NCERT Solutions for Class 12 Physics Chapter 2 - Electrostatic Potential And Capacitance

• Electric Potential: Electric potential at a point in an electric field is the amount of work done in bringing a unit positive charge from infinity to the point.   

(i) Electric potential is a scalar quantity.

(ii) S.I. unit: Volt(V).  

(iii) A positive charge in a field moves from higher potential to lower potential whereas an electron moves from lower potential to higher potential when left free.   

(iv) Work done in moving a charge q through a potential difference V is W = qV joule.

• If potential at infinity is chosen to be zero; potential at a point with position vector r due to a point charge Q placed at the origin is given is given by

$V(r)-\frac{1}{4\pi\varepsilon _0}\frac{Q}{r}$

For a charge configuration q1, q2, ..., qn with position vectors r1 , r2 , ... rn, the potential at a point P is given by the superposition principle

$V(r)-\frac{1}{4\pi\varepsilon _0}\left ( \frac{q_1}{r_{1P}}\frac{q_2}{r_{2P}}+...+\frac{q_n}{r_{nP}} \right )$

where r1P is the distance between q1 and P, and so on.

• Relation between electric field and potential:

$V=-\iint_E^{\amalg} \cdot d \stackrel{凶}{\mathrm{凶}}$

$\Rightarrow \mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dr}}$

• An equipotential surface is a surface over which potential has a constant value. For a point charge, concentric spheres centred at a location of the charge are equipotential surfaces. The electric field E at a point is perpendicular to the equipotential surface through the point. E is in the direction of the steepest decrease of potential.

• Electric potential at any point due to an electric dipole $V \equiv \frac{k p \cos \theta}{r^2}$; $\theta$ ; is the angle

• Between $\stackrel{\mathrm{w}}{\mathrm{p}} and\, \stackrel{\mathrm{w}}{\mathrm{w}}$

• Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations. Potential energy of two charges q1, q2 at r1, r2 is given by $U=\frac{1}{4\pi\varepsilon _0}\frac{q_1q_2}{r_{12}}$ where r12 is distance between q1 and q2. The potential energy of a charge q in an external potential V(r) is qV(r). The potential energy of a dipole moment p in a uniform electric field E is –p.E.

• Electrostatics field E is zero in the interior of a conductor; just outside the surface of a charged conductor, E is normal to the surface given by $E=\frac{\sigma}{\varepsilon _0}n$ where n is the unit vector along the outward normal to the surface and σ is the surface charge density. Charges in a conductor can reside only at its surface. Potential is constant within and on the surface of a conductor. In a cavity within a conductor (with no charges), the electric field is zero.

• A capacitor is a system of two conductors separated by an insulator. Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. C is determined purely geometrically, by the shapes, sizes and relative positions of the two conductors. The unit of capacitance is farad:, 1 F = 1 C V–1. For a parallel plate capacitor (with vacuum between the plates), $C=\varepsilon _0\frac{A}{d}$ where A is the area of each plate and d the separation between them.

• If the medium between the plates of a capacitor is filled with an insulating substance (dielectric), the electric field due to the charged plates induces a net dipole moment in the dielectric. This effect, called polarisation, gives rise to a field in the opposite direction. The net electric field inside the dielectric and hence the potential difference between the plates is thus reduced. Consequently, the capacitance C increases from its value C0 when there is no medium (vacuum), 

C = KC0

where K is the dielectric constant of the insulating substance.

• For capacitors in the series combination, the total capacitance C is given by $\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...$ In the parallel combination, the total capacitance C is: 

$C=C_1+C_2+C_3+...$

where C1, C2, C3 ... are individual capacitances.

• The energy U stored in a capacitor of capacitance C, with charge Q and voltage V is

$U=\frac{1}{2}QV=\frac{1}{2}CV^2=\frac{1}{2}\frac{Q^2}{C}$ 

The electric energy density (energy per unit volume) in a region with electric field is (1/2)ε0E2.

Mastering Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance - MCQs, QnA and Tips for Success

1. Two charges \[5\times {{10}^{-8}}C\] and \[-3\times {{10}^{-8}}C\]  are located \[16\text{ }cm\] apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Ans: It is provided that,

First charge, ${{q}_{1}}=5\times {{10}^{-8}}C$

Second charge, \[{{q}_{2}}=-3\times {{10}^{-8}}C\]

Distance between the two given charges, $d=16cm=0.16m$

Case 1. When point P is inside the system of two charges.

Consider a point named P on the line connecting the two charges.

(Image will Be Updated Soon)

r is the distance of point P from ${{q}_{1}}$.

Potential at point P will be,

$V=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}r}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(d-r)}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

But $V=0$so,

$0=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}r}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(d-r)}$

$\Rightarrow \frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}r}=-\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(d-r)}$

$\Rightarrow \frac{{{q}_{1}}}{r}=-\frac{{{q}_{2}}}{(d-r)}$

\[\Rightarrow \frac{5\times {{10}^{-8}}}{r}=-\frac{-3\times {{10}^{-8}}}{(0.16-r)}\]

\[\Rightarrow \frac{0.16}{r}=\frac{8}{5}\]

We get,

$r=0.1m=10cm$

Therefore, the potential is zero at \[10\text{ }cm\] distance from the positive charge.

Case 2. When point P is outside the system of two charges.

(Image will Be Updated Soon)

Potential at point P will be,

$V=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}s}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(s-d)}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

But $V=0$so,

$0=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}s}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(s-d)}$

$\Rightarrow \frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}s}=-\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}(s-d)}$

$\Rightarrow \frac{{{q}_{1}}}{s}=-\frac{{{q}_{2}}}{(s-d)}$

\[\Rightarrow \frac{5\times {{10}^{-8}}}{s}=-\frac{-3\times {{10}^{-8}}}{(s-0.16)}\]

\[\Rightarrow \frac{0.16}{s}=\frac{2}{5}\]

We get,

$s=0.4m=40cm$

Therefore, the potential is zero at \[40\text{ }cm\] distance from the positive charge.

2. A regular hexagon of side \[10\text{ }cm\] has a charge \[5\mu C\] at each of its vertices. Calculate the potential at the centre of the hexagon.

Ans: The given figure represents six equal charges, $q=5\times {{10}^{-6}}C$, at the hexagon's vertices.

(Image will Be Updated Soon)

Sides of the hexagon, $AB=BC=CD=DE=EF=FA=10cm$

The distance of O from each vertex, $d=10cm$

Electric potential at point O,

$V=\frac{6q}{4\pi {{\varepsilon }_{o}}d}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow V=\frac{6\times 9\times {{10}^{9}}\times 5\times {{10}^{-6}}}{0.1}$

$\Rightarrow V=2.7\times {{10}^{6}}V$

Clearly, the potential at the hexagon’s centre is $2.7\times {{10}^{6}}V$.

3. Two charges \[2\mu C\] and \[-2\mu C\] are placed at points A and B, \[\mathbf{6}\text{ }\mathbf{cm}\] apart. 

a) Identify an equipotential surface of the system. 

Ans: The given figure represents two charges.

(Image will Be Updated Soon)

An equipotential surface is defined as that plane on which electric potential is equal at every point. One such plane is normal to line AB. The plane is placed at the mid-point of line AB because the magnitude of charges is equal.

b) What is the direction of the electric field at every point on this surface?

Ans: The electric field's direction is perpendicular to the plane in the line AB direction at every location on this surface.

4. A spherical conductor of radius \[\mathbf{12}\text{ }\mathbf{cm}\] has a charge of $1.6\times {{10}^{-7}}C$ distributed uniformly on its surface. What is the electric field, 

a) inside the sphere?

Ans: It is provided that,

Spherical conductor’s radius, \[r=12cm=0.12m\]

The charge is evenly distributed across the conductor. The electric field within a spherical conductor is zero because the total net charge within a conductor is zero.

b) just outside the sphere?

Ans: Just outside the conductor, Electric field E is given by

$E=\frac{q}{4\pi {{\varepsilon }_{o}}{{r}^{2}}}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow E=\frac{1.6\times {{10}^{-7}}\times 9\times {{10}^{9}}}{{{(0.12)}^{2}}}$

$\Rightarrow E={{10}^{5}}N{{C}^{-1}}$

Clearly, the electric field just outside the sphere is ${{10}^{5}}N{{C}^{-1}}$.

c) at a point \[\mathbf{18}\text{ }\mathbf{cm}\] from the centre of the sphere?

Ans: Let electric field at a given point which is \[18\text{ c}m\] from the sphere centre = ${{E}_{1}}$

Distance of the given point from the centre, $d=\mathbf{18}\text{ }\mathbf{cm}=0.18m$

The formula for electric field is given by,

${{E}_{1}}=\frac{q}{4\pi {{\varepsilon }_{o}}{{d}^{2}}}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow E=\frac{1.6\times {{10}^{-7}}\times 9\times {{10}^{9}}}{{{(0.18)}^{2}}}$

$\Rightarrow E=4.4\times {{10}^{4}}N{{C}^{-1}}$

Therefore, the electric field at a given point \[18\text{ }cm\] from the sphere centre is $4.4\times {{10}^{4}}N{{C}^{-1}}$.

5. A parallel plate capacitor with air between the plates has a capacitance of \[~8pF\]. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant \[6\]?

Ans: It is provided that,

Capacitance between the capacitor’s parallel plates, \[C=8pF\]

Originally, the distance separating the parallel plates was d, and the air was filled in it.

Dielectric constant of air, \[k=1\]

The formula for Capacitance is given by:

$C=\frac{k{{\varepsilon }_{o}}A}{d}$

Here, \[k=1\], so,

$C=\frac{{{\varepsilon }_{o}}A}{d}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

A is the area of each plate.

If the distance separating the plates is decreased to half and the substance has a dielectric constant of \[6\] filled in between the plates.

Then,

$k'=6,d'=\frac{d}{2}$

Hence, capacitor’s capacitance becomes,

$C'=\frac{k'{{\varepsilon }_{o}}A}{d'}$

$\Rightarrow C'=\frac{6{{\varepsilon }_{o}}A}{\frac{d}{2}}$

$\Rightarrow C'=12C$

$\Rightarrow C'=12\times 8=96pF$

Therefore, the capacitance when the substance of dielectric constant $6$ is filled between the plates is \[96\text{ }pF\].

6. Three capacitors each of capacitance \[\mathbf{9}\text{ }\mathbf{pF}\] are connected in series. 

a) What is the total capacitance of the combination? 

Ans: It is provided that,

Capacitance of each three capacitors, \[C=9pF\]

The formula for equivalent capacitance $(~C')$ of the capacitors’ series combination is given by

$\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$

$\Rightarrow \frac{1}{C'}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}$

$\Rightarrow \frac{1}{C'}=\frac{3}{9}$

$\Rightarrow C'=3pF$

Clearly, total capacitance of the combination of the capacitors is \[3pF\].

b) What is the potential difference across each capacitor if the combination is connected to a \[\mathbf{120}\text{ }\mathbf{V}\] supply?

Ans: Provided that,

Supply voltage, \[V=120\text{ }V\]

Potential difference \[(V')\] across each capacitor will be one-third of the supply voltage.

$V'={}^{120}/{}_{3}=40V$

Clearly, the potential difference across each capacitor is \[40\text{ }V\].

7. Three capacitors of capacitance \[2pF,\text{ }3\text{ }pF\text{ }and\text{ }4pF\] are connected in parallel. 

a) What is the total capacitance of the combination?

Ans: Provided that,

Capacitances of the given capacitors are, 

\[{{C}_{1}}=2pF\text{ };\text{ }{{C}_{2}}=3pF\text{ };\text{ }{{C}_{3}}=4pF\]

The formula for equivalent capacitance $(~C')$ of the capacitors’ parallel combination is given by

\[C'={{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]

\[\Rightarrow C'=2+3+4=9pF\]

Therefore, total capacitance of the combination is \[9pF.\]

b) Determine the charge on each capacitor if the combination is connected to a \[100\text{ }V\] supply.

Ans: We have,

Supply voltage, \[V=100\text{ }V\]

Charge on a capacitor with capacitance C and potential difference V is given by,

$q=CV$……(i)

For $C=2pF$,

Charge $=VC=100\times 2=200pF$

For $C=3pF$,

Charge $=VC=100\times 3=300pF$

For $C=4pF$,

Charge $=VC=100\times 4=400pF$

8. In a parallel plate capacitor with air between the plates, each plate has an area of $6\times {{10}^{-3}}{{m}^{2}}$ and the distance between the plates is \[3\text{ }mm\]. Calculate the capacitance of the capacitor. If this capacitor is connected to a \[100\text{ }V\] supply, what is the charge on each plate of the capacitor?

Ans: It is provided that,

Area of parallel plate capacitor’s each plate, $A=6\times {{10}^{-3}}{{m}^{2}}$

Distance separating the plates, \[d=3mm=3\times {{10}^{-3}}m\]

Supply voltage, \[V=100\text{ }V\] 

The formula for parallel plate capacitor’s Capacitance is given by,

$C=\frac{{{\varepsilon }_{o}}A}{d}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

$\Rightarrow C=\frac{8.854\times {{10}^{-12}}\times 6\times {{10}^{-3}}}{3\times {{10}^{-3}}}$

\[\Rightarrow C=17.71\times {{10}^{-12}}F\]

\[\Rightarrow C=17.71pF\]

The formula for Potential V is related with charge q and capacitance C is given by,

$V=\frac{q}{C}$

$\Rightarrow q=CV=100\times 17.71\times {{10}^{-12}}$

$\Rightarrow q=1.771\times {{10}^{-9}}C$

Clearly, the capacitor's capacitance is \[17.71\text{ }pF\] and charge on each plate is $1.771\times {{10}^{-9}}C$.

9. Explain what would happen if in the capacitor given in Exercise 8, a \[3\text{ }mm\] thick mica sheet (of dielectric constant \[=6\]) were inserted between the plates, 

a) while the voltage supply remained connected.

Ans: It is provided that,

Mica sheet’s Dielectric constant, k = 6 

Initial capacitance, \[C=17.71\times {{10}^{-12}}F\]

New capacitance, \[C'=kC=6\times 17.71\times {{10}^{-12}}=106pF\]

Supply voltage, \[V=100V\]

New charge, $q'=C'V'=106\times 100pC=1.06\times {{10}^{-8}}C$

Potential across the plates will remain \[100\text{ }V\].

c) after the supply was disconnected.

Ans: It is provided that,

Mica sheet’s Dielectric constant, k = 6 

Initial capacitance, \[C=17.71\times {{10}^{-12}}F\]

New capacitance, \[C'=kC=6\times 17.71\times {{10}^{-12}}=106pF\]

If supply voltage is disconnected, then there will be no influence on the charge amount on the plates. 

The formula for potential across the plates is given by,

$V'=\frac{q}{C'}$

$V'=\frac{1.771\times {{10}^{-9}}}{106\times {{10}^{-12}}}=16.7V$

The potential across the plates when the supply was removed is $16.7V$.

10. A \[12pF\] capacitor is connected to a \[50\text{ }V\] battery. How much electrostatic energy is stored in the capacitor?

Ans: It is provided that,

Capacitance of the capacitor, \[C=12\times {{10}^{-12}}F\]

Potential difference, \[V=50\text{ }V\]

The formula for stored electrostatic energy in the capacitor is given by,

$E=\frac{1}{2}C{{V}^{2}}$

$\Rightarrow E=\frac{1}{2}\times 12\times {{10}^{-12}}\times {{50}^{2}}$

$\Rightarrow E=1.5\times {{10}^{-8}}J$

Therefore, the stored electrostatic energy in the capacitor is $1.5\times {{10}^{-8}}J$.

11. A \[600\text{ }pF\] capacitor is charged by a \[200\text{ }V\] supply. It is then disconnected from the supply and is connected to another uncharged \[600\text{ }pF\] capacitor. How much electrostatic energy is lost in the process?

Ans: It is provided that,

Capacitance of the capacitor, \[C=600\text{ }pF\]

Potential difference, \[V=200\text{ }V\]

The formula for stored electrostatic energy in the capacitor is given by,

$E=\frac{1}{2}C{{V}^{2}}$

$\Rightarrow E=\frac{1}{2}\times 600\times {{10}^{-12}}\times {{200}^{2}}$

$\Rightarrow E=1.2\times {{10}^{-5}}J$

If supply is removed from the capacitor and another capacitor of capacitance \[C=600\text{ }pF\] is joined to it, then equivalent capacitance $(C')$ of the series combination is given by

$\frac{1}{C'}=\frac{1}{C}+\frac{1}{C}$

$\Rightarrow \frac{1}{C'}=\frac{1}{600}+\frac{1}{600}$

$\Rightarrow \frac{1}{C'}=\frac{2}{600}$

$\Rightarrow C'=300pF$

New electrostatic energy will be,

$E'=\frac{1}{2}C'{{V}^{2}}$

$\Rightarrow E'=\frac{1}{2}\times 300\times {{10}^{-12}}\times {{200}^{2}}$

$\Rightarrow E'=0.6\times {{10}^{-5}}J$

Loss in electrostatic energy $=E-E'$

$\Rightarrow E-E'=1.2\times {{10}^{-5}}-0.6\times {{10}^{-5}}=0.6\times {{10}^{-5}}J$

$\Rightarrow E-E'=6\times {{10}^{-6}}J$

Clearly, the lost electrostatic energy in the process is $6\times {{10}^{-6}}J$.

12. A charge of \[8\text{ }mC\] is located at the origin. Calculate the work done in taking a small charge of $-2\times {{10}^{-9}}C$ from a point \[P\left( 0,0,3\text{ }cm \right)\] to a point \[Q\left( 0,4\text{ }cm,0 \right)\], via a point \[R(0,6\text{ }cm,9\text{ }cm)\].

Ans: Charge located at the origin, $q=8\text{ }mC=8\times {{10}^{-3}}C$

A small charge is moved from a point P to point R to point Q, ${{q}_{1}}=-2\times {{10}^{-9}}C$

The figure given below represents all points.

(Image will Be Updated Soon)

Potential at point P, \[{{V}_{1}}=\frac{q}{4\pi {{\varepsilon }_{o}}{{d}_{1}}}\]

Where ${{d}_{1}}=3cm=0.03m$

Potential at point Q, \[{{V}_{2}}=\frac{q}{4\pi {{\varepsilon }_{o}}{{d}_{2}}}\]

Where ${{d}_{2}}=4cm=0.04m$

Work done by the electrostatic force does not depend on the path.

$W={{q}_{1}}\left[ {{V}_{2}}-{{V}_{1}} \right]$

$\Rightarrow W={{q}_{1}}\left[ \frac{q}{4\pi {{\varepsilon }_{o}}{{d}_{2}}}-\frac{q}{4\pi {{\varepsilon }_{o}}{{d}_{1}}} \right]$

$\Rightarrow W=\frac{{{q}_{1}}q}{4\pi {{\varepsilon }_{o}}}\left[ \frac{1}{{{d}_{2}}}-\frac{1}{{{d}_{1}}} \right]$

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow W=9\times {{10}^{9}}\times 8\times {{10}^{-3}}\times (-2\times {{10}^{-9}})\left[ \frac{1}{0.04}-\frac{1}{0.03} \right]$

$\Rightarrow W=1.27J$

Clearly, work done during the process is $1.27J$.

13. A cube of side \[b\] has a charge \[q\] at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Ans: Length of each cube’s side $=l$

Charge at each of cube’s vertices $=q$

This figure shows the cube of side b.

(Image will Be Updated Soon)

d is the length of the diagonal of face of the cube.

$d=\sqrt{{{b}^{2}}+{{b}^{2}}}=\sqrt{2{{b}^{2}}}$

$d=\sqrt{2}b$

l is the length of the diagonal of the cube.

$l=\sqrt{3}b$

r is the distance between the cube’s centre and one of the eight vertices of the cube.

$r=\frac{\sqrt{3}b}{2}$

The electric potential (V) at the cube’s centre is due to the presence of eight charges present at the vertices of the cube is given by,

$V=\frac{8q}{4\pi {{\varepsilon }_{o}}r}$

$\Rightarrow V=\frac{8q}{4\pi {{\varepsilon }_{o}}\left( \frac{\sqrt{3}b}{2} \right)}$

$\Rightarrow V=\frac{4q}{\sqrt{3}\pi {{\varepsilon }_{o}}b}$

Clearly, the potential at the cube’s centre is $\frac{4q}{\sqrt{3}\pi {{\varepsilon }_{o}}b}$.

The electric field at the cube's centre is due to the eight charges, which get cancelled because the charges are divided symmetrically concerning the cube's centre. Therefore, the electric field is zero at the centre.

14. Two tiny spheres carrying charges \[1.5\mu C\] and \[2.5\mu C\] are located \[30\text{ }cm\] apart. Find the potential and electric field: 

a) at the mid-point of the line joining the two charges, and 

Ans: Two charges placed at points A and B are represented in the given figure. 

O is the mid-point of the line connecting the two charges.

(Image will Be Updated Soon)

Magnitude of charge located at A, \[{{q}_{1}}=1.5\mu C\] 

Magnitude of charge located at B, \[{{q}_{2}}=2.5\mu C\]

Distance between the two charges, \[d=30cm=0.3m\]

Let \[{{V}_{1}}\] and ${{E}_{1}}$ are the electric potential and electric field respectively at O point. 

\[{{V}_{1}}\] is the sum of potential due to charge at A and potential due to charge at B.

\[{{V}_{1}}=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}\left( \frac{d}{2} \right)}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}\left( \frac{d}{2} \right)}=\frac{1}{4\pi {{\varepsilon }_{o}}\left( \frac{d}{2} \right)}\left( {{q}_{1}}+{{q}_{2}} \right)\]

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

\[\Rightarrow {{V}_{1}}=\frac{9\times {{10}^{9}}\times {{10}^{-6}}}{\left( \frac{0.3}{2} \right)}\left( 1.5+2.5 \right)=2.4\times {{10}^{5}}V\]

And, ${{E}_{1}}$is the electric field due to \[{{q}_{2}}\] $-$ electric field due to \[{{q}_{1}}\]

${{E}_{1}}=\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}{{\left( \frac{d}{2} \right)}^{2}}}-\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}{{\left( \frac{d}{2} \right)}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{o}}{{\left( \frac{d}{2} \right)}^{2}}}\left( {{q}_{2}}-{{q}_{1}} \right)$

\[\Rightarrow {{E}_{1}}=\frac{9\times {{10}^{9}}\times {{10}^{-6}}}{{{\left( \frac{0.3}{2} \right)}^{2}}}\left( 2.50-1.5 \right)=4\times {{10}^{5}}V{{m}^{-1}}\]

Therefore, the potential at mid-point is \[2.4\times {{10}^{5}}V\] and the electric field at mid-point is \[4\times {{10}^{5}}V{{m}^{-1}}\]. The field is pointed from the greater charge to the smaller charge.

b) at a point \[10\text{ }cm\] from this midpoint in a plane normal to the line and passing through the mid-point.

Ans: Consider a point Z such that \[OZ=10\text{ }cm=0.1\text{ }m\], as shown in figure,

(Image will Be Updated Soon)

\[{{V}_{2}}\] and ${{E}_{2}}$ are the electric potential and electric field respectively at Z point.

$BZ=AZ=\sqrt{{{0.1}^{2}}+{{0.15}^{2}}}=0.18m$

\[{{V}_{2}}\] is the sum of potential due to charge at A and potential due to charge at B.

\[{{V}_{2}}=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}\left( AZ \right)}+\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}\left( BZ \right)}=\]

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

\[\Rightarrow {{V}_{2}}=\frac{9\times {{10}^{9}}\times {{10}^{-6}}}{0.18}\left( 1.5+2.5 \right)=2\times {{10}^{5}}V\]

Electric field due to \[{{q}_{1}}\] at Z,

${{E}_{A}}=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{o}}{{\left( AZ \right)}^{2}}}$

\[\Rightarrow {{E}_{A}}=\frac{9\times {{10}^{9}}\times 1.5\times {{10}^{-6}}}{{{(0.18)}^{2}}}=0.416\times {{10}^{6}}V{{m}^{-1}}\]along AZ

Electric field due to \[{{q}_{2}}\] at Z,

${{E}_{B}}=\frac{{{q}_{2}}}{4\pi {{\varepsilon }_{o}}{{\left( BZ \right)}^{2}}}$

\[\Rightarrow {{E}_{B}}=\frac{9\times {{10}^{9}}\times 2.5\times {{10}^{-6}}}{{{(0.18)}^{2}}}=0.69\times {{10}^{6}}V{{m}^{-1}}\] along BZ.

The resultant field intensity at Z,

$E=\sqrt{E_{A}^{2}+E_{B}^{2}+2{{E}_{A}}{{E}_{B}}\cos 2\theta }$

From figure,

$\cos \theta =\frac{0.10}{0.18}=0.5556$

$\Rightarrow \theta ={{\cos }^{-1}}(0.5556)=56.25{}^\circ $

$\Rightarrow 2\theta =112.5{}^\circ $

$\Rightarrow \cos 2\theta =-0.38$

Now, 

\[E=\sqrt{{{\left( 0.416\times {{10}^{6}} \right)}^{2}}+{{\left( 0.69\times {{10}^{6}} \right)}^{2}}+2\times 0.416\times {{10}^{6}}\times 0.416\times {{10}^{6}}\times (-0.38)}\]

\[E=6.6\times {{10}^{5}}V{{m}^{-1}}\]

Therefore, the potential at a point \[10\text{ }cm\] (perpendicular to the mid-point) is \[2\times {{10}^{5}}V\] and electric field is \[6.6\times {{10}^{5}}V{{m}^{-1}}\].

15. A spherical conducting shell of inner radius ${{r}_{1}}$ and outer radius ${{r}_{2}}$ has a charge Q. 

a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?

Ans: Provided that,

Charge located at the centre of the shell is \[+q\]. Hence, a charge of magnitude \[-q\] will be induced to the inner surface of the shell. Therefore, net charge on the shell’s inner surface is \[-q\]. 

Surface charge density at the shell’s inner surface is given by the relation,

${{\sigma }_{1}}=\frac{-q}{4\pi r_{1}^{2}}$

A charge of \[+q\] is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, the total charge on the outer surface of the shell is $Q+q$. Surface charge density at the shell’s outer surface is,

${{\sigma }_{2}}=\frac{-q}{4\pi r_{2}^{2}}$

b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Ans: Yes, the electric field intensity is zero inside a cavity, even if the shell is not spherical and has any random shape. Take a closed circle such that a part of it is inside the hole along a field line while the rest is within the conductor. The network performed by the field in taking a test charge over a closed circuit is zero because the field is zero inside the conductor. Hence, the electric field is zero, whatever the frame is.

16. 

a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by \[\left( {{{\vec{E}}}_{2}}-{{{\vec{E}}}_{1}} \right)\centerdot \hat{n}=\frac{\sigma }{{{\varepsilon }_{o}}}\]

Where \[\hat{n}\] is a unit vector normal to the surface at a point and \[\sigma \] is the surface charge density at that point (The direction of \[\hat{n}\] is from side 1 to side 2). Hence show that just outside a conductor, the electric field is \[\frac{\hat{n}\sigma }{{{\varepsilon }_{o}}}\].

Ans: Electric field on the charged body’s one side is ${{E}_{1}}$ and the electric field on the same body’s other side is ${{E}_{2}}$. 

If infinite plane body has a uniform thickness, then electric field due to one surface is given by,

\[{{\vec{E}}_{1}}=-\frac{\sigma }{2{{\varepsilon }_{o}}}\hat{n}\]……(1)

Where, \[\hat{n}\] is the unit vector normal to the surface at a point

\[\sigma \] is the surface charge density at that point 

Electric field due to the other surface of the charged body,

\[{{\vec{E}}_{2}}=-\frac{\sigma }{2{{\varepsilon }_{o}}}\hat{n}\] ……(2)

Electric field at any point due to the two surfaces

\[\left( {{{\vec{E}}}_{2}}-{{{\vec{E}}}_{1}} \right)=\frac{\sigma }{2{{\varepsilon }_{o}}}\hat{n}+\frac{\sigma }{2{{\varepsilon }_{o}}}\hat{n}=\frac{\sigma }{{{\varepsilon }_{o}}}\hat{n}\]

\[\Rightarrow \left( {{{\vec{E}}}_{2}}-{{{\vec{E}}}_{1}} \right)\centerdot \hat{n}=\frac{\sigma }{{{\varepsilon }_{o}}}\hat{n}\]

Inside a closed conductor, \[{{\vec{E}}_{1}}=0\]

\[{{\vec{E}}_{2}}={{\vec{E}}_{1}}=-\frac{\sigma }{2{{\varepsilon }_{o}}}\hat{n}\]

Clearly, the electric field just outside the conductor is\[\frac{\sigma }{{{\varepsilon }_{o}}}\hat{n}\].

b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

Ans: When a charged particle is transferred from one location to the other on a closed-loop, the work performed by the electrostatic field is zero. Hence, the tangential segment of the electrostatic field is continuous from the charged surface's one side.

17. A long charged cylinder of linear charged density \[\lambda \] is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Ans: Provided that,

There is a long-charged cylinder of length L and radius r and having charge density is \[\lambda \]. 

Another cylinder of equal length encloses the previous cylinder. The radius of this cylinder is R. 

Let E be the electric field generated in the space between the two cylinders. 

According to Gauss’s theorem, electric flux is given by,

$\phi =E(2\pi d)L$

d is the distance of a point from the cylinder’s common axis. 

Let q be the cylinder’s total charge.

So, $\phi =E(2\pi d)L=\frac{q}{{{\varepsilon }_{o}}}$

Where, q  is the Charge on the outer cylinder’s inner sphere.

${{\varepsilon }_{o}}$is the Permittivity of free space

$\Rightarrow E(2\pi d)L=\frac{\lambda L}{{{\varepsilon }_{o}}}$

$\Rightarrow E=\frac{\lambda }{2\pi d{{\varepsilon }_{o}}}$

Clearly, the electric field in the space between the two cylinders is $\frac{\lambda }{2\pi d{{\varepsilon }_{o}}}$.

18. In a hydrogen atom, the electron and proton are bound at a distance of about $d=0.53\overset{o}{\mathop A}\,$: 

a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. 

Ans: Provided that,

The distance separating electron-proton of a hydrogen atom, $d=0.53\overset{o}{\mathop A}\,$

Charge on an electron, ${{q}_{1}}=-1.6\times {{10}^{-19}}C$

Charge on a proton, ${{q}_{2}}=1.6\times {{10}^{-19}}C$

The value of potential is zero at infinity. 

Potential energy of the system is,

$U=0-\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{o}}d}$

$\Rightarrow U=0-\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{0.53\times {{10}^{-10}}}$

$\Rightarrow U=-43.7\times {{10}^{-19}}J$

$\Rightarrow V=-27.2eV$

Clearly, the potential energy of the system is $-27.2eV$.

b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? 

Ans: It is mentioned in the question that, 

Kinetic energy is half of the potential energy by magnitude. 

Kinetic energy \[=13.6eV\]

Total energy \[=13.6-27.2=-\text{ }13.6eV\]

Therefore, the minimum work needed to free the electron is \[13.6\text{ }eV.\]

c) What are the answers to (a) and (b) above if the zero of potential energy is taken at \[1.06\overset{o}{\mathop A}\,\] separation?

Ans: When potential energy is taken as zero, ${{d}_{1}}=1.06\overset{o}{\mathop A}\,$

Potential energy of the system is,

$U=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{o}}{{d}_{1}}}-27.2$

$\Rightarrow U=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{1.06\times {{10}^{-10}}}-27.2$

$\Rightarrow U=21.7.3\times {{10}^{-19}}-27.2$

$\Rightarrow U=-13.6eV$

Clearly, the potential energy of the system is $-13.6eV$.

19. If one of the two electrons of a ${{H}_{2}}$ molecule is removed, we get a hydrogen molecular ion ${{H}_{2}}^{+}$. In the ground state of an ${{H}_{2}}^{+}$, the two protons are separated by roughly \[1.5\overset{o}{\mathop A}\,\], and the electron is roughly \[1\overset{o}{\mathop A}\,\] from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Ans: The figure shows two protons and one electron.

(Image will Be Updated Soon)

Charge on proton 1, ${{q}_{1}}=1.6\times {{10}^{-19}}C$

Charge on proton 2, ${{q}_{2}}=1.6\times {{10}^{-19}}C$

Charge on electron, ${{q}_{3}}=-1.6\times {{10}^{-19}}C$

Distance between protons 1 and 2, ${{d}_{1}}=1.5\times {{10}^{-10}}m$

Distance between proton 1 and electron, ${{d}_{2}}=1\times {{10}^{-10}}m$

Distance between proton 2 and electron, ${{d}_{3}}=1\times {{10}^{-10}}m$

There is zero potential energy at infinity.

The formula for potential energy of the system is given by:

$U=\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{o}}{{d}_{1}}}+\frac{{{q}_{2}}{{q}_{3}}}{4\pi {{\varepsilon }_{o}}{{d}_{3}}}+\frac{{{q}_{1}}{{q}_{3}}}{4\pi {{\varepsilon }_{o}}{{d}_{2}}}$

\[\Rightarrow U=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{9}} \right)}^{2}}}{{{10}^{-10}}}\left[ -1+\frac{1}{1.5}-1 \right]\]

\[\Rightarrow U=-30.72J\]

\[\Rightarrow U=-19.2eV\]

Clearly, the potential energy of the system is \[-19.2eV\].

20. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Ans: Let a be the radius of a sphere A, \[{{Q}_{A}}\] be the charge on the sphere A, and \[{{C}_{A}}\] be the capacitance of the sphere A. 

Let b be the radius of sphere B, \[{{Q}_{B}}\] be the charge on the sphere B, and \[{{C}_{B}}\] be the capacitance of the sphere B. 

The two spheres are joined with a wire, their potential will be equal. 

Let ${{E}_{A}}$ and \[{{E}_{B}}\] are the electric field of sphere A and sphere B respectively. 

Now,

$\frac{{{E}_{A}}}{{{E}_{B}}}=\frac{{{Q}_{A}}}{4\pi {{\varepsilon }_{o}}{{a}^{2}}}\times \frac{4\pi {{\varepsilon }_{o}}{{b}^{2}}}{{{Q}_{B}}}$

$\Rightarrow \frac{{{E}_{A}}}{{{E}_{B}}}=\frac{{{Q}_{A}}}{{{a}^{2}}}\times \frac{{{b}^{2}}}{{{Q}_{B}}}$……(1)

And, $\frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{{{C}_{A}}V}{{{C}_{B}}V}$, $\frac{{{C}_{A}}}{{{C}_{B}}}=\frac{a}{b}$

$\Rightarrow \frac{{{Q}_{A}}}{{{Q}_{B}}}=\frac{a}{b}$……(2)

Now from (1) and (2),

$\Rightarrow \frac{{{E}_{A}}}{{{E}_{B}}}=\frac{a}{{{a}^{2}}}\times \frac{{{b}^{2}}}{b}=\frac{b}{a}$

Hence, at surface, the ratio of the electric fields is $\frac{b}{a}$.

A pointed and sharp end can be arranged as a sphere of a minimum radius, and a flat portion acts as a sphere of a much greater radius. Therefore, the charge density on pointed and sharp ends of a conductor is much higher than on its flatter portions.

21. Two charges \[-q\] and \[+q\] are located at points \[(0,0,-a)\] and \[(0,0,a)\], respectively. 

a) What is the electrostatic potential at the points \[\left( 0,0,z \right)\] and \[\left( x,y,0 \right)?\]

Ans: Charge \[-q\] is placed at \[(0,0,-a)\] and charge \[+q\] is placed at \[(0,0,a)\]. Hence, they make a dipole. Point \[\left( 0,0,z \right)\] is on the dipole axis and point \[\left( x,y,0 \right)\] is perpendicular to the axis of the dipole. So, electrostatic potential at point \[\left( x,y,0 \right)\] is zero.

Electrostatic potential at point \[\left( 0,0,z \right)\] is given by,

$V=\frac{1}{4\pi {{\varepsilon }_{o}}}\left[ \frac{q}{z-a}-\frac{q}{z+a} \right]$

$\Rightarrow V=\frac{2qa}{4\pi {{\varepsilon }_{o}}\left( {{z}^{2}}-{{a}^{2}} \right)}$

$\Rightarrow V=\frac{p}{4\pi {{\varepsilon }_{o}}\left( {{z}^{2}}-{{a}^{2}} \right)}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

p is the dipole moment of two charges systems.

b) Obtain the dependence of potential on the distance \[r\] of a point from the origin when \[{}^{r}/{}_{a}\text{ }>>1\]. 

Ans: Distance \[r\] is much larger than half of the distance separating the two charges. Hence, the potential (V) at a distance \[r\] is inversely proportional to the distance’s square.

$V\propto \frac{1}{{{r}^{2}}}$

c) How much work is done in moving a small test charge from the point \[\left( 5,\text{ }0,\text{ }0 \right)\] to \[(-7,0,0)\] along the x-axis? Does the answer change if the path the test charge between the same points is not along the x-axis?

Ans: The answer does not change if the path of the test is not along the x-axis. A test charge is moved from point \[\left( 5,0,0 \right)\] to point \[(-7,0,0)\] along the x-axis. 

Electrostatic potential \[({{V}_{1}})\] at point \[\left( 5,0,0 \right)\] is given by

${{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{o}}}\left[ \frac{-q}{\sqrt{{{(5-0)}^{2}}+{{(-a)}^{2}}}}+\frac{q}{\sqrt{{{(5-0)}^{2}}+{{(a)}^{2}}}} \right]$

$\Rightarrow {{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{o}}}\left[ \frac{-q}{\sqrt{25+{{a}^{2}}}}+\frac{q}{\sqrt{25+{{a}^{2}}}} \right]=0$

Electrostatic potential \[({{V}_{2}})\] at point \[\left( -7,0,0 \right)\] is given by

${{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{o}}}\left[ \frac{-q}{\sqrt{{{(-7)}^{2}}+{{(-a)}^{2}}}}+\frac{q}{\sqrt{{{(-7)}^{2}}+{{(a)}^{2}}}} \right]$

$\Rightarrow {{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{o}}}\left[ \frac{-q}{\sqrt{49+{{a}^{2}}}}+\frac{q}{\sqrt{49+{{a}^{2}}}} \right]=0$

Hence, zero work is done in taking a small test charge from point \[\left( 5,0,0 \right)\] to point \[\left( -7,0,0 \right)\] along the x-axis because work performed by the electrostatic field in moving a test charge between the two locations is path independent connecting the two points.

22. Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on $r$ for \[{}^{r}/{}_{a}\text{ }>>1\] and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

(Image will Be Updated Soon)

Ans: Four charges of equal magnitude are located at points \[X,\text{ }Y,\text{ Y},\text{ }and\text{ }Z\] respectively, as shown in the below figure,

(Image will Be Updated Soon)

A point P is r distance far away from point Y. 

The system of charges makes an electric quadrupole. It can be taken that the electric quadrupole has three charges, that is, charge \[+q\] located at point X, charge \[-2q\] located at point Y and charge \[+q\] located at point Z.

Now,

\[XY=YZ=a\]

\[YP=r\]

\[PX\text{ =r+a}\]

\[PZ\text{ =r-a}\]

The Electrostatic potential created by the three charges system at point P is given by,

$V=\frac{1}{4\pi {{\varepsilon }_{o}}}\left[ \frac{q}{XP}-\frac{2q}{YP}+\frac{q}{ZP} \right]$

$\Rightarrow V=\frac{1}{4\pi {{\varepsilon }_{o}}}\left[ \frac{q}{r+a}-\frac{2q}{r}+\frac{q}{r-a} \right]$

$\Rightarrow V=\frac{q}{4\pi {{\varepsilon }_{o}}}\left[ \frac{2{{a}^{2}}}{r\left( {{r}^{2}}-{{a}^{2}} \right)} \right]$

$\Rightarrow V=\frac{q}{4\pi {{\varepsilon }_{o}}{{r}^{3}}}\left[ \frac{2{{a}^{2}}}{\left( 1-\frac{{{a}^{2}}}{{{r}^{2}}} \right)} \right]$

Since, \[{}^{r}/{}_{a}\text{ }>>1\]

$\Rightarrow {}^{a}/{}_{r}\text{ }1$

We can take $\frac{{{a}^{^{2}}}}{{{r}^{2}}}$ as negligible.

$\Rightarrow V=\frac{q2{{a}^{2}}}{4\pi {{\varepsilon }_{o}}{{r}^{3}}}$

It can be said that $V\propto \frac{1}{{{r}^{3}}}$.

We know for dipole’s potential, $V\propto \frac{1}{{{r}^{2}}}$ and for a monopole, $V\propto \frac{1}{r}$.

23. An electrical technician requires a capacitance of \[2\mu F\] in a circuit across a potential difference of \[1\text{ }kV.\] A large number of \[1\mu F\] capacitors are available to him each of which can withstand a potential difference of not more than \[400\text{ }V\]. Suggest a possible arrangement that requires the minimum number of capacitors.

Ans: It is provided that,

Total required capacitance, \[C=2\mu F\]

Potential difference, \[V=1kV=1000V\]

Capacitance of each capacitor, \[{{C}_{1}}=1\mu F\]

Each capacitor can withstand a potential difference, \[{{V}_{1}}=400V\]

Suppose a number of capacitors are joined in series and these series circuits are joined in parallel to each other. The potential difference in each row must be \[1000\text{ }V\] and potential difference across each capacitor must be \[400\text{ }V.\] Clearly, the capacitors’ numbers in each row is given by,

$\frac{1000}{400}=2.5$

Hence, there are three capacitors across each row. Capacitance of each row,

$\frac{1}{1+1+1}=3\mu F$

There are n rows and each row has three capacitors, which are joined in parallel. Clearly, circuit’s equivalent capacitance is given by,

$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}......\text{n terms=}\frac{n}{3}$

However, circuit’s capacitance is given as \[2\mu F.\]

$\Rightarrow \frac{n}{3}=2$

$\Rightarrow n=6$

Clearly, there are 6 rows and each row has three capacitors. A minimum of \[6\times 3=18\] capacitors are needed for the given arrangement.

24. What is the area of the plates of a \[2\text{ }F\] parallel plate capacitor, given that the separation between the plates is \[0.5\text{ }cm\]?

Ans: It is provided that,

Capacitance of a parallel capacitor, \[C=2\text{ }F\]

Distance separating the two plates, $d=0.5\text{ }cm=0.5\times {{10}^{-2}}m$

The formula for parallel plate capacitor’s capacitance is given by,

$C=\frac{{{\varepsilon }_{o}}A}{d}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

We get,

$A=\frac{Cd}{{{\varepsilon }_{o}}}$

$A=\frac{2\times 0.5\times {{10}^{-2}}}{8.854\times {{10}^{-12}}}=1130K{{m}^{2}}$

25. Obtain the equivalent capacitance of the network in Figure. For a \[\mathbf{300}\text{ }\mathbf{V}\] supply, determine the charge and voltage across each capacitor.

(Image will Be Updated Soon)

Ans: It is provided that,

Capacitance of capacitor ${{C}_{1}}$ is \[100\text{ }pF.\]

Capacitance of capacitor ${{C}_{2}}$  is \[200\text{ }pF.\]

Capacitance of capacitor ${{C}_{3}}$ is \[200\text{ }pF.\]

Capacitance of capacitor ${{C}_{4}}$ is \[100\text{ }pF.\]

Supply potential, \[V=300\text{ }V\]

Capacitors ${{C}_{2}}$ and ${{C}_{3}}$ are connected in series, therefore, their equivalent capacitance be \[C'\].

$\frac{1}{C'}=\frac{1}{200}+\frac{1}{200}$

$\Rightarrow \frac{1}{C'}=\frac{2}{200}$

$\Rightarrow C'=100pF$

Capacitors ${{C}_{1}}$ and \[C'\] are in parallel, therefore their equivalent capacitance be \[C''.\]

$C''={{C}_{1}}+C'$

$\Rightarrow C''=100+100=200pF$

\[C''\] and ${{C}_{4}}$ are connected in series, therefore, their equivalent capacitance be \[C\].

$\frac{1}{C'}=\frac{1}{200}+\frac{1}{100}$

$\Rightarrow \frac{1}{C'}=\frac{3}{200}$

$\Rightarrow C'=\frac{200}{3}pF$

Clearly, the equivalent capacitance of the circuit is $\frac{200}{3}pF$.

Potential difference across $C''=V''$

Potential difference across ${{C}_{4}}={{V}_{4}}$

$\Rightarrow {{V}_{4}}+V''=V=300V$

Charge on ${{C}_{4}}$ is given by,

${{Q}_{4}}=VC$

$\Rightarrow {{Q}_{4}}=300\times \frac{200}{3}\times {{10}^{-12}}$

\[\Rightarrow {{Q}_{4}}=2\times {{10}^{-8}}C\]

$\Rightarrow {{V}_{4}}=\frac{{{Q}_{4}}}{{{C}_{4}}}$

$\Rightarrow {{V}_{4}}=\frac{2\times {{10}^{-8}}}{100\times {{10}^{-12}}}=200V$

Voltage across ${{C}_{1}}$is given by,

\[{{V}_{1}}=V-{{V}_{4}}\]

\[\Rightarrow {{V}_{1}}=200-100=100V\]

Hence, the potential difference, ${{V}_{1}}$, across ${{C}_{1}}$ is \[100\text{ }V\].

Charge on ${{C}_{1}}$ is given by,

${{Q}_{1}}={{V}_{1}}{{C}_{1}}$

$\Rightarrow {{Q}_{1}}=100\times 100\times {{10}^{-12}}$

\[\Rightarrow {{Q}_{1}}={{10}^{-8}}C\]

${{C}_{2}}$ and ${{C}_{3}}$ having the same capacitances have a \[100\text{ }V\] potential difference together. Since ${{C}_{2}}$ and ${{C}_{3}}$ are in series, the potential difference across ${{C}_{2}}$ and ${{C}_{3}}$ is given by,

${{V}_{2}}={{V}_{3}}=50V$

Charge on ${{C}_{2}}$ is given by,

${{Q}_{2}}={{V}_{2}}{{C}_{2}}$

$\Rightarrow {{Q}_{2}}=50\times 200\times {{10}^{-12}}$

\[\Rightarrow {{Q}_{2}}={{10}^{-8}}C\]

Charge on ${{C}_{3}}$ is given by,

${{Q}_{3}}={{V}_{3}}{{C}_{3}}$

$\Rightarrow {{Q}_{3}}=50\times 200\times {{10}^{-12}}$

\[\Rightarrow {{Q}_{3}}={{10}^{-8}}C\]

Clearly, the equivalent capacitance of the given circuit is $\frac{200}{3}pF$and,

\[{{Q}_{1}}={{10}^{-8}}C\], \[{{Q}_{2}}={{10}^{-8}}C\], \[{{Q}_{3}}={{10}^{-8}}C\], \[{{Q}_{4}}=2\times {{10}^{-8}}C\],

\[{{V}_{1}}=100V\], ${{V}_{2}}=50V$,  ${{V}_{3}}=50V$,  ${{V}_{4}}=200V$.

26. The plates of a parallel plate capacitor have an area of $90c{{m}^{2}}$ each and are separated by \[2.5\text{ }mm\]. The capacitor is charged by connecting it to a \[400\text{ }V\] supply. 

a) How much electrostatic energy is stored by the capacitor? 

Ans: It is provided that,

Area of the parallel capacitor’s plates, $A=90c{{m}^{2}}=90\times {{10}^{-4}}{{m}^{2}}$

Distance separating the plates, $d=2.5mm=2.5\times {{10}^{-3}}m$

Potential difference across the pates, \[V=400V\]

The formula for capacitance will be,

$C=\frac{{{\varepsilon }_{o}}A}{d}$

Electrostatic energy stored in capacitor is given by,

${{E}_{1}}=\frac{1}{2}C{{V}^{2}}$

$\Rightarrow {{E}_{1}}=\frac{1}{2}\frac{{{\varepsilon }_{o}}A}{d}{{V}^{2}}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

$\Rightarrow {{E}_{1}}=\frac{1}{2}\frac{8.854\times {{10}^{-12}}\times 90\times {{10}^{-4}}}{2.5\times {{10}^{-3}}}{{400}^{2}}$

$\Rightarrow {{E}_{1}}=2.55\times {{10}^{6}}J$

The stored electrostatic energy inside the capacitor is, $2.55\times {{10}^{6}}J$.

b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Ans: Volume of the given capacitor, $V'=A\times d$

$\Rightarrow V'=90\times {{10}^{-4}}\times 2.5\times {{10}^{-3}}=2.25\times {{10}^{-5}}{{m}^{3}}$

Energy density in the capacitor is given by,

$u=\frac{{{E}_{1}}}{V'}$

$u=\frac{2.55\times {{10}^{-6}}}{2.25\times {{10}^{-5}}}=0.113J{{m}^{-3}}$

Also, $u=\frac{{{E}_{1}}}{V'}$

$\Rightarrow u=\frac{\frac{1}{2}\frac{{{\varepsilon }_{o}}A}{d}{{V}^{2}}}{Ad}=\frac{1}{2}{{\varepsilon }_{o}}{{\left( \frac{V}{d} \right)}^{2}}$

Where, $\frac{V}{d}=E$, E is electric field.

$\Rightarrow u=\frac{1}{2}{{\varepsilon }_{o}}{{E}^{2}}$

Hence, derived.

27. A \[4\mu F\] capacitor is charged by a \[200\text{ }V\] supply. It is then disconnected from the supply, and is connected to another uncharged \[2\mu F\] capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Ans: It is provided that,

A charged capacitor has capacitance, \[{{C}_{1}}=4\mu F=4\times {{10}^{-6}}F\]

Supply voltage, \[{{V}_{1}}=200V\]

Electrostatic energy stored in ${{C}_{1}}$capacitor is given by,

${{E}_{1}}=\frac{1}{2}{{C}_{1}}V_{1}^{2}$

$\Rightarrow {{E}_{1}}=\frac{1}{2}\times 4\times {{10}^{-6}}\times {{200}^{2}}$

$\Rightarrow {{E}_{1}}=8\times {{10}^{-2}}J$

An uncharged capacitor’s capacitance, \[{{C}_{2}}=2\mu F=2\times {{10}^{-6}}F\]

When ${{C}_{2}}$ is joined to the circuit, the potential attained by it is ${{V}_{2}}$. 

According to the conservation of charge,

${{V}_{2}}\left( {{C}_{1}}+{{C}_{2}} \right)={{V}_{1}}{{C}_{1}}$

$\Rightarrow {{V}_{2}}\left( 4+2 \right)\times {{10}^{-6}}=200\times 4\times {{10}^{-6}}$

$\Rightarrow {{V}_{2}}=\frac{400}{3}V$

The formula for electrostatic energy for the two capacitors combination is given by,

${{E}_{2}}=\frac{1}{2}{{V}_{2}}^{2}\left( {{C}_{1}}+{{C}_{2}} \right)$

$\Rightarrow {{E}_{2}}=\frac{1}{2}{{\left( \frac{400}{3} \right)}^{2}}\left( 4+2 \right)\times {{10}^{-6}}$

$\Rightarrow {{E}_{2}}=5.33\times {{10}^{-2}}J$

The amount of lost electrostatic energy by capacitor is,

$={{E}_{1}}-{{E}_{2}}=0.08-0.0533=0.0267J$

Therefore, the lost electrostatic energy is $0.0267J$.

28. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to $\frac{1}{2}QE$ where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor $\frac{1}{2}$.

Ans: Let F be the applied force to separate the parallel plates of a capacitor by a distance of \[\Delta x\]. Hence, work done by the force \[=F\Delta x\]

As a result, the capacitor’s potential energy rises by an amount given as \[uA\Delta x\]

Where, u is the energy density,

 A is the area of each plate,

d is the Distance separating the plates,

V is the difference in potential across the plates. 

The work done will be equal to the rise in the potential energy i.e.,

\[F\Delta x=uA\Delta x\]

\[\Rightarrow F=uA=\frac{1}{2}{{\varepsilon }_{o}}{{E}^{2}}A\]

The formula for electric intensity is given by,

$E=\frac{V}{d}$

 \[\Rightarrow F=\frac{1}{2}{{\varepsilon }_{o}}E\left( \frac{V}{d} \right)A\]

Hence, capacitance will be,

$C=\frac{{{\varepsilon }_{o}}A}{d}$

\[\Rightarrow F=\frac{1}{2}CVE\]

The formula for charge in the capacitor is,

$Q=CV$

\[\Rightarrow F=\frac{1}{2}QE\]

The actual origin of the force formula's half factor is that just outside the conductor, the field is E, and it is zero inside it. Henceforth, it is the average amount of the field that contributes to the force.

29. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Figure).

(Image will Be Updated Soon)

Show that the capacitance of a spherical capacitor is given by

$C=\frac{4\pi {{\varepsilon }_{o}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}$

Where ${{r}_{1}}$ and ${{r}_{2}}$ are the radii of outer inner spheres, respectively.

Ans: It is provided that,

Outer shell’s radius $={{r}_{1}}$

Inner shell’s radius $={{r}_{2}}$

The outer shell’s inner surface has charge $+Q$.

The inner shell’s outer surface has induced charge $-Q$.

The difference in potential between the two shells is given by,

$V=\frac{Q}{4\pi {{\varepsilon }_{o}}{{r}_{2}}}-\frac{Q}{4\pi {{\varepsilon }_{o}}{{r}_{1}}}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

$\Rightarrow V=\frac{Q}{4\pi {{\varepsilon }_{o}}}\left[ \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right]$

\[\Rightarrow V=\frac{Q\left( {{r}_{1}}-{{r}_{2}} \right)}{4\pi {{\varepsilon }_{o}}{{r}_{1}}{{r}_{2}}}\]

The formula for capacitance is given by,

$C=\frac{Q}{V}$

$\Rightarrow C=\frac{4\pi {{\varepsilon }_{o}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}$

This proved.

30. A spherical capacitor has an inner sphere of radius \[\mathbf{12}\text{ }\mathbf{cm}\] and an outer sphere of radius \[\mathbf{13}\text{ }\mathbf{cm}\]. The outer sphere is earthed and the inner sphere is given a charge of \[\mathbf{2}.\mathbf{5\mu C}\]. The space between the concentric spheres is filled with a liquid of dielectric constant \[\mathbf{32}\]. 

a) Determine the capacitance of the capacitor. 

Ans: It is provided that,

Outer cylinder’s radius, \[{{r}_{1}}=13cm=0.13m\]

Radius of inner cylinder, \[{{r}_{2}}=12cm=0.12m\]

Charge on the inner cylinder, $q=2.5\mu C=2.5\times {{10}^{-6}}C$

The formula for capacitor’s capacitance is given by,

$C=\frac{4\pi {{\varepsilon }_{o}}{{\varepsilon }_{r}}{{r}_{1}}{{r}_{2}}}{{{r}_{1}}-{{r}_{2}}}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

Value of \[\frac{1}{4\pi {{\varepsilon }_{o}}}=9\times {{10}^{9}}N{{C}^{-2}}{{m}^{-2}}\]

$\Rightarrow C=\frac{32\times 0.12\times 0.13}{9\times {{10}^{9}}(0.13-0.12)}$

$\Rightarrow C\approx 5.5\times {{10}^{-9}}F$

Hence, the capacitor’s capacitance is approximately $5.5\times {{10}^{-9}}F$.

b) What is the potential of the inner sphere? 

Ans: The inner sphere’s potential is given by,

$V=\frac{q}{C}$

$\Rightarrow V=\frac{2.5\times {{10}^{-6}}}{5.5\times {{10}^{-9}}}=4.5\times {{10}^{2}}V$

Hence, the inner sphere’s potential is $4.5\times {{10}^{2}}V$.

c) Compare the capacitance of this capacitor with that of an isolated sphere of radius \[\mathbf{12}\text{ }\mathbf{cm}\]. Explain why the latter is much smaller.

Ans: Isolated sphere’s radius, $r=12\times {{10}^{-2}}m$

The formula for sphere’s capacitance is given by,

$C'=4\pi {{\varepsilon }_{o}}r$

$\Rightarrow C'=4\pi \times 8.854\times {{10}^{-12}}\times 12\times {{10}^{-2}}$ 

$\Rightarrow C'=1.33\times {{10}^{-11}}F$

The isolated sphere's capacitance is less in contrast to the concentric spheres because the concentric spheres' outer sphere is earthed. Hence, the difference in potential is minor, and the capacitance is higher than the isolated sphere.

31. Answer carefully: 

a) Two large conducting spheres carrying charges ${{Q}_{1}}$ and ${{Q}_{2}}$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $\frac{{{Q}_{1}}{{Q}_{2}}}{4\pi {{\varepsilon }_{o}}{{r}^{2}}}$, where r is the distance between their centres? 

Ans: The expression does not precisely give the force between two conducting spheres, $\frac{{{Q}_{1}}{{Q}_{2}}}{4\pi {{\varepsilon }_{o}}{{r}^{2}}}$ Because there is an irregular charge distribution on the spheres, hereafter, Coulomb’s law is not valid.

b) If Coulomb’s law involved $\frac{1}{{{r}^{3}}}$ dependence (instead of $\frac{1}{{{r}^{2}}}$), would Gauss’s law be still true?

Ans: Gauss’s law would not be valid if Coulomb’s law included dependency on $\frac{1}{{{r}^{3}}}$ instead of $\frac{1}{{{r}^{2}}}$.

c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point? 

Ans: Yes, suppose a small test charge is discharged at rest at an electrostatic field configuration location. In that case, it will move along the field lines crossing through the point, mainly if the field lines are straight because the field lines provide the acceleration's direction and not of velocity.

d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical? 

Ans: Whenever the electron transits in a circular orbit, the work performed by the force of the nucleus is zero, making an orbit either circular or elliptical; the work done by the field of a nucleus is zero.

e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there? 

Ans: No, the electric field is not continuous across the covering of a charged conductor. However, an electric potential is continuous.

f) What meaning would you give to the capacitance of a single conductor? 

Ans: A single conductor's capacitance is considered a parallel plate capacitor with one of the two plates at infinity.

f) Guess a possible reason why water has a much greater dielectric constant \[\left( =80 \right)\] than say, mica \[\left( =6 \right)\].

Ans: Water has an irregular space as contrasted to mica. Since it has a persistent dipole moment, it has a higher dielectric constant than mica.

32. A cylindrical capacitor has two co-axial cylinders of length \[15\text{ }cm\] and radii \[1.5\text{ }cm\] and \[1.4\text{ }cm\]. The outer cylinder is earthed and the inner cylinder is given a charge of \[3.5\mu C\]. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Ans: It is provided that,

Co-axial cylinder’s length, \[l=15cm=0.15m\]

Outer cylinder’s radius, \[{{r}_{1}}=1.5cm=0.015m\]

Radius of inner cylinder, \[{{r}_{2}}=1.4cm=0.014m\]

Charge on the inner cylinder, $q=3.5\mu C=3.5\times {{10}^{-6}}C$

The formula for co-axial cylinder’s capacitance of radii ${{r}_{1}}$ and ${{r}_{2}}$ is given by,

$C=\frac{2\pi {{\varepsilon }_{o}}l}{{{\log }_{e}}\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}$

Where, ${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

$\Rightarrow C=\frac{2\pi \times 8.854\times {{10}^{-12}}\times 0.15}{2.303{{\log }_{10}}\left( \frac{0.015}{0.014} \right)}$

$\Rightarrow C=1.2\times {{10}^{-10}}F$

The difference in potential of the inner cylinder is given by,

$V=\frac{q}{C}$

$\Rightarrow V=\frac{3.5\times {{10}^{-6}}}{1.2\times {{10}^{-10}}}=2.92\times {{10}^{4}}V$

The difference in potential will be $2.92\times {{10}^{4}}V$.

33. A parallel plate capacitor is to be designed with a voltage rating \[1\text{ }kV\], using a material of dielectric constant \[3\] and dielectric strength about \[{{10}^{7}}\text{ }V{{m}^{-1}}\]. (Dielectric strength is the maximum electric filed a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionization.) For safety, we should like the field never to exceed, say \[10%\] of the dielectric strength. What minimum area of the plates is required to have a capacitance of \[50\text{ }pF\]?

Ans: It is provided that,

Parallel plate capacitor’s potential, \[V=1kV=1000V\]

Material’s dielectric constant, ${{\varepsilon }_{r}}=3$

Dielectric strength \[={{10}^{7}}\text{ }V{{m}^{-1}}\]

For safety, the electric field intensity never exceeds 10% of the dielectric strength. 

Electric field intensity, $E=0.1\times {{10}^{7}}={{10}^{6}}V{{m}^{-1}}$

Parallel plate capacitor’s capacitance, \[C=50\text{ }pF=50\times {{10}^{-12}}F\]

The formula for distance separating the plates is given by,

$d=\frac{V}{E}$

$\Rightarrow d=\frac{1000}{{{10}^{6}}}=10{}^{-3}m$

The formula for capacitance is given by,

$C=\frac{{{\varepsilon }_{o}}{{\varepsilon }_{r}}A}{d}$

A is the area of each plate,

${{\varepsilon }_{o}}$is the Permittivity of free space

${{\varepsilon }_{o}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$

$A=\frac{Cd}{{{\varepsilon }_{o}}{{\varepsilon }_{r}}}$

$A=\frac{50\times {{10}^{-12}}\times {{10}^{-3}}}{8.85\times {{10}^{-12}}\times 3}\approx 19c{{m}^{2}}$

Clearly, the area of each plate is about $19c{{m}^{2}}$.

34. Describe schematically the equipotential surface corresponding to 

a) A constant electric field in the z-direction, 

Ans: Equipotential surfaces are the equidistant planes parallel to the x-y plane for the constant electric field in the z-direction.

b) A field that uniformly increases in magnitude but remains in a constant (say, z) direction, 

Ans: Equipotential surfaces are the parallel planes to the x-y plane, except that the field increases when the planes get closer.

c) A single positive charge at the origin, and 

Ans: Equipotential surfaces are the concentric spheres centered at the origin.

d) A uniform grid consisting of long equally spaced parallel charged wires in a plane.

Ans: A periodically changing shape near the provided grid is the equipotential surface. This shape slowly reaches the shape of planes that are parallel to the grid at a greater distance.

35. In a Van de Graaff type generator a spherical metal shell is to be a \[15\times {{10}^{6}}V\] electrode. The dielectric strength of the gas surrounding the electrode is \[5\times {{10}^{7}}\text{ }V{{m}^{-1}}\]. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential).

Ans: It is provided that,

Potential difference, \[V=15\times {{10}^{6}}V\]

surroundings gas’s dielectric strength$=5\times {{10}^{7}}\text{ }V{{m}^{-1}}$

Electric field intensity is equal to the dielectric strength,  $E=5\times {{10}^{7}}\text{ }V{{m}^{-1}}$

The formula for spherical shell’s minimum radius required for the purpose is given by,

$r=\frac{V}{E}$

$\Rightarrow r=\frac{15\times {{10}^{6}}}{5\times {{10}^{7}}}=0.3m=30cm$

Clearly, the required minimum radius of the spherical shell is $30cm$.

36. A small sphere of radius ${{r}_{1}}$ and charge ${{q}_{1}}$ is enclosed by a spherical shell of radius \[{{r}_{2}}\] and charge \[{{q}_{2}}\]. Show that if \[{{q}_{1}}\] is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge \[{{q}_{2}}\] on the shell is.

Ans: According to Gauss's law, the electric field between a sphere and a shell is concluded by the charge \[{{q}_{1}}\] on a small sphere. Hence, the potential difference between the sphere and the shell does not depend on the charge \[{{q}_{2}}\]. For positive charge \[{{q}_{1}}\], the potential difference V is always positive.

37. Answer the following: 

a) The top of the atmosphere is at about \[400\text{ }kV\] with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about \[100V{{m}^{-1}}\] . Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

Ans: When we step out of our house, we do not get an electric shock because the primary equipotential surfaces of open-air change, putting our body and the ground at the same potential.

b) A man fixes outside his house one evening a two-metre-high insulating slab carrying on its top a large aluminium sheet of area \[1{{m}^{2}}\]. Will he get an electric shock if he touches the metal sheet in the morning?

Ans: Yes, the man gets an electric shock if he touches the metal slab the following day. The constant discharging current in the atmosphere charges up the aluminium sheet. As an outcome, its voltage increases gradually. The increment in the voltage depends on the capacitance produced by the aluminium slab and the ground.

c) The discharging current in the atmosphere due to the small conductivity of air is known to be \[\mathbf{1800}\text{ }\mathbf{A}\] on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?

Ans: The existence of thunderstorms and lightning energizes the atmosphere continuously. Therefore, indeed with a discharging current of \[1800\text{ }A\], the atmosphere is not discharged entirely. The two reversing currents are in equilibrium, and the atmosphere persists in neutral.

d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during lightning?

Ans: Light energy, sound energy, and heat energy are wasted in the atmosphere during lightning and thunderstorms.

NCERT Solutions for Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

NCERT Solutions for Class 12 Physics - Free PDF Download

Students can gain huge benefits with Vedantu's NCERT solution for Class 12 Physics Chapter 2. It will give the students a proper theoretical explanation of this chapter, enabling the students to solve the questions by themselves. The critical questions list; questions which are often asked in the exams; is enlisted, described and explained comprehensively by the experts.

This chapter's important topics are more or less, the introduction to electrostatic potential and capacitance, Dielectrics and polarisation, the potential energy of a system charge, the relationship between field and potential, etc. It will enable the students to understand the chapter entirely and allow them to solve question answers independently. Students can understand & learn all the questions easily by understanding all the basic concepts of electrostatic potential and capacitance Class 12 from Vedantu's NCERT solution.

NCERT Solutions for Class 12 Physics Chapter 2 free pdf is available below and can be accessed offline.. Download the NCERT electrostatics Class 12 Ncert pdf link given below on this page.

Topics Covered in Chapter 2 Electrostatic Potential and Capacitance

Subtopics of Class 12 NCERT Physics Chapter 2 – Electrostatic Potential and Capacitance

Benefits of NCERT Solutions for Class 12 Physics

Our experienced and well-learned professionals have prepared the solutions in a structured manner which is easy to grasp and retain. Here are some of the benefits that you will gain from NCERT solutions of Class 12 Physics - 

• All the solutions provided in the  PDF are completely correct & of very high quality. 

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• All the answers in the PDF are written while keeping CBSE guidelines in mind. 

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Important Formulas Covered in the NCERT Solutions for Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

Students who are preparing for the Physics 12th board exams can go through the below given important formulas for brushing up their concepts. 

1. Electric Potential Energy ΔU = W  

Here, ΔU represents changes in potential energy and W represents the work done by electric lines of force. 

2. Electric dipole P = qd

Here, q is the test charge and d represents the vector joining two charge points.

3. Capacitance of the capacitor C = \[\frac{q}{v}\] 

The unit of capacitance is farad.

4. Energy stored in the capacitor E = \[\frac{QV}{2}\] or \[\frac{CV^{2}}{2}\] 

Students can download the free PDF of formulas used in the NCERT Class 12 Physics Chapter 2 from Vedantu for last-minute revision.

• List of Important Formulas of Electrostatic Potential and Capacitance 

NCERT Class 12 Physics Solutions

• Chapter 1 - Electric Charges and Fields

• Chapter 2 - Electrostatic Potential and Capacitance

• Chapter 3 - Current Electricity

• Chapter 4 - Moving Charges and Magnetism

• Chapter 5 - Magnetism And Matter

• Chapter 6 - Electromagnetic Induction

• Chapter 7 - Alternating Current

• Chapter 8 - Electromagnetic Waves

• Chapter 9 - Ray Optics and Optical Instruments

• Chapter 10 - Wave Optics

• Chapter 11 - Dual Nature of Radiation and Matter

• Chapter 12 - Atoms

• Chapter 13 - Nuclei

• Chapter 14 - Semiconductor Electronic: Material, Devices And Simple Circuits

• Chapter 15 - Communication Systems