}} - \operatorname{Vpg} ) = {{\text{T}}^\prime }\cos \theta $

$\therefore \quad \tan \theta  = \dfrac{{{\text{F}}{{\text{e}}^\prime }}}{{({\text{mg}} - {\text{Vpg}})}} = \dfrac{{{{\text{Q}}^2}}}{{4\pi {\varepsilon _0}\;{\text{K}}({\text{mg}} - {\text{V\backslash \rho g}}){{\text{x}}^2}}}$

When this system was in air

$\tan \theta  = \dfrac{{{\text{Fe}}}}{{{\text{mg}}}} = \dfrac{{{{\text{Q}}^2}}}{{4\pi {\varepsilon _0}{\text{mg}}{{\text{x}}^2}}}$

So equating these two gives us

$\dfrac{1}{m} = \dfrac{1}{{k(m - Vp)}} \Rightarrow K = \dfrac{m}{{m - V\rho }} = \dfrac{1}{{\left( {1 - \dfrac{V}{m}\rho } \right)}}$

If $\sigma $ is the density of material of ball then

$K = \dfrac{1}{{\left( {1 - \dfrac{\rho }{\sigma }} \right)}} = \dfrac{\sigma }{{\sigma  - \delta }}$

8.4 Time Period of Oscillation of a Charged Body

(a) Simple pendulum based: If a simple pendulum having length $l$ and mass of bob $m$ oscillates about it's mean position than it's time period of oscillation ${\text{T}} = 2\pi \sqrt {\ell /{\text{g}}} $.

Case-1: If some charge say $ + Q$ is given to bob and an electric field ${\text{E}}$ is applied in the direction as shown in figure then equilibrium position of charged bob (point charge) changes from ${\text{O}}$ to ${\text{O}}$'.

(Image will be uploaded soon)

On displacing the bob from it's equilibrium position ${{\text{O}}^r}$. It will oscillate under the effective acceleration ${g^\prime }$, where

${\text{m}}{{\text{g}}^\prime } = \sqrt {{{({\text{mg}})}^2} + {{({\text{QE}})}^2}}  \Rightarrow {g^\prime } = \sqrt {{{\text{g}}^2} + {{({\text{QE}}/{\text{m}})}^2}} $

Hence the new time period is ${T_1} = 2\pi \sqrt {\dfrac{\ell }{{{{\text{g}}^\prime }}}} $

$ \Rightarrow \quad {T_1} = 2\pi \sqrt {\dfrac{\ell }{{{{\left( {{g^2} + {{(QE/m)}^2}} \right)}^{\dfrac{1}{2}}}}}} $

Since ${g^\prime } > {\text{g}}$, hence ${{\text{T}}_1} < {\text{T}}$

i.e. time period of pendulum will decrease.

Case-2: If electric field is applied in the downward direction then.

Effective acceleration ${{\text{g}}^\prime } = {\text{g}} + {\text{QE}}/{\text{m}}$

So new time period ${{\text{T}}_{\text{2}}}{\text{ =  2\pi }}\sqrt {\dfrac{l}{{{\text{g  + }}\left( {{\text{QE/m}}} \right)}}} $

${T_2} < {T_1}$

Case -3: In case 2, if electric field is applied in upward direction then, effective acceleration.

${{\text{g}}^\prime } = {\text{g}} - {\text{QE}}/{\text{m}}$

So new time period

${{\text{T}}_3} = 2\pi \sqrt {\dfrac{\ell }{{{\text{g}} - ({\text{QE}}/{\text{m}})}}} {{\text{T}}_3} > {\text{T}}$

(b) Charged circular ring: A thin stationary ring of radius $R$ has a positive charge $ + Q$ unit. If a negative charge $ - q$ (mass $m$ ) is placed at a small distance $x$ from the centre. Then motion of the particle will be simple harmonic motion.

Electric field at the location of $ - q$ charge

${\text{E}} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{{\text{Q}}x}}{{{{\left( {{{\text{x}}^2} + {{\text{R}}^2}} \right)}^{\dfrac{3}{2}}}}}$

Since $x \ll R$, So ${x^2}$ neglected hence $E = \dfrac{1}{{4\pi {\varepsilon _0}}},\dfrac{{Qx}}{{{R^3}}}$

Force experienced by charge $ - q$ is $F =  - q\dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{Qx}}{{{R^3}}}$

$ \Rightarrow {\text{F}} \propto  - {\text{x}}$ hence motion is simple harmonic

Having time period ${\text{T}} = 2\pi \sqrt {\dfrac{{4\pi {\varepsilon _0}{\text{m}}{{\text{R}}^3}}}{{{\text{Q}}q}}} $

(c) Spring mass system: A block of mass $m$ containing a negative charge $ - Q$ is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant $k$ as shown. If electric field $B$ applied as shown in figure the block experiences an electric force, hence spring compress and block comes in new position. This is called the equilibrium position of block. under the influence of electric field. If block compressed further or stretched, it execute oscillation having time period ${\text{T}} = 2\pi \sqrt {{\text{m}}/{\text{k}}} $. Maximum compression in the spring due to electric field $ = {\text{QE}}/{\text{k}}$.

9. Electric Potential Energy

For the expression of total potential energy of a system of n charges consider $\dfrac{{n\left( {n - 1} \right)}}{2}$ number of pairs of charges.

Using Work Energy theorem:

10. Electric Potential

10.1 Potential Due to Charge Distribution

${{\text{W}}_{{\text{Ext}}}} = \vartriangle {\text{KE}} + \Delta {\text{U}}$

If only conservative forces are there (e.g. gravity / spring / coulomb force), then ${{\text{W}}_{{\text{ext}}}} = 0$

$\Delta {\text{KE}} + \Delta {\text{U}} = 0$ or, ${\text{K}}{{\text{E}}_{\text{f}}} + {{\text{U}}_{\text{f}}} = {\text{K}}{{\text{E}}_{\text{i}}} + {{\text{U}}_{\text{i}}}$

Work $ = \Delta {\text{KE}} + \Delta {\text{U}}$

If $\Delta {\text{KE}} = 0\quad \Delta {\text{U}} = {W_{ext}} =  - {W_{{\text{coulamb force }}}}$

If charges are assembled from infinity:

$\Delta {\text{U}} = {\text{U}}({\text{r}}) - {\text{U}}(\infty ) = $$U(r)[{\text{U}}(\infty ) = 0]$

We know, $\Delta {\text{U}} = {{\text{W}}_{{\text{ext }}}}\quad $ [when $\Delta {\text{K}}.{\text{E}} = 0$]

$\operatorname{If} U(\infty ) = 0 \Rightarrow {\text{U}}({\text{r}}) = {{\text{W}}_{ext}}$

If ${{\text{W}}_{{\text{ext }}}} = 0\quad \Delta {\text{KE}} + \Delta {\text{U}} = 0$ or ${\text{K}}{{\text{E}}_{\text{j}}} + {{\text{U}}_{\text{i}}} = {\text{K}}{{\text{E}}_{\text{f}}} + {{\text{U}}_{\text{f}}}$

10.2 Zero Potential due to a System of Two Point Charge

I. For internal point : ${Q_1}$ and ${Q_2}$ (opposite signs)

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$\dfrac{{{Q_1}}}{{{{\text{x}}_1}}} = \dfrac{{{{\text{Q}}_2}}}{{\left( {{\text{x}} - {{\text{x}}_1}} \right)}} \Rightarrow {{\text{x}}_1} = \dfrac{{\text{x}}}{{\left( {\left| {{{\text{Q}}_2}} \right|/\left| {{{\text{Q}}_1}} \right| + 1} \right)}}$

II. For External point : 

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${{\mathbf{Q}}_1}$ and ${Q_2}$ (opposite signs) $\left. {\left| {{Q_1}} \right| < \mid {Q_2}} \right)$

$\dfrac{{\left| {{Q_1}} \right|}}{{{x_2}}} = \dfrac{{\left| {{Q_2}} \right|}}{{\left( {x + {x_2}} \right)}} \Rightarrow {x_2} = \dfrac{x}{{\left( {\left| {{Q_2}} \right|/\left| {{Q_1}} \right| - 1} \right)}}$

11. Electric Dipole

11.1 Electric Field Due to a Dipole

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Using the concept that if we know potential electric field can be calculated we have already calculated

${{\text{V}}_{\text{p}}} = \dfrac{{{\text{kp}}\cos \theta }}{{{{\text{r}}^2}}}$

To Calculate net electric field at P we need ${\text{E}}$ (Radial Component) & ${E_1}$ (tangential component) of electric field at P.

${{\text{E}}_{\text{r}}} = \dfrac{{ - {\text{dV}}}}{{{\text{dr}}}}.$ (When we travel in the radial direction).

${{\text{E}}_{\text{t}}} =  - \dfrac{{{\text{dV}}}}{{{\text{rd}}\theta }}$  (When we travel in the tangential direction).  

\[{{\text{V}}_{\text{p}}} = \dfrac{{{\text{kP}}\cos \theta }}{{{{\text{r}}^2}}} \]

\[ {{\text{E}}_{\text{r}}} = \dfrac{{ - {\text{d}}}}{{{\text{dr}}}}\left( {\dfrac{{{\text{kP}}\cos \theta }}{{{{\text{r}}^2}}}} \right) = \dfrac{{2{\text{kP}}\cos \theta }}{{{{\text{r}}^3}}}\]

${{\text{E}}_{\text{t}}} = \dfrac{{ - {\text{d}}}}{{{\text{rd}}\theta }}\left( {\dfrac{{{\text{kP}}\cos \theta }}{{{{\text{r}}^2}}}} \right) = \dfrac{{ - {\text{kP}}}}{{{{\text{r}}^3}}}\dfrac{{\text{d}}}{{{\text{d}}\theta }}\cos \theta  = \dfrac{{{\text{kP}}\sin \theta }}{{{{\text{r}}^3}}}$

${E_{{\text{net }}}} = \sqrt {E_r^2 + E_t^2}  = \sqrt {{{\left( {\dfrac{{kP}}{{{r^3}}}} \right)}^2}\left[ {4{{\cos }^2}\theta  + {{\sin }^2}\theta } \right]} $

${E_{net}} = \sqrt {{{\left( {\dfrac{{kP}}{{{r^3}}}} \right)}^2}\left[ {1 + 3{{\cos }^2}\theta } \right]} $

${E_{net}} = \dfrac{{kP}}{{{r^3}}}\sqrt {1 + 3{{\cos }^2}\theta }  $

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$\tan \alpha  = \dfrac{{{E_t}}}{{{E_r}}} = \dfrac{{\dfrac{{kP}}{{{r^3}}}\sin \theta }}{{2\dfrac{{kP\cos \theta }}{{{r^3}}}}} = \dfrac{{\tan \theta }}{2}\quad \alpha  = {\tan ^{ - 1}}\left[ {\dfrac{{\tan \theta }}{2}} \right]$

(Note: $\alpha $ is the angle with the radial direction)

11.2 Equilibrium of Dipole

We know that, for any equilibrium net torque and net force on a particle (or system) should be zero.

We already discussed when a dipole is placed in an uniform electric field net force on dipole is always zero. But net torque will be zero only when $\theta  = {0^o }$ or ${180^0 }$.

When $\theta  = {0^o }$ i.e., dipole is placed along the electric field it is said to be in stable equilibrium, because after turning it through a small angle, dipole tries to align itself again in the direction of electric field.

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When $\theta  = {180^o}$ i.e., dipole is placed opposite to electric field, it is said to be in unstable equilibrium.

Stable equilibrium Unstable equilibrium

$\tau  = 0$ ${\tau _{\max }} = pE$ $\tau  = 0$

$W = 0$ $W = pE$ ${W_{\max }} = 2pE$

${U_{\min }} =  - pE$ $U = 0$ ${U_{\max }} = pE$

11.3 Angular SHM

In a uniform electric field (intensity $E$ ) if a dipole (electric) is slightly displaced from it's stable equilibrium position it executes angular SHM having period of oscillation. If $I = $ moment of inertia of dipole about the axis passing through it's centre and perpendicular to it's length.

For electric dipole : ${\text{T}} = 2\pi \sqrt {{\text{I}}/{\text{pE}}} $

11.4 Dipole-Point Charge Interaction

If a point charge is placed in dipole field at a distance $r$. from the mid point of dipole then force experienced by point charge varies according to the relation ${\text{F}} \propto \dfrac{1}{{{{\text{r}}^3}}}$

11.5 Electric Dipole in Non-Uniform Electric Field

When an electric dipole is placed in a non-uniform field, the two charges of dipole experiences unequal forces, therefore the net force on the dipole is not equal to zero.

Due to two unequal forces, a torque is produced which rotate the dipole so as to align it in the direction of field.

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So in non-uniform electric field.

(i) Motion of the dipole is translatory and rotatory

(ii) Torque on it may be zero.

Gauss’s Law

1. Electric Flux

1.1 Definition

Electric flux is defined as proportional to number of field lines crossing or cutting any area of cross section in space.

'The number of field lines passing through perpendicular unit area will be prop ortional to the magnitude of Electric Field there" (Theory of Field Lines)

$\dfrac{{\text{N}}}{{{{\text{A}}_ \bot }}} \propto {\text{E}} \Rightarrow {\text{N}} \propto {{\text{E}}_ \bot }$

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$\therefore \quad $ Electric Flux, ${\Phi _{{A_ \bot }}} = {\text{E}}{{\text{A}}_ \bot }$

As $\theta $ increases, flux through area A decreases. If we draw a vector of magnitude ${\text{A}}$ along the positive normal, it is called the area vector, $\vec A$ corresponding to the area $A$.

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$\therefore \quad $ Electric Flux, ${\Phi _A} = {\text{EA}}\cos \theta  = \overrightarrow {\text{E}}  \cdot \overrightarrow {\text{A}} $

(Assuming Electric Field is uniform over whole area)


If Electric field is not constant over the area of cross section, then

$\Phi  = \int_A {\overrightarrow {\text{E}} }  \cdot {\text{d}}\overrightarrow {\text{A}} $

1.2 Unit and Dimension

Flux is a scalar quantity.

S.I. unit : (volt $ \times {\text{m}})$ or $\dfrac{{{\text{N}} \cdot {{\text{m}}^2}}}{{\text{C}}}$

It 's Dimensional formula: $\left( {{\text{M}}{{\text{L}}^3}\;{{\text{T}}^{ - 3}}\;{{\text{A}}^{ - 1}}} \right)$

1.3 Types of Flux

For a closed body outward flux is taken to be positive, while inward flux is taken to be negative.

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2. Gauss's Law

2.1 Definition

According to Gauss's law, total electric flux through a closed surface enclosing a charge is $\dfrac{1}{{{\varepsilon _0}}}$ times the magnitude of the charge enclosed.

i.e., ${\phi _{{\text{not}}}} = \dfrac{1}{{{\varepsilon _0}}}\left( {{Q_{{\text{cc}}}}} \right)$

i.e., $\oint {\vec E}  \cdot d\vec A = \dfrac{{{Q_{en}}}}{{{E_0}}}$.


Gauss's law is only applicable for a closed surface.

2.2 Gaussian Surface

The closed surface on which Gauss law is applicable is defined as a Gaussian surface.


  • Gaussian surface can be of any shape \& size, only condition is that it should be closed.

  • Gaussian surface is hypothetical in nature. It does not have a physical existence.

2.3 Deriving Gauss's Law from Coulomb's Law

Let’s take a spherical gaussian surface with charge '$ + Q$' kept at the centre.

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We know field lines for a +ve charge are always radially outward.

Angle between $d\vec A$ and $\vec E$ is zero.

${\text{E}} = \dfrac{{{\text{kQ}}}}{{{{\text{r}}^2}}} = \dfrac{{\text{Q}}}{{4\pi { \in _0}{{\text{r}}^2}}}$

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Hence Net flux $ = Q/{\varepsilon _0}$.

Although we derived gauss law for a spherical surface it is valid for any shape of gaussian surface and for any charge kept anywhere inside the surface.

2.4 Coulomb's Law from Gauss's Law

We choose an imaginary sphere (Gaussian surface) of radius ${\text{r}}$ centred on the charge $ + {\text{q}}$. Due to symmetry, ${\text{E}}$ must have the same magnitude at any point on the surface, and $\vec E$ points radially outward, parallel to $\overrightarrow {{\text{dA}}} $. Hence we write the integral in Gauss's law as:

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${\phi _{{\text{net }}}} = \oint {\vec E}  \cdot \overrightarrow {dA}  = \oint E dA = E\oint d A = E\left( {4\pi {r^2}} \right)$

${Q_{{\text{enclosed }}}} = q$

Thus, $E\left( {4\pi {r^2}} \right) = \dfrac{q}{{{\varepsilon _0}}}$ or $E = \dfrac{q}{{4\pi {\varepsilon _0}{r^2}}}$

From the definition of the electric field, the force on a point charge ${{\text{q}}_0}$ located at a distance ${\text{r}}$ from the charge q is ${\text{F}} = {{\text{q}}_0}{\text{E}}$. Therefore,

${\text{F}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{\text{q}}{{\text{q}}_0}}}{{{{\text{r}}^2}}}$

which is Coulomb's law.

3. Applications of Gauss's Law

Using Gauss's law to derive ‘E’ due to various charge distributions.

3.1 Electric Field Due to a Line Charge

Consider an infinite line which has a linear charge density $\lambda $. Using Gauss's law, let us find the electric field at a distance '${\text{r}}$' from the line charge.

The cylindrical symmetry tells us that the field strength will be the same at all points at a fixed distance $r$ from the line. Thus, if the charges are positive. The field lines are directed radially outwards, perpendicular to the line charge.

The appropriate choice of Gaussian surface is a cylinder of radius $r$ and length ${\text{L}}$. On the flat end faces, ${{\text{S}}_2}$ and ${{\text{S}}_3},\overrightarrow {\text{E}} $ is perpendicular ${\text{d}}\overline {\text{S}} $, which means flux is zero on them. On the curved surface ${S_1},\vec E$ is parallel to $d\bar S$, so that $\overrightarrow {\text{E}}  \cdot \overrightarrow {{\text{dS}}}  = {\text{EdS}}$. The charge enclosed by the cylinder is $Q = \lambda L$. Applying Gauss's law to the curved surface, we have

${\text{E}}\oint {{\text{dS}}}  = {\text{E}}(2\pi {\text{rL}}) = \dfrac{{\lambda {\text{L}}}}{{{\varepsilon _0}}}{\text{ or E}} = \dfrac{\lambda }{{2\pi {\varepsilon _0}{\text{r}}}} = \dfrac{{{\text{2k\lambda }}}}{{\text{r}}}$


This is the field at a distance ${\text{r}}$ from the line. It is directed away from the line if the charge is positive and towards the line if the charge is negative.

3.2 Electric Field Due to a Plane Sheet of Charge

Consider a large plane sheet of charge with surface charge density (charge per unit area) $\sigma $. We have to find the electric field ${\text{E}}$ at a point ${\text{P}}$ in front of the sheet.

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If the charge is positive, the field is away from the plane.

To calculate the field ${\text{E}}$ at ${\text{P}}$. Choose a cylinder of area of cross-section A through the point P as the Gaussian surface. The flux due to the electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.

According to Gauss law $\oint {\vec E}  \cdot d\bar S = {q_{{\text{in }}}}/{\varepsilon _0}$

\[\mathop {\int {\overrightarrow E .\overline {dS} } }\limits_{{\text{I circular surface}}}  + \mathop {\int {\overrightarrow E .\overline {dS} } }\limits_{{\text{II circular surface}}}  + \mathop {\int {\overrightarrow E .\overrightarrow {dS} } }\limits_{{\text{cylindricalr surface}}}  = \dfrac{{\sigma A}}{{{\varepsilon _0}}}\]

or ${\text{EA}} + {\text{EA}} + 0 = \dfrac{{\sigma {\text{A}}}}{{{\varepsilon _0}}}$

or $E = \dfrac{\sigma }{{2{\varepsilon _0}}}$


We see that the field is uniform and does not depend on the distance from the charge sheet. This is true as long as the sheet is large as compared to its distance from ${\text{P}}$.

3.3 Uniform Spherical Charge Distribution

3.3.1 Outside the Sphere

P is a point outside the sphere at a distance $r$ from the centre.

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According to Gauss law, $\oint {\overrightarrow {\text{E}} }  \cdot {\text{d}}\overrightarrow {\text{s}}  = \dfrac{{\text{Q}}}{{{{\text{E}}_{\text{D}}}}}$ or ${\text{E}}\left( {4\pi {{\text{r}}^2}} \right) = \dfrac{{\text{Q}}}{{{\varepsilon _{\text{D}}}}}$

Electric field at P (Outside sphere)

${E_{{\text{out }}}} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{{{r^2}}} = \dfrac{{\sigma {R^2}}}{{{\varepsilon _0}{r^2}}}$ and

${{\text{V}}_{{\text{out }}}} =  - \int_\infty ^{\text{r}} {\overrightarrow {\text{E}} } {\text{d}}\overrightarrow {\text{r}}  = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{\text{Q}}}{{\text{r}}} = \dfrac{{\sigma {{\text{R}}^2}}}{{{\varepsilon _0}{\text{r}}}}$


${Q = 0 \times A}$  

${ = \sigma  \times 4\pi {R^2}}$

3.3.2 At the surface of sphere

At surface $r = R$

So, ${E_e} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{{{R^2}}} = \dfrac{\sigma }{{{\varepsilon _0}}}$ and ${V_\varepsilon } = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{R} = \dfrac{{\sigma R}}{{{\varepsilon _0}}}$

3.3.3 Inside the Sphere

Inside the conducting charged sphere electric field is zero and potential remains constant everywhere and equals to the potential at the surface.

Graphical Variation of Electric Field and Potential With Distance

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3.4 Uniform Spherical Volume Charge Distribution

We consider a spherical uniformly charge distribution of radius $R$ in which total charge $Q$ is uniformly distributed throughout the volume.

The charge density

$\rho  = \dfrac{{{\text{ total ch arge }}}}{{{\text{ total volume }}}} = \dfrac{Q}{{\dfrac{4}{3}\pi {R^3}}} = \dfrac{{3Q}}{{4\pi {R^3}}}$

3.4.1 Outside the Sphere at $P(r \geqslant R)$

According to Gauss law $\oint {\overrightarrow {\text{E}} }  \cdot \overrightarrow {{\text{ds}}}  = \dfrac{{\text{Q}}}{{{\varepsilon _0}}}$ or ${\text{E}}\left( {4\pi {r^2}} \right) = \dfrac{{\text{Q}}}{{{\varepsilon _{\text{O}}}}}$

${{\text{E}}_{{\text{out }}}} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{\text{Q}}}{{{{\text{r}}^2}}}{\text{ and }}{{\text{V}}_{{\text{out}}}} =  - \int_\infty ^{\text{I}} {\overrightarrow {\text{E}} } {\text{d}}\overrightarrow {\mathbf{r}}  = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{\text{Q}}}{{\text{r}}}$

using $\rho  = \dfrac{Q}{{\dfrac{4}{3}\pi {R^3}}}$

${E_{{\text{out }}}} = \dfrac{{\rho {{\text{R}}^3}}}{{3{\varepsilon _0}{r^2}}}{\text{ and }}{V_{0ut}} = \dfrac{{\rho {R^3}}}{{3{\varepsilon _{\text{J}}}{\mathbf{r}}}}({\text{V}}(\infty ) = 0)$

3.4.2 At the Surface of Sphere

At surface $r = R$

${E_e} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{{{R^2}}} = \dfrac{{\rho R}}{{3{\varepsilon _0}}}{\text{ and }}{V_s} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{R} = \dfrac{{\rho {R^2}}}{{3{\varepsilon _0}}}$

3.4.3 Inside the Sphere

At a distance $r$ from the centre. $(r \leqslant R)$

$\oint {{{\vec E}_{in}}}  \cdot \overrightarrow {ds}  = \dfrac{{{q_{in}}}}{{{\varepsilon _0}}} = \dfrac{{Q{r^3}}}{{{\varepsilon _0}{R^3}}}{\text{ or }}{E_{jn}}\left( {4\pi {r^2}} \right) = \dfrac{{Q{r^3}}}{{{G_0}{R^3}}}$

${E_{in}} = \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{Qr}}{{{R^3}}} = \dfrac{{\rho r}}{{3{\varepsilon _0}}}\left\{ {{E_{in}} \propto r} \right\}$ and

${V_{{\text{in}}{\text{. }}}} =  - \int_k^r {\vec E}  \cdot d\vec r = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Q\left[ {3{R^2} - {r^2}} \right]}}{{2{R^3}}} = \dfrac{{\rho \left( {3{R^2} - {r^2}} \right)}}{{6{\varepsilon _a}}}$


At centre $(r = 0)$. ${V_{{\text{centre }}}} = \dfrac{3}{2} \times \dfrac{1}{{4\pi {\varepsilon _0}}} \cdot \dfrac{Q}{R} = \dfrac{3}{2}{V_o}$

 i.e., ${V_{{\text{centre }}}} > {V_{{\text{surface }}}} > {V_{{\text{out }}}}$

Graphical Variation of Electric Field and Potential With Distance.

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4. Properties of Conductors

1. Inside a Conductor, Electrostatic Field Is Zero

Consider a conductor, neutral or charged. There may also be an external electrostatic field. In the static situation the electric field is zero everywhere inside the conductor. As long as electric field is not zero, the free charge carriers would experience force and drift. In the static situation, the free charges have so distributed themselves that the electric field is zero everywhere inside. Electrostatic field is zero inside a conductor.

2. At the Surface of a Charged Conductor, Electrostatic Field Must Be Normal to the Surface at Every Point.

If E were not normal to the surface, it would have some nonzero component along the surface. Free charges on the surface of the conductor would then experience force and move. In the static situation, therefore, ${\text{E}}$ should have no tangential component. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface).

3. The Charge Kept in the Material of a Conductor Will Come to Its Outermost Surface.

We know electric field at all points inside the material of a conductor is zero. This means ' ${{\text{E}}^\prime }$ at all points on the Gaussian surface is zero.

$\int {\overrightarrow {\text{E}} }  \cdot \overrightarrow {{\text{dA}}}  = \dfrac{{{{\text{Q}}_{{\text{e1}}}}}}{{{\varepsilon _0}}} \Rightarrow {\text{E}} = 0 \Rightarrow {{\text{Q}}_{e{\text{n}}}} = 0$

Charge cannot remain inside so it comes outside dotted surface.

4. Electrostatic Potential is Constant Throughout the Volume of the Conductor and Has the Same Value (As Inside) on Its Surface.

This follows from results 1 and 2 above. Since $E = 0$ inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on the surface of the conductor Hence, the result. If the conductor is charged, electric field normal to the surface exists; this means potential will be different for the surface and a point just outside the surface.

In a system of conductors of arbitrary size, shape and charge configuration, each conductor is characterised by a constant value of potential, but this constant may differ from one conductor to the other.

5. Electric Field at the Surface of a Charged Conductor

${\text{E}} = \dfrac{\sigma }{{{\varepsilon _0}}}{\text{n}}$

where $\sigma $ is the surface charge density and $\hat n$ is a unit vector normal to the surface in the outward direction.

For $\sigma  > 0$,electric field is normal to the surface outward; for $\sigma  > 0$, electric field is normal to the surface inward.

5. Gauss Law

5.1 Some Important Points About Gauss Law

1. In the above expression, charge enclosed is $\left( {{Q_1}{\text{ and }}{Q_2}} \right)$. Net flux will only depend on ${Q_1}{\text{ and }}{Q_2}$.

But 'E' in the Gauss Law will be due to all the charges ${{\text{Q}}_1}$, ${{\text{Q}}_2},{{\text{Q}}_3}{\text{ and }}{{\text{Q}}_4}$. Hence electric field calculated through Gauss law is not just due to enclosed charges but due to all the charges.

2. If ${Q_{en}}$ by a gaussian surface is given to be zero then it does not necessarily imply that ${\text{E}} = 0$. It may or may not be zero. For example:

${Q_{en}} = 0$

but electric field on the Gaussian surface is present.

3. If $E$ at all points on the gaussian surface is zero then it mean ${Q_{en}}$ has to be zero.

Because $\oint {\overrightarrow {\text{E}} } \overrightarrow {{\text{dA}}}  = \dfrac{{{{\text{Q}}_{{\text{en}}}}}}{{{\varepsilon _{\text{D}}}}}$.

5.2 Zero Flux

(i) Net charge equals to zero

If a dipole is enclosed by a surface,

$\phi  = 0;{Q_{enc}} = 0$

If the magnitude of positive and negative charges are equal inside a closed surface.

(ii) Charges are absent

If a closed body (not enclosing any charge) is placed in an electric field (either uniform or non-uniform) total flux linked with it will be zero.

5.3 Observe Flux through Common Geometrical Figures

5.3.1 Cube

(i) Charge at the centre of cube.

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${\phi _{{\text{total }}}} = \dfrac{1}{{{\varepsilon _0}}} \cdot (Q)$

Flux through each face, ${\phi _{{\text{face }}}} = \dfrac{Q}{{6{\varepsilon _0}}}$

(ii) Charge situated at the corner of a cube.

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For the charge at the corner, we require eight cube to symmetrically enclose it in a Gaussian surface. The total flux ${\phi _{\text{T}}} = \dfrac{Q}{{{\varepsilon _0}}}.$ Therefore the flux through one cube will be

${\phi _{{\text{cube }}}} = \dfrac{Q}{{8{\varepsilon _0}}}$. The cube has six faces and flux linked with three faces (through A) is zero $({\text{ABCD}},{\text{AHED}},{\text{ABGH}})$, so flux linked with remaining three faces will $\dfrac{{\text{Q}}}{{8{\varepsilon _0}}}$. Now as the remaining three are identical so flux linked with each of the three faces will be $ = \dfrac{1}{3} \times \left[ {\dfrac{1}{8}\left( {\dfrac{Q}{{{\varepsilon _0}}}} \right)} \right] = \dfrac{1}{{24}}\dfrac{Q}{{{\varepsilon _0}}}$.

Charge Position

\[{\phi _r}\]

Cube centre

\[q/{\varepsilon _0}\]

Face centre

\[q/2{\varepsilon _0}\]

At corner

\[q/8{\varepsilon _0}\]

At centre of edge

\[q/4{\varepsilon _0}\]

5.3.2 Hemisphere

\[{\phi _{{\text{out}}}} = {\phi _{{\text{in}}}} = \pi {{\text{R}}^2}{\text{E}}\]

${\phi _{\text{T}}} = 0$

${\phi _{\text{T}}} = \dfrac{q}{{{\varepsilon _0}}},\,\,{\phi _{{\text{hemisphere}}}} = \dfrac{q}{{2{\varepsilon _0}}}$

(dotted part shows imaginary part to enclose the charge completely)

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5.3.3 Cylinder

${\phi _T} = \dfrac{q}{{{\varepsilon _0}}},{\phi _{cy1}} = \dfrac{q}{{2{\varepsilon _0}}}$

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6. Cases of Concentric Shells

\[{{\text{V}}_{\text{a}}} = \dfrac{{{\text{k}}{{\text{Q}}_1}}}{{{{\text{r}}_1}}} + \dfrac{{{\text{k}}\left( { - {{\text{Q}}_1}} \right)}}{{{{\text{r}}_2}}} + \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} + {{\text{Q}}_2}} \right)}}{{{{\text{r}}_2}}} \]

 \[\;{{\text{V}}_{\text{b}}} = \dfrac{{{\text{k}}{{\text{Q}}_1}}}{{{{\text{r}}_2}}} + \dfrac{{{\text{k}}\left( { - {{\text{Q}}_1}} \right)}}{{{{\text{r}}_2}}} + \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} + {{\text{Q}}_2}} \right)}}{{{{\text{r}}_2}}} \]

7. Cases of Earthing a Conductor

1. Firstly do charge distribution before earthing.

2. After charge distribution, assume some '${\text{x}}$' charge flown to ground (after earthing).

3. Do re-distribution of charge.

4. Take a point on the conductor (which is earthed) & do net potential of it equals 0. Calculate $x$.

${V_P} = 0 = \dfrac{{k{Q_1}}}{{{r_2}}} + \dfrac{{k\left( { - {Q_1}} \right)}}{{{r_2}}} + \dfrac{{k\left( {{Q_1} + {Q_2} - x} \right)}}{{{r_2}}} \Rightarrow {\text{x}} = {Q_1} + {Q_2}$

Charge is flown from outer surface because as long as ${Q_1}$ remains on inner shell, ‘$ - {{\text{Q}}_1}$’ will be induced on inner shell.

${{\text{V}}_{\text{P}}} = 0 = \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\text{r}}_1}}} + \dfrac{{{\text{k}}\left( { - \left( {{{\text{Q}}_1} - {\text{x}}} \right)} \right)}}{{{{\text{r}}_2}}} + \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} + {{\text{Q}}_2} - {\text{x}}} \right)}}{{{{\text{r}}_2}}}$

$\dfrac{{{{\text{Q}}_1} - {\text{x}}}}{{{{\text{r}}_1}}} - \dfrac{{\left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\text{r}}_2}}} + \dfrac{{{Q_1} + {{\text{Q}}_2} - {\text{x}}}}{{{{\text{r}}_2}}} = 0{\text{ }}$

$ \Rightarrow \quad \dfrac{{{{\text{Q}}_1}}}{{{{\text{r}}_1}}} + \dfrac{{{{\text{Q}}_2}}}{{{{\text{r}}_2}}} = \dfrac{{\text{x}}}{{\text{r}}} = {\text{x}} = {{\text{Q}}_1} + {{\text{Q}}_2}\dfrac{{{{\text{r}}_1}}}{{{{\text{r}}_2}}}$


As it can be seen not all charge on the surface flows to ground. When the outermost conductor is earthed then the charge residing on the outermost surface of outer conductor will flow to ground.

8. Connection of Charged Conductors


1. Do charge distribution before connection.

2. Assume ‘x’ charge flows from one conductor to another.

3. Do redistribution of charges.

4. Equate net potential of conductor (1) equal to net potential of conductor (2).

Assumption: Distance between them is very large.

${{\text{V}}_{\text{P}}} = {{\text{V}}_{\text{R}}}$$ \Rightarrow \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\text{r}}_1}}} = \dfrac{{{\text{k}}\left( {{{\text{Q}}_2} + {\text{x}}} \right)}}{{{{\text{r}}_2}}} = {\text{x}} = \dfrac{{{{\text{Q}}_1}{{\text{r}}_2} - {Q_2}{{\text{r}}_1}}}{{{{\text{r}}_{\text{l}}} + {{\text{r}}_2}}}$

Final Charges

Final common potential $k = \left( {\dfrac{{{Q_1} + {Q_2}}}{{{r_1} + {r_2}}}} \right)$

${V_{\text{P}}} = {{\text{V}}_{\text{R}}}$

${{\text{V}}_{\text{P}}} = \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\text{r}}_1}}} + \dfrac{{{\text{k}} - \left( {{{\text{Q}}_2} - {\text{x}}} \right)}}{{{{\text{r}}_2}}} + \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} + {{\text{Q}}_2}} \right)}}{{{{\text{r}}_2}}}$

$ \Rightarrow {{\text{V}}_{\text{R}}} = \dfrac{{{\text{K}}\left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\mathbf{r}}_2}}} + \dfrac{{{\text{k}} - \left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\mathbf{r}}_2}}} + \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} + {{\text{Q}}_2}} \right)}}{{{{\mathbf{r}}_2}}}$

$\;{{\text{V}}_{\text{P}}} = {{\text{V}}_{\text{R}}}$

$ \Rightarrow \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\text{r}}_1}}} = \dfrac{{{\text{k}}\left( {{{\text{Q}}_1} - {\text{x}}} \right)}}{{{{\text{r}}_2}}} \Rightarrow \left( {{{\text{Q}}_1} - {\text{x}}} \right)\left[ {\dfrac{1}{{{{\text{r}}_1}}} - \dfrac{1}{{{{\text{r}}_2}}}} \right] = 0$

$ \Rightarrow \quad {\text{x}} = {{\text{Q}}_1}$

This indicates that all the charge on shell (1) will flow to shell(2).

9. Self Energy of Charged Sphere

Consider a uniformly charged sphere of radius R having a total charge Q. The electric potential energy of this sphere is equal to the work done in bringing the charges from infinity to assemble the sphere.

${\text{U}} = \dfrac{{3{Q^2}}}{{20\pi {\varepsilon _0}{\text{R}}}}\quad {\text{U}} = \dfrac{{{{\text{Q}}^2}}}{{8\pi {\varepsilon _0}{\text{R}}}}$

10. Energy Density

The energy stored per unit volume around a point in an electric field is given by ${u_{\text{e}}} = \dfrac{{\text{U}}}{{{\text{ Volume }}}} = \dfrac{1}{2}{\varepsilon _0}{{\text{E}}^2}.$ If in place of vacuum some medium of dielectric constant K is present then ${u_e} = \dfrac{1}{2}K{\varepsilon _0}{E^2}$

11. Plate Theory

11.1 Charged Conducting Pate

Net Flux $ = \dfrac{{{Q_{en}}}}{{{\varepsilon _0}}} = \dfrac{{\sigma \pi {r^2}}}{{{\varepsilon _0}}}$ (cylindrical Gaussian surface)

${\text{E}}\pi {{\text{r}}^2} = \dfrac{{\sigma \pi {{\text{r}}^2}}}{{{\varepsilon _{\text{o}}}}}$

$E = \dfrac{\sigma }{{{\varepsilon _0}}}$

Net electric field at point $P$, near a conducting surface, ${\sigma ^\prime }$ is given by $\left[ {\sigma /{\varepsilon _0}} \right]$

(Image will be uploaded soon)

11.2 Parallel Plate Theory

To find charge distribution on each surface of plates.

Two conducting plates having area ‘A’ (area is large as compared to distance, so that field is uniform) and the thickness of plates is small so that charge only appears on parallel faces.

Since the field lines are parallel, the net flux through the gaussian surface will be zero, surface (1) and (2) be inside the material of the conductor.

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Hence it can be said that net charge enclosed will be zero which implies the charges appearing on the facing surfaces are equal & opposite to each other.

Net electric field at any point 'P' or 'R' has to be zero.

${\left( {{E_{{\text{net }}}}} \right)_P} = 0$

There are 4 distributions, the net field at ${\text{P}}$ should be zero.

${({E_P})_1} = \dfrac{\sigma }{{2{\varepsilon _0}}}( \to ) = \dfrac{{\left( {{Q_1} - q} \right)}}{{2A{\varepsilon _0}}}( \to ){\left( {{E_P}} \right)_2} = \dfrac{q}{{2A{\varepsilon _0}}}( \leftarrow )$

${\left( {{E_P}} \right)_3} = \dfrac{q}{{2A{\varepsilon _0}}}( \to )\quad {\left( {{E_P}} \right)_4} = \dfrac{{{Q_2} + q}}{{2A{\varepsilon _0}}}( \leftarrow )$

$ \Rightarrow \left| {{{\left( {{E_P}} \right)}_1}} \right| + \left| {{{\left( {{E_P}} \right)}_3}} \right| = \left| {{{\left( {{E_P}} \right)}_2}} \right| + \left| {{{\left( {{E_P}} \right)}_4}} \right|$

This shows $\dfrac{{{Q_1} - q}}{{2A{\varepsilon _0}}} + \dfrac{q}{{2A{\varepsilon _0}}} = \dfrac{q}{{2A{\varepsilon _0}}} + \dfrac{{{Q_2} + q}}{{2A{\varepsilon _0}}}$

$q = \dfrac{{{Q_1} - {Q_2}}}{2}$ so final distributions would be,


When charged conducting plates are placed parallel to each other, the two outermost surfaces get equal charges and the facing surfaces get equal and opposite charges.

11.3 Force on a Charged conductor

To find force on a charged conductor (due to repulsion of like charges) imagine a small part ${\text{XY}}$ to be cut and just separated from the rest of the conductor ${\text{MLN}}$. The field in the cavity due to the rest of the conductor is ${E_2}$, while field due to small part is $E$. Then

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Inside the conductor

$\therefore {{\text{E}}_1} - {{\text{E}}_2} = 0{\text{ or }}{{\text{E}}_1} = {{\text{E}}_2}$

Outside the conductor ${\text{E}} = {{\text{E}}_1} + {{\text{E}}_2} = \dfrac{\sigma }{{{\varepsilon _0}}}$. Thus, ${E_1} = {E_2} = \dfrac{\sigma }{{2{\varepsilon _0}}}$

To find force, imagine charged part ${\text{XY}}$ (having charge $\sigma dA$ placed in the cavity MN having field ${{\text{E}}_2}$ ). Thus force ${\text{dF}} = (\sigma {\text{dA}}){{\text{E}}_2}$ or ${\text{dF}} = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}}\;{\text{d}}\;{\text{A}}$. The force per unit area or electric pressure is ${\text{P}} = \dfrac{{{\text{dF}}}}{{{\text{dA}}}} = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}}}$. (Electrostatic pressure) The force is always outwards as ${( \pm \sigma )^2}$ is positive i.e., whether charged positively or negatively, this force will try to expand the charged body.

A soap bubble or rubber balloon expands on given charge to it (charge of any kind $ + $ or $ - $ ).


1. Capacitance

1.1 Definition

We know that charge given to a conductor increases it's potential i.e., ${\text{Q}} \propto {\text{V}} \Rightarrow {\text{Q}} = {\text{CV}}$

Where $C$ is a proportionality constant, called capacity or capacitance of conductor. Hence capacitance is the ability of conductor to hold the charge (and associated electrical energy).

1.2 Unit and Dimensional Formula

S.I. unit is $\dfrac{{{\text{ Coulomb }}}}{{{\text{ Volt }}}} = $ Farad (F)

Smaller S.I. units are ${\text{mF}},\mu {\text{F}},{\text{nF}}$ and ${\text{pF}}$

$\left( {1{\text{mF}} = {{10}^{ - 3}}\;{\text{F}},1\mu {\text{F}} = {{10}^{ - 6}}\;{\text{F}},1{\text{nF}} = {{10}^{ - 9}}\;{\text{F}},1{\text{pF}} = {{10}^{ - 12}}\;{\text{F}}} \right)$

C.G.S. unit is Stat Farad. $1\;{\text{F}} = 9 \times {10^{11}}$ Stat Farad.

Dimension: $[{\text{C}}] = \left[ {{{\text{M}}^{ - 1}}\;{{\text{L}}^{ - 2}}\;{{\text{T}}^4}\;{{\text{A}}^2}} \right]$.

2. Capacitor

2.1 Definition

A capacitor is a device that stores electric energy. It is also named condenser.


A capacitor is a pair of two conductors of any shape, which are close to each other and have equal and opposite charge.

2.2 Symbol

The symbol of capacitor are shown below:

(Image will be uploaded soon)

2.3 Capacitance

The capacitance of a capacitor is defined as the magnitude of the charge $Q$ on the positive plate divided by the magnitude of the potential difference $V$ between the plates i.e., ${\text{C}} = {\text{Q}}/{\text{V}}$.


Capacitance of a capacitor is constant for the given dimensions & medium.

2.4 Charge on Capacitor

Net charge on a capacitor is always zero, but when we speak of the charge $Q$ on a capacitor, we are referring to the magnitude of the charge on each plate.

2.5 Energy Stored

When a capacitor is charged by a voltage source (say battery) it stores the electric energy.

Energy density $ = \dfrac{U}{{{\text{ vol }}}} = \dfrac{1}{2}{\varepsilon _0}{{\text{E}}^2}$

If ${\text{C}} = $ Capacitance of capacitor; $Q = $ Charge on capacitor and $V = $ Potential difference across capacitor then energy stored in capacitor ${\text{U}} = \dfrac{1}{2}{\text{C}}{{\text{V}}^2} = \dfrac{1}{2}{\text{QV}} = \dfrac{{{{\text{Q}}^2}}}{{2{\text{C}}}}$.


In charging capacitor by battery half the energy supplied is stored in the capacitor and remaining half energy $(1/2QV)$ is lost in the form of heat.

2.6 Types of Capacitors

Capacitors are of mainly three types as described in given table:

Parallel Plate Capacitor

Spherical Capacitor

Cylindrical Capacitor

It consists of two parallel metallic plates (may be circular, rectangular, square) separated by a small distance

$A = $ area of plate

$Q = $ Magnitude of charge

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$C = \dfrac{{{\varepsilon _0}A}}{d}$

$\sigma  = $ Surface charge density

${\text{V}} = $ potential difference

${\text{E}} = $ Electric field between the plates

the plates $\left( { = \sigma /{\varepsilon _0}} \right)$

In the presence of dielectric

between plates ${\text{C}} = \dfrac{{{\text{K}}{\varepsilon _{\text{O}}}{\text{A}}}}{{\text{d}}}$

It consists of two concentric conducting spheres of radii $a$ and $b(a < b)$. Inner sphere is given charge $ + Q$, while outer sphere is given charge $ - {\text{Q}}$ [by battery]

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Capacitance: $C = 4\pi {\varepsilon _0}\dfrac{{ab}}{{b - a}}$

in C.GS. ${\text{C}} = \dfrac{{{\text{ab}}}}{{{\text{b}} - {\text{a}}}}$.

In the presence of dielectric medium (dielectric constant ${\text{K}})$ between the spheres

${{\text{C}}^\prime } = 4\pi {\varepsilon _0}\;{\text{K}}\dfrac{{{\text{ab}}}}{{{\text{b}} - {\text{a}}}}$

It consists of two concentric cylinders of radii $a$ and $b(a < b)$, inner cylinder is given charge $ + Q$ while outer cylinder is given charge -Q. Common length of the cylinders is $l$ then

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$C = \dfrac{{2\pi {\varepsilon _0}l}}{{\ln \left( {\dfrac{b}{a}} \right)}}$

In the presence of dielectric medium (dielectric constant K) capacitance increases by ${\text{K}}$ times and

${C^{\prime \prime }} = \dfrac{{2\pi {\varepsilon _0}Kl}}{{\ln \left( {\dfrac{b}{a}} \right)}}$

2.7 Capacity of an Isolated Spherical Conductor

When charge $Q$ is given to a spherical conductor of radius $R$, then potential at the surface of sphere is,

$V = \dfrac{1}{{4\pi {\varepsilon _0}}}.\dfrac{Q}{R}$

Hence it's capacity ${\text{C}} = \dfrac{Q}{{\;{\text{V}}}} = 4\pi {\varepsilon _0}{\text{R}}$

$\Rightarrow {\text{C}} = 4\pi {\varepsilon _0}{\text{R}} = \dfrac{1}{{9 \times {{10}^9}}} \cdot {\text{R}} $

$\,\,\,\,\,{\text{ in C}}{\text{.G}}{\text{.S}}{\text{. C}} = {\text{R}}$

2.8 Force Between the Plates of a Parallel Plate Capacitor

Field due to charge on one plate on the other is ${\text{E}} = \dfrac{\sigma }{{2{\varepsilon _0}}}$, hence the force $F = QE$

3. Properties of an Ideal Battery

(a) A battery has two terminals.

(b) The potential difference $V$ between the terminals is constant for a given battery. The terminal with higher potential is called the positive terminal and that with lower potential is called the negative terminal.

(c) The value of this fixed potential difference is equal to the electromotive force or emf of the battery. If a conductor is connected to a terminal of a battery, the potential of the conductor becomes equal to the potential of the terminal. When the two plates of a capacitor are connected to the terminals of a battery, the potential difference between the plates of the capacitor becomes equal to the emf of the battery.

(d) The total charge in a battery always remains zero. If its positive terminal supplies a charge $Q$, its negative terminal supplies an equal, negative charge $ - {\text{Q}}$.

(e) When a charge Q passes through a battery of emf $\mathcal{E}$ from the negative terminal to the positive terminal, an amount QE of work is done by the battery.

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An ideal battery is represented by the symbol shown in figure. The potential difference between the facing parallel lines is equal to the emf $\mathcal{E}$ of the battery. The longer line is at the higher potential.

4. Grouping of Capacitors

4.1 Series grouping

  • Charge on each capacitor remains same and equals to the main charge supplied by the battery.

  • Equivalent capacitance $\dfrac{1}{{{C_{{\text{eq }}}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}} + \dfrac{1}{{{C_3}}}$

  • If two capacitors having capacitances ${C_1}$ and ${C_2}$ are connected in series then

  • ${{\text{V}}_1} = \left( {\dfrac{{{{\text{C}}_2}}}{{{{\text{C}}_1} + {{\text{C}}_2}}}} \right) \cdot {\text{V and }}{{\text{V}}_2} = \left( {\dfrac{{{{\text{C}}_1}}}{{{{\text{C}}_1} + {{\text{C}}_2}}}} \right){\text{V}}$

  • If $n$ identical capacitors each having capacitances $C$ are connected in series with supply voltage $V$ then

  • Equivalent capacitance ${{\text{C}}_{eq}} = \dfrac{{\text{C}}}{{\text{n}}}$ and Potential difference across each capacitor ${{\text{V}}^\prime } = \dfrac{{\text{V}}}{{\text{n}}}$.


Two capacitors are in series when charge leaving one capacitor directly enters into another capacitor, undivided and undisturbed. In series combination equivalent capacitance is always lesser than that of either of the individual capacitors.

4.2 Parallel grouping

  • Potential difference across each capacitor remains same and equal to the applied potential difference.

  • ${{\text{C}}_{ - {\text{eq}}}} = {{\text{C}}_1} + {{\text{C}}_2} + {{\text{C}}_3}$

  • If two capacitors having capacitance ${{\text{C}}_1}$ and ${{\text{C}}_2}$ respectively are connected in parallel then

  • ${{\text{Q}}_1} = \left( {\dfrac{{{{\text{C}}_1}}}{{{{\text{C}}_1} + {{\text{C}}_2}}}} \right) \cdot {\text{Q and }}{{\text{Q}}_2} = \left( {\dfrac{{{{\text{C}}_2}}}{{{{\text{C}}_1} + {{\text{C}}_2}}}} \right) \cdot {\text{Q}}$

  • If $n$ identical capacitors are connected in parallel Equivalent capacitance ${{\text{C}}_{{\text{eq }}}} = n{\text{C}}$ and Charge on each capacitor ${Q^\prime } = \dfrac{Q}{n}$.


Two capacitors are in parallel when their positive plates are connected and negative plates are also connected with each other. In parallel combination, equivalent capacitance is always greater than the individual capacitance.

5. Simple Circuits (series and parallel)

Suppose equivalent capacitance is to be determined in the following networks between points A and B.

6. Dielectric

Dielectrics are insulating (non-conducting) materials which transmits electric effect without conducting. We know that in every atom, there is a positively charged nucleus and a negatively charged electron cloud surrounding it. The two oppositely charged regions have their own centres of charge. The centre of positive charge is the centre of mass of positively charged protons in the nucleus. The centre of negative charge is the centre of mass of negatively charged electrons in the atoms/molecules.

6.1 Polarization of a Dielectric Slab

It is the process of inducing equal and opposite charges on the two faces of the dielectric on the application of electric field.

Suppose a dielectric slab is inserted between the plates of a capacitor. As shown in the figure.

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Induced electric field inside the dielectric is ${{\text{E}}_{\text{i}}}$, hence this induced electric field decreases the main field ${\text{E}}$ to ${\text{E}} - {{\text{E}}_{\text{i}}}$ i.e., New electric field between the plates will be ${E_{{\text{net }}}} = E - {E_1}$.

6.2 Dielectric Constant

After placing a dielectric slab in an electric field. The net field is decreased in that region hence,

If ${\text{E}} = $ Original electric field and ${{\text{E}}_{{\text{net }}}} = $ Net electric field. Then $\dfrac{E}{{{E_{{\text{nut }}}}}} = K$ where $K$ is called dielectric constant $K$ is also known as relative permittivity $\left( {{\varepsilon _r}} \right)$ of the material.

The value of ${\text{K}}$ is always greater than one. For vacuum there is no polarization and hence $E = {E^\prime }$ and $K = 1$

${{\text{E}}_{\text{i}}} = {\text{E}}\left[ {1 - \dfrac{1}{{\text{k}}}} \right]{\sigma _{\text{i}}} = \sigma \left[ {1 - \dfrac{1}{{\text{k}}}} \right]$

6.3 Dielectric Breakdown and Dielectric Strength

If a very high electric field is created in a dielectric, the outer electrons may get detached from their parent atoms. The dielectric then behaves like a conductor. This phenomenon is known as dielectric breakdown.

The maximum value of electric field (or potential gradient) that a dielectric material can tolerate without it's electric breakdown is called it's dielectric strength.

S.I. unit of dielectric strength of a material is${\text{V}}/{\text{m}}$ but practical unit is ${\text{kV/mm}}$.

6.4 Variation of Different Variables (Q,C, V, E and U) of Parallel Plate Capacitor

Suppose we have an air filled charged parallel plate capacitor having variables as follows:

Charge: Q

Surface charge density: $\sigma  = \dfrac{Q}{A}$,

Capacitance: $C = \dfrac{{{\varepsilon _0}A}}{d}$

Potential difference across the plates: ${\text{V}} = {\text{E}}.{\text{d}}$

Electric field between the plates: $E = \dfrac{\sigma }{{{\varepsilon _0}}} = \dfrac{Q}{{A{z_0}}}$

Energy stored: ${\text{L}} = \dfrac{1}{2}C\;{{\text{V}}^2} = \dfrac{{{{\text{Q}}^2}}}{{2C}} = \dfrac{1}{2}{\text{QV}}$


Battery is removed

Battery remains connected

Capacity charge potential

Intensity Energy

\[{\text{C'  =  KC}} \]

\[{\text{Q'  =  Q}}\,\,\,\,\left( {{\text{Charge is conserved}}} \right) \]

\[{\text{V'  =  V/K}} \]

\[{\text{E' = E/K}} \]

 \[{\text{U'  =  U/K}} \]

\[{\text{C'  =  KC}} \]

 \[{\text{Q'  =  KQ}} \]

${\text{V'  =  V}}$ (since battery maintains the potential difference)

 \[{\text{E'  =  E}} \]

 \[{\text{U'  =  KU}} \]


If nothing is said it is to be assumed that the battery is disconnected.

7. Van De Graff Electrostatic Generator

A Van de graff generator is a device used for building up high potential differences of the order of a few million volts Such high potential differences are used to accelerate charged particles like electrons, protons, ions etc. needed for various experiments of Nuclear Physics.

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It was designed by Van de graff in the year 1931.

Principle: This generator is based on

(i) the action of sharp points, i.e., the phenomenon of corona discharge.

(ii) the property that charge given to a hollow conductor is transferred to outer surface and is distributed uniformly over it.

Construction: The essential parts of Vat de graff generator are shown in fig. $S$ is large spherical conducting shell of radius equal to a few meters. This is supported by a conducting shell of radius equal to a few metres. This is supported at a suitable height (of several metres above the ground) over the insulating pillars ${p_1},{{\text{P}}_2}.$ A long narrow belt of insulating material like, silk, rubber or rayon is wrapped around two pulleys ${P_1}$ and ${P_2} \cdot {P_1}$ is at the ground level and ${{\text{P}}_2}$ is at the centre of ${\text{S}}$. The belt is kept moving continuously over the pulleys with the help of a motor (not shown). ${{\text{B}}_1}$ and ${{\text{B}}_2}$ are two sharply pointed metal combs fixed as shown. ${B_1}$ is called the spray comb and ${B_3}$ is called the collecting comb.

The positive ions to be accelerated are produced in a discharge tube $D$. The ion source lies at the head of the tube inside the spherical shell. The other end of the tube carrying the target nucleus is earthed.

The generator is enclosed in a steel chamber C filled with nitrogen or methane at high pressure in order to minimise leakage in a steel spherical conductor.

Working: The spray comb is given a positive potential $\left( { \approx {{10}^4}} \right.$ volt) w.r.t. the earth by high tension source H.T. Due to discharging action of sharp points, a positively charged electric wind is set up, which sprays positive charge on the belt (corona discharge). As the belt moves, and reaches the comb., a negative charge is induced on the sharp ends of collecting comb ${{\text{B}}_2}$ and an equal positive charge is induced on the farther end of ${B_2}$. This positive charge shifts immediately to the outer surface of $S$. Due to discharging action of sharp points of ${B_2}$, a negatively charged electric wind is set up. This neutralises the positive charge on the belt. The uncharged belt returns down, collects the positive charge from ${{\text{B}}_1}$, which in turn is collected by ${{\text{B}}_2}$. This is repeated. Thus, the positive charge on S goes on accumulating.

Now, the capacity of spherical shell $c = 4\pi {\varepsilon _0}R$, where $R$ is radius of the shell.

As $\quad V = \dfrac{Q}{C}\therefore V = \dfrac{Q}{{4\pi { \in _0}R}}$

Hence the potential $V$ of the spherical shell goes on increasing with increase in $Q$.

The breakdown field of air is about $3 \times {10^6}\;{\text{V}}/{\text{m}}$. The moment the potential of spherical shell exceeds this value, air around $S$ is ionised and leakage of charge starts. The leakage is minimised by housing the generator assembly inside a steel chamber filled with nitrogen or methane at high pressures.

If ${\text{q}}$ is the charge on the ion to be accelerated and ${\text{V}}$ is the potential difference developed across the ends of the discharge tube, then energy acquired by the ions $ = {\text{qV}}$. The ions hit the target with this energy and carry out the artificial transmutation etc.

8. Combination of Drops

Suppose we have $n$ identical drops each having - Radius $ - {\text{r}}$, Capacitance $ - {\text{c}}$, Charge $ - {\text{q}}$, Potential - ${\text{v}}$ and Energy $ - {\text{u}}$.

If these drops are combined to form a big drop of Radius $ - {\text{R}}$, Capacitance $ - {\text{C}}$, Charge $ - {\text{Q}}$, Potential $ - {\text{V}}$ and Energy - U then -

(i) Charge on big drop:

$Q = nq$

(ii) Radius of big drop:

Volume of big drop$ = n \times $ volume of a single drop i.e.,

$\dfrac{4}{3}\pi {{\text{R}}^3} = {\text{n}} \times \dfrac{4}{3}\pi {{\text{r}}^3},{\text{R}} = {{\text{n}}^{1/3}}{\text{r}}$

(iii) Capacitance of big drop:

${\text{C}} = {{\text{n}}^{1/3}}{\text{c}}$

(iv) Potential of big drop:

${\text{V}} = \dfrac{{\text{Q}}}{{\text{C}}} = \dfrac{{{\text{nq}}}}{{{{\text{n}}^{1/3}}{\text{c}}}}$

${\text{V}} = {{\text{n}}^{2/3}}{\text{v}}$

(v) Energy of big drop:

${\text{U}} = \dfrac{1}{2}{\text{C}}{{\text{V}}^2} = \dfrac{1}{2}\left( {{{\text{n}}^{1/3}}{\text{c}}} \right){\left( {{{\text{n}}^{2/3}}{\text{v}}} \right)^2}$

${\text{U}} = {{\text{n}}^{5/3}}{\text{u}}$


  • It is a very common misconception that a capacitor stores charge but actually a capacitor stores electric energy in the electrostatic field between the plates.

  • Two plates of unequal area can also form a capacitor because effective overlapping area is considered.

  • Capacitance of a parallel plate capacitor depends up on the effective overlapping area of plates $({\text{C}} \propto {\text{A}})$, separation between the plates $({\text{C}} \propto 1/{\text{d}})$ and dielectric medium filled between the plates. While it is independent of charge given, potential raised or nature of metals and thickness of plates.

  • The distance between the plates is kept small to avoid fringing or edge effect (non-uniformity of the field) at the bounderies of the plates.

  • Spherical conductor is equivalent to a spherical capacitor with it's outer sphere of infinite radius.

  • A spherical capacitor behaves as a parallel plate capacitor if it's spherical surfaces have large radii and are close to each other.

  • The intensity of electric field between the plates of a parallel plate capacitor $\left( {{\text{E}} = \sigma /{\varepsilon _0}} \right)$ does not depends upon the distance between them.

  • Radial and non-uniform electric field exists between the spherical surfaces of spherical capacitor.


(CircuitSolving method)

Sometimes it may not be easy to find the equivalent capacitance of a combination using the equations for series-parallel combinations. We may then use the general method as follows:

Step 1:

Identify the two points between which the equivalent capacitance is to be calculated. Call any one of them as A and the other as ${\text{B}}$.

Step 2:

Connect (mentally) a battery between ${\text{A}}$ and ${\text{B}}$ with the positive terminal connected to ${\text{A}}$ and the negative terminal to ${\text{B}}$. Send a charge $ + {\text{Q}}$ from the positive terminal of the battery and $ - {\text{Q}}$ from the negative terminal of the battery.

Step 3:

Write the charges appearing on each of the plates of the capacitors. The charge conservation principle may be used. The facing surfaces of a capacitor will always have equal and opposite charges. Assume variables ${Q_1},{Q_2} \ldots .$ etc. for charges wherever needed. Mark the polarity across each circuit element correspond ing to higher (+) and lower (-) potential ends.

Step 4:

The algebraic sum of all the potential differences along a closed loop in a circuit is zero.

While using this rule, one starts from a point on the loop and goes along the loop, either clockwise or anticlockwise, to reach the same point again. Any potential difference encountered (from -ve to +ve) is taken to be positive and any potential drop (from + ve to -ve) is taken to be negative. The net sum of all these potential differences should be zero.

The loop law follows directly from the fact that electrostatic force is a conservative force and the work done by it in any closed path is zero.

Step 5:

Number of variables ${Q_1},{Q_2}$, etc. must be the same as the number of equations obtained (loop equation). The equivalent capacitance ${C_{ \oplus q}} = \dfrac{Q}{V}$, where $V$ is the potential difference across the assumed battery terminals.

10. Wheatstone Bridge Based Circuit

If in a network five capacitors are arranged as shown in following figure, the network is called wheatstone bridge type circuit. If it is balanced then

(Image will be uploaded soon) 

$\dfrac{{{C_1}}}{{{C_2}}} = \dfrac{{{C_3}}}{{{C_4}}}$ hence ${C_5}$ is removed and equivalent capacitance between ${\text{A}}$ and ${\text{B}}$

${C_{AB}} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} + \dfrac{{{C_3}{C_4}}}{{{C_3} + {C_4}}}$

11. Extended Wheatstone Bridge

The given figure consists of two wheatstone bridge connected together. One bridge is connected between points AEGHFA and the other is connected between points EGBHFE.

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This problem is known as extended wheatstone bridge problem, it has two branches ${\text{EF}}$ and ${\text{GH}}$ to the left and right of which symmetry in the ratio of capacities can be seen.

It can be seen that ratio of capacitances in branches ${\text{AE}}$ and ${\text{EG}}$ is same as that between the capacitances of the branches ${\text{AF}}$ and ${\text{FH}}$. Thus, in the bridge AEGHFA; the branch EF can be removed. Similarly in the bridge EGBHFE branch ${\text{GH}}$ can be removed

${C_{SE}} = \dfrac{{2C}}{3}$

12. Infinite Network of Capacitors

(i) Suppose the effective capacitance between $A$ and $B$ is ${C_{{\text{P: }}}}$ Since the network is infinite, even if we remove one pair of capacitors from the chain, remaining network would still have infinite pair of capacitors, i.e, effective capacitance between ${\text{X}}$ and I would also be ${{\text{C}}_{\text{R}}}$.

Hence equivalent capacitance between $A$ and $B$

${{\text{C}}_{{\text{AB}}}} = \dfrac{{{{\text{C}}_1}\left( {{{\text{C}}_2} + {C_{\text{f}}}} \right)}}{{{{\text{C}}_1} + {{\text{C}}_2} + {{\text{C}}_{\text{R}}}}} = {{\text{C}}_{\text{R}}}$

$ \Rightarrow \quad {{\text{C}}_{{\text{AB}}}} = \dfrac{{{{\text{C}}_2}}}{2}\left[ {\sqrt {\left( {1 + 4\dfrac{{{{\text{C}}_1}}}{{{{\text{C}}_2}}}} \right)}  - 1} \right]$

(ii) For what value of${C_0}$ in the circuit shown below will the net effective capacitance between $A$ and $B$ be independent of the number of sections in the chain.

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Suppose there are n sections between A and B and the network is terminated by ${{\text{C}}_0}$ with equivalent capacitance ${{\text{C}}_{{{\text{R}}^\prime }}}$ Now if we add one more sections to the network between ${\text{D}}$ and ${\text{C}}$ (as shown in the following figure), the equivalent capacitance of the network ${C_R}$ will be independent of number of sections if the capacitance between $D$ and $C$ still remains ${C_0}$ i.e.,

Hence ${{\text{C}}_0} = \dfrac{{{{\text{C}}_1} \times \left( {{{\text{C}}_2} + {{\text{C}}_0}} \right)}}{{{{\text{C}}_1} + {{\text{C}}_2} + {{\text{C}}_0}}} \Rightarrow {\text{C}}_0^2 + {{\text{C}}_2}{{\text{C}}_0} - {{\text{C}}_1}{{\text{C}}_2} = 0$

On simplification ${{\text{C}}_0} = \dfrac{{{{\text{C}}_2}}}{2}\left[ {\sqrt {\left( {1 + 4\dfrac{{{{\text{C}}_1}}}{{{{\text{C}}_2}}}} \right)}  - 1} \right]$

13. Circuits With Extra Wire (plate numbering method)

If there is no capacitor in any branch of a network then every point of this branch will be at same potential. Suppose equivalent capacitance is to be determine in following cases.

14. Using Symmetry Between Two Points

1. Symmetry is always defined between 2 points.

2 Equivalent (symmetric) paths have same number, value and order of circuit elements along it.

3. When two or more paths in any network are equivalent, then charges flowing through those paths will be same. This technique makes the circuit easy to comprehend.

15. Battery Superposition Method15. Battery Superposition Method

In a circuit involving multiple batteries, the charge flowing will be the superposition effect of each battery.

Effect of $10\;{\text{V}}$ battery

Effect of 5 V battery

Combined effect


When more than 2 batteries are present, take the individual effect of the battery, assuming other batteries are absent. And then superimpose to get the total effect.

16. Dielectric

16.1 Series and Parallel (With Dielectrics)

(a) When dielectric is partially filled between the plates

If a dielectric slab of thickness $t(t < d)$ is inserted between the plates as shown below, then ${\text{E}} = $ Main electric field between the plates, ${{\mathbf{E}}_3} = $ Induced electric field in dielectric. ${E^\prime } = \left( {{\text{E}} - {{\text{E}}_1}} \right) = $ The reduced value of electric field in the dielectric. Potential difference between the two plates of capacitor is given by

\[{V^\prime } = {\text{E}}({\text{d}} - {\text{t}}) + {{\text{E}}^\prime }{\text{t}} = {\text{E}}({\text{d}} - {\text{t}}) + \dfrac{{\text{E}}}{{\text{K}}} \cdot {\text{t}} \]

\[\Rightarrow \quad {{\text{V}}^\prime } = {\text{E}}\left( {{\text{d}} - {\text{t}} + \dfrac{{\text{t}}}{{\text{K}}}} \right) = \dfrac{\sigma }{{{\varepsilon _0}}}\left( {\;{\text{d}} - {\text{t}} + \dfrac{{\text{t}}}{{\text{K}}}} \right) = \dfrac{Q}{{\;{\text{A}}{{\bar \varepsilon }_0}}}\left( {{\text{d}} - {\text{t}} + \dfrac{{\text{t}}}{{\text{K}}}} \right) \]

Now capacitance of the capacitor:

${C^\prime } = \dfrac{Q}{{{V^\prime }}} \Rightarrow {C^\prime } = \dfrac{{{\varepsilon _0}A}}{{d - t + \dfrac{t}{K}}}$

${C^\prime } = \dfrac{{{\varepsilon _0}A}}{{d - \left( {{t_1} + {t_2} + {t_3} +  \ldots  \ldots } \right) + \left( {\dfrac{{{t_1}}}{{{K_1}}} + \dfrac{{{t_2}}}{{{K_2}}} + \dfrac{{{t_3}}}{{{K_3}}} +  \ldots  \ldots } \right)}}$

(b) When a metallic slab is inserted between the plates

${{\text{C}}^\prime } = \dfrac{{{\varepsilon _0}A}}{{(\;{\text{d}} - {\text{t}})}}$

${C^\prime } = \infty $ (In this case capacitor is said to be short circuited)

(c) Advance case of compound dielectrics

If several dielectric medium filled between the plates of a parallel plate capacitor in different ways as shown.

(i) The system can be assumed to be made up of two capacitors ${C_1}$ and ${C_2}$ which may be said to connected in series.

${C_1} = \dfrac{{{K_1}{\varepsilon _0}A}}{{\dfrac{d}{2}}},{C_2} = \dfrac{{{K_2}{\varepsilon _0}A}}{{\dfrac{d}{2}}}$ and $\dfrac{1}{{{C_{eq}}}} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$

$ \Rightarrow \quad {C_{eq}} = \left( {\dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}} \right) \cdot \dfrac{{{\varepsilon _0}A}}{d}$ Also ${K_{eq}} = \dfrac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}}$

(ii) In this case these two capacitors are in parallel and

${{\text{C}}_1} = \dfrac{{{{\mathbf{K}}_1}{\varepsilon _0}\;{\text{A}}}}{{2\;{\text{d}}}},{{\text{C}}_2} = \dfrac{{{{\text{K}}_2}{\varepsilon _0}\;{\text{A}}}}{{2\;{\text{d}}}}$

Hence, ${C_{eq}} = {C_1} + {C_2} \Rightarrow {C_{eq}} = \left( {\dfrac{{{K_1} + {K_2}}}{2}} \right) \cdot \dfrac{{{\varepsilon _0}A}}{d}$

Also ${{\mathbf{K}}_{{\text{eq }}}} = \dfrac{{{{\mathbf{K}}_1} + {{\mathbf{K}}_2}}}{2}$

(iii) In this case ${{\text{C}}_1}$ and ${{\text{C}}_2}$ are in series while this combination is in parallel with ${C_3}$

${C_1} = \dfrac{{{{\text{K}}_1}{\varepsilon _0}\dfrac{{\text{A}}}{2}}}{{\dfrac{{\text{d}}}{2}}} = \dfrac{{{{\text{K}}_1}{\varepsilon _0}\;{\text{A}}}}{{\;{\text{d}}}},{{\text{C}}_2} = \dfrac{{{{\text{K}}_1}{\varepsilon _0}\dfrac{{\text{A}}}{2}}}{{\dfrac{{\text{d}}}{2}}} = \dfrac{{{{\text{K}}_2}{\varepsilon _0}\;{\text{A}}}}{{\;{\text{d}}}}$

and ${C_3} = \dfrac{{{K_3}{\varepsilon _0}\dfrac{A}{2}}}{d} = \dfrac{{{K_3}{\varepsilon _0}A}}{{2d}}$

${\text{ Hence, }}{C_{eq}} = \dfrac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} + {C_3} = \dfrac{{\dfrac{{{K_1}{\varepsilon _0}A}}{d} \times \dfrac{{{K_2}{\varepsilon _0}A}}{d}}}{{\dfrac{{{k_1}{\varepsilon _0}A}}{d} + \dfrac{{{k_2}{\varepsilon _0}A}}{d}}} + \dfrac{{{k_3}{\varepsilon _0}A}}{{2d}}$

$ \Rightarrow \quad {C_{?q}} = \left( {\dfrac{{{k_1}{k_2}}}{{{k_1} + {k_2}}} + \dfrac{{{k_3}}}{2}} \right) \cdot \dfrac{{{\varepsilon _0}A}}{d}$

Also ${{\text{k}}_{eq}} = \left( {\dfrac{{{{\text{k}}_3}}}{2} + \dfrac{{{{\text{k}}_{\text{e}}}{{\text{k}}_2}}}{{{{\text{k}}_{\text{t}}} + {{\text{k}}_2}}}} \right)$

16.2 When Separation Between the Plates Is Changing

If separation between the plates changes then it's capacitance also changes according to ${\text{C}} \propto \dfrac{1}{{\;{\text{d}}}}$. The effect on other variables depends on the fact that whether the charged capacitor is disconnected from the battery or battery is still connected.

(i) Separation is increasing


Battery is removed

Battery remains connected




Potential difference

Electric field


Decreases because ${\text{C}} \propto \dfrac{1}{{\;{\text{d}}}}$ i.e., ${{\text{C}}^\prime } < {\text{C}}$

Remains constant because a battery is not present i.e., ${Q^\prime } = Q$

Increases because ${\text{v}} = \dfrac{{\text{Q}}}{{\text{C}}} \Rightarrow {\text{V}} \propto \dfrac{1}{{\text{C}}}$ i.e., ${{\text{V}}^\prime } > {\text{V}}$ difference)

Remains constant because ${\text{E}} = \dfrac{\sigma }{{{\varepsilon _{\text{D}}}}} = \dfrac{{\text{Q}}}{{{\text{A}}{\varepsilon _{\text{D}}}}}$ i.e., ${{\text{E}}^\prime } = {\text{E}}$

Increases because ${\text{U}} = \dfrac{{{{\text{Q}}^2}}}{{2{\text{C}}}} \Rightarrow {\text{U}} \propto \dfrac{1}{{\text{C}}}$ i.e., ${{\text{U}}^\prime } > {\text{U}}$

Decreases i.e., $\quad {C^\prime } < {\text{C}}$

Decreases because battery is present i.e., ${Q^\prime } < Q$ Remaining charge $\left( {Q - {Q^\prime }} \right)$ goes back to the

${V^\prime } = V$ (Since Battery maintains the potential

Decrease because ${\text{E}} = \dfrac{Q}{{\;{\text{A}}{\varepsilon _0}}} \Rightarrow {\text{E}} \propto {\text{Q}}$ i.e., ${E^{\prime}}< E$

Decreases because ${\text{U}} = \dfrac{1}{2}{\text{C}}{{\text{V}}^2} \Rightarrow {\text{U}} \propto C$ i.e., ${U^\prime } < {\text{U}}$

(ii) Separation is decreasing


Battery is removed

Battery remains connected




Potential difference

Electric field


Increases because ${\text{C}} \propto \dfrac{1}{{\;{\text{d}}}}$ i.e., ${C^\prime } > {\text{C}}$

Remains constant because battery is not present i.e., ${Q^\prime } = Q$

Decreases because ${\text{V}} = \dfrac{{\text{Q}}}{{\text{C}}} \Rightarrow {\text{V}} \propto \dfrac{1}{{\text{C}}}$ i.e., ${\text{V}} < {\text{V}}$

Remains constant because ${\text{E}} = \dfrac{\sigma }{{{\varepsilon _0}}} = \dfrac{{\text{Q}}}{{{\text{A}}{\varepsilon _0}}}$ i.e., ${{\text{E}}^\prime } = {\text{E}}$

Decreases because ${\text{U}} = \dfrac{{{{\text{Q}}^2}}}{{2{\text{C}}}} \Rightarrow {\text{U}} \propto \dfrac{1}{{\text{C}}}$ i.e.,${\text{U}} < {\text{U}}$

Increases i.e., ${C^\prime } > C$

Increases because battery is present i.e., ${Q^\prime } > Q$ Remaining charge $\left( {{Q^\prime } - Q} \right)$ supplied from the 

${V^\prime } = V$ (Since Battery maintains the potential difference)

Increases because $E = \dfrac{Q}{{A{\varepsilon _0}}} \Rightarrow E \propto Q$ i.e., ${E^\prime } > E$ 

Increases because ${\text{U}} = \dfrac{1}{2}{\text{C}}{{\text{V}}^2} \Rightarrow {\text{U}} \propto {\text{C}}$ i.e., ${{\text{U}}^\prime } > {\text{U}}$

16.3 Force on Dielectric

When dielectric is placed near the charged capacitor (rectangular plates), it experiences force towards the capacitor, due to fringing field just outside the plates.

(Image will be uploaded soon)

(a) Battery connected (V remains same)

$F = \dfrac{1}{2}\dfrac{{{\varepsilon _0}b{V^2}(K - 1)}}{d}$ (towards capacitor)

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  • Force doesn't depend on the amount of dielectric inside the plates.

  • Force becomes zero when Dielectric is in middle of plates.

(b) Battery disconnected (Q remains same)

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\[F=\dfrac {{Q^2}d\left ( k-1 \right )}{2{\varepsilon_0}b\left ( l+x\left ( k-1 \right ) \right )^{2}}\]


Force depends on ${\text{x}}$ (amount of dielectric inside the capacitor plates).

17. Redistribution of Charge Between Two Capacitors

When a charged capacitor is connected across an uncharged capacitor, then redistribution of charge occur to equalize the potential difference across each capacitor. Some energy is also wasted in the form of heat.

Suppose we have two charged capacitors ${{\text{C}}_1}$ and ${{\text{C}}_2}$ after disconnecting these two from their respective batteries. These two capacitors are connected to each other as shown below (positive plate of one capacitor is connected to positive plate of other while negative plate of one is connected to negative plate of other).

Charge on capacitors redistributed and new charge on them will be $Q_i^\prime  = Q\left( {\dfrac{{{C_1}}}{{{C_1} + {C_2}}}} \right),Q_2^\prime  = Q\left( {\dfrac{{{C_2}}}{{{C_1} + {C_2}}}} \right)$ The common potential ${\text{V}} = \dfrac{{{{\text{Q}}_1} + {{\text{Q}}_2}}}{{{{\text{C}}_1} + {{\text{C}}_2}}} = \dfrac{{{{\text{C}}_1}\;{{\text{V}}_1} + {{\text{C}}_2}\;{{\text{V}}_2}}}{{{{\text{C}}_1} + {{\text{C}}_2}}}$ and loss of energy $\Delta U = \dfrac{{{C_1}{C_2}}}{{2\left( {{C_l} + {C_2}} \right)}}{\left( {{V_1} - {V_2}} \right)^2}$.


Two capacitors of capacitances ${{\text{C}}_1}$ and ${{\text{C}}_2}$ are charged to potential of ${{\text{V}}_{\text{L}}}$ and ${{\text{V}}_2}$ respectively. After disconnecting from batteries they are again connected to each other with reverse polarity i.e., positive plate of a capacitor connected to negative plate of other. So common potential ${\text{V}} = \dfrac{{{{\text{Q}}_1} - {{\text{Q}}_2}}}{{{{\text{C}}_1} + {{\text{C}}_2}}} = \dfrac{{{{\text{C}}_1}\;{{\text{V}}_1} - {{\text{C}}_2}\;{{\text{V}}_2}}}{{{{\text{C}}_1} + {{\text{C}}_2}}}$.

Electric Field and Charges Class 12 Notes for Chapter 1

To assist you with that, Vedantu is here with Class 12 Physics Chapter 1 Notes. Class 12 Physics Notes prepared by our subject experts will help you understand the important topics and remember the key points from the exam point of view. Below we have provided the class 12 physics chapter 1 electric charges and field notes. 


Physics Class 12 Chapter 1 Notes

As per the latest information, Unit-1 has a weightage of 9 % in the NEET exam. In the JEE Main exam, 2 to 3 questions with 8-12 % weightage are likely to come. This topic carries a greater weightage in board exams and the competitive entrance exams.The , electric charges and fields class 12 notes PDF will help to learn this topic more effectively.

So, Class 12 Physics Chapter 1 is divided into two parts, i.e., Electric Charges and Electric Fields.


Electric Charges and Fields Class 12 Notes

  • Unlike Charges: These charges repel each other.

  • Like Charges: These charges attract each other.


Quantization of Charge

Quantization of electric charge is an attribute in which all free charges are an integral multiple of a basic unit of charge represented by e. The charge (q)  of a body is given by:

 q =± ne              

Here, + e is for proton & - e is for electron.    


Additivity of Charges

The total charge in the system = The algebraic sum of all the individual charges present in the system.


Conservation of Charge

The total charge of an isolated system remains conserved.

For example, on rubbing the rod with silk, silk gains the charge, while the rod loses it.


Coulomb’s Law

Coulomb’s law tells about the force of attraction or repulsion between any two point charges. This force is directly proportional to their product and inversely proportional to the square of the distance between them.


This force acts along the line joining the center of these two charges. It is given by:

                   F α \[\frac{|q_{1}q_{2}|}{r^{2}}\]

                   F = k \[\frac{|q_{1}q_{2}|}{r^{2}}\]

k = electrostatic force constant, which is equal to\[\frac{1}{4 \pi \epsilon _{0}}\].

F =  \[\frac{1}{4 \pi \epsilon _{0}}\] \[\frac{|q_{1}q_{2}|}{r^{2}}\] 

Here, ε0 = Permittivity of free space, with a value of 8.85 x 10-12C2N-1m-2and dimension of [M-1L-3T4A2].

Forces Between Multiple Charges

Coulomb’s Law helps to find the electrostatic force between two point charges, but how to find the force between multiple charges?

Experiments show that force on a charge due to other charges around it, taken one at a time, is the vector sum of all the forces exerted by all the charges on that charge. This is given by the superposition principle, stated below.

Superposition Principle

The superposition principle states that the individual forces of two charges remain unaffected due to the presence of other charges.

Let us consider a system of three charges q1, q2 and q3. The force on q1 due to q2 is denoted by F12 and due to q3 is  F13. 

The force on a charge, say q1 due to two other charges q2 and q3 can be obtained by vector addition of the forces due to q2 on q1 and q3 on q1. 


$F_{12} = 14 \pi \epsilon q_1 q_2 r_{212} r_{12}$      


$F_{13} = 14 \pi \epsilon q_1 q_3 r_{213} r_{13}$      

Thus, the total force F1 on q1 due to q2 and q3 is,

F1 =  F12 + F13

$F_1 = 14 \pi \epsilon q_1 q_2 r_{212} r_{12} + 14 \pi \epsilon q_1 q_3 r_{213} r_{13}$


Properties of Electric Field Lines

  1. Electric Field lines are endless curves, but they can’t form closed loops.

  2. They Intersect each other at infinity.

  3. Electric field lines initiate from positive charges and terminate at negative charges.


Electric Field Due to Dipole

\[|\overrightarrow{E}|\] = \[\frac{\overrightarrow{p}}{4 \pi \epsilon _{0}}\] \[\frac{2r}{(r^{2}-a^{2})}\]

If the dipole is short, and  2a < < r, then the formula becomes:

\[|\overrightarrow{E}|\] = \[\frac{2|\overrightarrow{p}|}{4 \pi \epsilon _{0}r^{3}}\]


Gauss Law

Gauss law states that the total electric flux out of the Gaussian surface equals the integration of net charge enclosed in this surface divided by the permittivity of free space. Its formula is:

ФE = ∮S \[|\overrightarrow{E}|\] . \[| \overrightarrow{ds} | \]  \[\frac{Q}{ \epsilon _{0}}\]

(Image to be added soon)


Solved Sample Numerical Problem

How should we divide a charge ‘Q’ to get the maximum force of repulsion between them?

Let q, Q - q be two point charges, then,

               F =\[\frac{1}{4 \pi \epsilon _{0}r^{2}}\] \[\frac{ |q||Q-q|}{r^{2}}\]

For F to be maximum, dF/dq should be = 0

So,\[\frac{dF}{dq}\]=\[\frac{1}{4 \pi \epsilon _{0}r^{2}}\] \[\frac{d}{dq}\] ( \[\frac{ |q||Q-q|}{r^{2}}\] )= 0

=> \[\frac{1}{4 \pi \epsilon _{0}r^{2}}\] \[\frac{ |Q-2q|}{r^{2}}\]= 0 

q = Q/2

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Notes of Physics Class 12 – Chapter 1 Electric Charges and Fields

Electric Charges and Fields is a fundamental concept of Physics. The chapter explains the basic concepts like Electric Charge, its properties and electric field. Electric Charge is the physical property of matter when it comes in the electromagnetic field due to which other matters experience a force. There are two types of Electric Charges:

  • Positive Electric Charge

  • Negative Electric Charge

Electric Field: Electric field is a region around a matter due to which another matter experiences a force.

The formula of Electric Field is: E= F/q

where E- Electric Field, F- Force and q- Charge

Based on Electric Charge and Electric Field, there are various equations, and laws are explained in Class 12 Chapter 1 Physics.

CBSE Class 12 Physics Notes Chapter 1 Electric Charges and Fields provide you with a base for practical application questions along with the proper explanation of these equations and laws.

While reading the CBSE Chapter 1 Physics Class 12 Notes, students will come across various fundamentals topics like

  • Electric Charge

  • Electric potential energy

  • The relationship between the electric field and potential

  • Electric dipole

  • Gauss's Law and its application

  • Properties of a conductor

  • Plate theory

  • Properties of Charge 

  • Electroscope 

  • Coulomb’s Law

  • Electric Potential Energy

While Chapter 1 Physics Class 12 Notes are invaluable, only reading the notes won't help, you will need to practice for easy implementation of the theoretical principles. 



Class12 Physics Ch1 Notes have been written concisely, but the lucid explanation helps for clear interpretation.  CBSE Class 12 Physics Revision Notes Chapter 1 can be downloaded and read offline as well. The PDF will help students cover their syllabus well in time. Students must make sure to start reading these notes 3-4 months before the exams. This will help students to stay calm during the exam.

FAQs on Electric Charges and Fields Class 12 Notes CBSE Physics Chapter 1 (Free PDF Download)

1. Force F is acting between two point charges. If a sheet of metal ( εr = 8) is placed between these charges, then what will be the force?

Let’s use the formula:  F =   ( 1/(4 𝜋 𝜖0) ) ( |q1| |q2| / r2 )

F’ = ( 1/(4 𝜋 𝜖0K) ) ( |q1| |q2| / r2 ) = F/K = F/6

 2. Does a Moving Charge Have an Electric Field?

Yes, a moving charge has both electric and magnetic fields.

3. Write Applications of Gauss Law.

The applications of Gauss law are as follows.

  • Gauss law helps solve complex electrostatic problems involving specific symmetries like cylindrical, spherical, or planar symmetries. 

  • This law simplifies the evaluation of electric fields.

  • The importance of this law comes when it determines the number of electric field lines passing through a closed surface without taking into consideration the radius and its inner surface.

4. How Can I Develop Numerical Solving Ability in Physics?

A subject like physics demands a great hold on reasoning and mathematics. To develop the numerical solving ability, you need to brush up your conceptual understanding. We have provided you with Electric Field and Charges Class 12 Notes. For all your doubts, get in touch with our experts.

5. Write properties of Electric field lines Class12 Physics Ch1 Notes?

Electric field lines have several properties such as - 

  • The field lines have a curve. 

  • There is no break where no charge is present. 

  • Another important property of electric field lines is that the two lines are always separated from each other and never intersect with each other. 

  • The electric field lines never form closed loops and they begin from the positive charge and stop at the negative side.

These properties have been explained in brief in the Electric Field and Charges Class 12 Notes.

6. What are the important topics covered in Class 12 Physics Chapter 1?

Class 12 Physics Chapter 1 is mainly about electric charges and electric fields. The chapter deals with the properties of electric charges, the difference between conductors and insulators, activity of charges, transfer of electrons, etc. It is an important chapter for the board exams. Therefore, students should focus on the main topics given in the chapter. They can refer to the Class 12 Physics Chapter 1 Revision Notes for understanding the concepts.

7. What do you understand by quantization of charge Class 12 Physics?

An object will either have a positive or a negative charge. A charge is present in discrete amounts. When a quantity is present in discrete amounts and not in continuous amounts, it means the quantity is quantized. Therefore, the charge is also present in discrete amounts and is not continuous, it is also said to be quantized. Students will study the different concepts of charge in Class 12 Physics Chapter 1.

8. What is the benefit of studying Chapter 1 of Class 12 Physics from Vedantu?

Students of Class 12 can get help in understanding the concepts of Chapter 1 of Physics if they study from Vedantu’s Revision Notes for this chapter. All important topics of Class 12 Physics Chapter 1 are explained here in simple and easy language. Class 12 Physics Chapter 1 notes available on the Vedantu website and the Vedantu app for free, are prepared by professional and expert teachers who have proper knowledge of the concepts and help students to understand the important concepts without any difficulty.

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Students can find many websites offering Class 12 Physics notes Chapter 1. You can visit Vedantu to download the relevant and authentic Class 12 Physics notes. Vedantu hires experienced scholars for preparing the notes. The main aim of making Class 12 Physics notes is to help students understand the Physics concepts and score high marks in their board exams and ace competitive exams. Visit the page Class 12 Physics Revision Notes for Chapter 1 to download the Class 12 Physics Chapter 1 Revision Notes at free of cost.