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NCERT Solutions for Class 12 Physics Chapter 1 - Electric Charges And Fields

Last updated date: 19th Jun 2024
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NCERT Class 12 Physics Chapter 1: Electric Charges And Fields

The very first chapter of Class 12 NCERT Physics introduces students to the concept of electric charges and how it exerts a force on other materials by attracting or repelling them. The NCERT Solutions for Class 12 Chapter 1 are available on Vedantu for students preparing for the CBSE 12th board examinations along with other competitive exams. Skilled teachers with years of experience have prepared these solutions to facilitate a strong understanding of the topics and the best techniques for fetching more marks. Students can download the NCERT Class 12 Physics Chapter 1 PDF for free from Vedantu.

 Class: NCERT Solutions for Class 12 Subject: Class 12 Physics Chapter Name: Chapter 1 - Electric Charges And Fields Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes

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List of Important Topics Covered under NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges And Fields

Below are The Important Topics Discussed in The Chapter 1 of Class 12 Physics

 Electric Charge Conductors and Insulators Charging by Induction (Deleted from the current academic year) Basic Properties of Electric Charge Coulomb’s Law Forces between Multiple Charges Electric Field Electric Dipole Electric Flux Dipole in a Uniform External Field Continuous Charge Distribution Gauss’s Law Application of Gauss Law

Deleted Syllabus of Class 12 Electromagnetic Waves (2024-25)

 Nature of Electromagnetic Waves Deleted only about ether and page 277 Examples 8.1; 8.3.2; 8.4; 8.5 Exercises 8.11–8.15

List of Important Formulas in Class 12 Physics Chapter 1 Electric Charges And Fields

While preparing a chapter, it is important for students to memorize the formula of a particular topic. This will help them to boost their score. Some important formulas are listed below.

• Coulomb’s Law

The electric force between the two-point charges are directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

F = K$\frac{q_{1}q_{2}}{r2}$

Here,

• F represents the electric force.

• K represents Coulomb’s constant.

• $q_{1}q_{2}$ represents the two charges.

$\lambda$ represents the distance between two charges.

• Electric field intensity

• Electric field intensity is the vector quantity.

• E $\frac{F}{q_{1}}$

Here,

• F is the force experienced by the test charge.

• $q_1$ is the test charge.
• Electric Flux

dФ = E.da

dФ = E.da cos𝛉

Ф = ഽ E.da

Go through the below link to recap all the important formulas in the Class 12 Physics Chapter 1 Electric Charges And Fields.

Would you like to view a summarized version of this chapter? Check out the 'Chapter at a glance' section below the PDF of NCERT Solutions.

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Chapter at a Glance for Electric Charges & Fields NCERT Solutions

• From simple experiments on frictional electricity, one can infer that there are two types of charges in nature; and that like charges repel and unlike charges attract. By convention, the charge on a glass rod rubbed with silk is positive; that on a plastic rod rubbed with fur is then negative.

• Charge is not only a scalar (or invariant) under rotation; it is also invariant for frames of reference in relative motion. This is not always true for every scalar. For Example-, kinetic energy is a scalar under rotation, but is not invariant for frames of reference in relative motion.

• Conductors allow movement of electric charge through them, insulators do not. In metals, the mobile charges are electrons; in electrolytes both positive and negative ions are mobile.

• Coulomb’s Law: The mutual electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 and inversely proportional to the square of the distance r21 separating them.

$F_{21}=\text{force}\;q_{2} \;\text{du to}\;q_{1}\;=\frac{k(q_{1}q_{2})}{r^2_{21}}\;r_{21}$

Where $r_{21}$ is a unit vector in the direction from q1 to q2 and $k=\frac{1}{4\pi \varepsilon _{0}}$ is the constant of proportionality.

In SI units, the unit of charge is coulomb. The experimental value of the constant $\varepsilon _{0}$ is

$\varepsilon _{0}=8.854\times 10^{-12}C^{2}N^{-}M^{-2}$

The approximate value of k is- 9\times $10^{9}Nm^2C^{-2}$

• Superposition Principle: The principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of a third (or more) additional charge(s). For an assembly of charges Q1, Q2, Q3..., the force on any charge, say Q1, is the vector sum of the force on  Q1 due to Q2, the force on Q1 due to Q3, and so on. For each pair, the force is given by Coulomb's law for two charges stated earlier.

• An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of the electric field at that point. The relative closeness of field lines indicates the relative strength of electric field at different points; they crowd near each other in regions of strong electric field and are far apart where the electric field is weak. In regions of constant electric field, the field lines are uniformly spaced parallel straight lines.

• Some of the important properties of field lines are: (i) Field lines are continuous curves without any breaks. (ii) Two field lines cannot cross each other. (iii) Electrostatic field lines start at positive charges and end at negative charges (iv) They cannot form closed loops.

• Electric flux is the measure of the field lines crossing a surface. It is scalar quantity, with SI unit $\frac{N}{C}$-m2 or V-m. “The number of field lines passing through perpendicular unit area will be proportional to the magnitude of Electric Field there”

• Gauss’s law: The flux of electric field through any closed surface S is $1/\varepsilon _0$ times the total charge enclosed by S. The law is especially useful in determining electric field E, when the source distribution has simple symmetry.

• An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a. Its dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis from –q to q.

• Field of an electric dipole in its equatorial plane (i.e., the plane perpendicular to its axis and passing through its centre) at a distance r from the centre:

$E=\frac{-P}{4\pi \varepsilon _0}\frac{1}{(\alpha^2+r^2)^{3/2}}\cong \frac{-P}{4\pi \varepsilon _0r^3}.\,\text{for}\,r\gg \alpha$

Dipole electric field on the axis at a distance r from the centre:

$E=\frac{2pr}{4\pi\varepsilon_0(r^2-\alpha^2)^3}\cong \frac{2{p}}{4\pi\varepsilon_0{r}^3}\,\,\text{for}\,\,r\gg\alpha$

The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r2 dependence of electric field due to a point charge.

• In a uniform electric field E, a dipole experiences a torque  τ given by: τ = p × E

but experiences no net force.

• Electric Field Due to Various Uniform Charge Distribution

 (i) At the centre of circular arc $E=\frac{kQ}{R^2}\frac{\text{sin}(\theta/2)}{\theta/2}$ (ii) At a point on the axis of ring $E=\frac{kQx}{(R^2+x^2)^{3/2}}$ (iii) At a point on the axis of disc $E=\frac{2kQ}{R^2}\left [ 1-\frac{x}{(R^2+x^2)^{3/2}} \right ]$ (iv) Hollow sphere For x < R: E = 0For x  ≥  R: $E=\frac{kQ}{x^2}$ (v) Non conducting solid sphere For x < R: $E=\frac{kQx}{R^3}$For x ≥ R: $E=\frac{kQ}{x^2}$ (vi) Infinite thin sheet $E=\frac{\sigma }{2\varepsilon _0}$ (vii) Infinite wire $E=\frac{2k\lambda}{x}$

Mastering Class 12 Physics Chapter 8: Electromagnetic Waves - MCQs, Question and Answers, and Tips for Success

1. What is the force between two small charged spheres of charges

$2\times {{10}^{-7}}C$ and $3\times {{10}^{-7}}C$ placed $30cm$ apart in air?

Ans: We are given the following information:

Repulsive force of magnitude,$F=6\times {{10}^{-3}}N$

Charge on the first sphere, ${{q}_{1}}=2\times {{10}^{-7}}C$

Charge on the second sphere, ${{q}_{2}}=3\times {{10}^{-7}}C$

Distance between the two spheres, $r=30cm=0.3m$

Electrostatic force between the two spheres is given by Coulomb’s law as,

$F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

Where, ${{\varepsilon }_{0}}$is the permittivity of free space and,

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

Now on substituting the given values, Coulomb’s law becomes,

$F = \frac{9\times 10^{9}\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^{2}}$

Therefore, we found the electrostatic force between the given charged spheres to be $F=6\times {{10}^{-3}}N$. Since the charges are of the same nature, we could say that the force is repulsive.

2. The electrostatic force on a small sphere of charge $0.4\mu C$ due to another small sphere of charge $-0.8\mu C$ in air is 0.2N.

a) What is the distance between the two spheres?

Ans: Electrostatic force on the first sphere is given to be, $F=0.2N$

Charge of the first sphere is, ${{q}_{1}}=0.4\mu C=0.4\times {{10}^{-6}}C$

Charge of the second sphere is, ${{q}_{2}}=-0.8\mu C=-0.8\times {{10}^{-6}}C$

We have the electrostatic force given by Coulomb’s law as,

$F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$

$\Rightarrow r=\sqrt{\frac{{{q}_{1}}{{q}_{2}}}{4\pi {{\varepsilon }_{0}}F}}$

Substituting the given values in the above equation, we get,

$\Rightarrow r=\sqrt{\frac{0.4\times {{10}^{-6}}\times 8\times {{10}^{-6}}\times 9\times {{10}^{9}}}{0.2}}$

$\Rightarrow r=\sqrt{144\times {{10}^{-4}}}$

$\therefore r=0.12m$

Therefore, we found the distance between charged spheres to be $r=0.12m$.

b) What is the force on the second sphere due to the first?

Ans: From Newton’s third law of motion, we know that every action has an equal and opposite reaction.

Thus, we could say that the given two spheres would attract each other with the same force.

So, the force on the second sphere due to the first sphere will be $0.2N$.

3. Check whether the ratio $\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}$is dimensionless. Look up a table of physical constants and hence determine the value of the given ratio. What does the ratio signify?

Ans: We are given the ratio, $\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}$.

Here, G is the gravitational constant which has its unit $N{{m}^{2}}k{{g}^{-2}}$;

${{m}_{e}}$and ${{m}_{p}}$ are the masses of electron and proton in $kg$ respectively;

$e$ is the electric charge in $C$;

$k$ is a constant given by $k=\frac{1}{4\pi {{\varepsilon }_{0}}}$

In the expression for k, ${{\varepsilon }_{0}}$ is the permittivity of free space which has its unit $N{{m}^{2}}{{C}^{-2}}$.

Now, we could find the dimension of the given ratio by considering their units as follows:

$\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}=\frac{\left[ N{{m}^{2}}{{C}^{-2}} \right]{{\left[ C \right]}^{2}}}{\left[ N{{m}^{2}}k{{g}^{-2}} \right]\left[ kg \right]\left[ kg \right]}={{M}^{0}}{{L}^{0}}{{T}^{0}}$

Clearly, it is understood that the given ratio is dimensionless.

Now, we know the values for the given physical quantities as,

$e=1.6\times {{10}^{-19}}C$

$G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$

${{m}_{e}}=9.1\times {{10}^{-31}}kg$

${{m}_{p}}=1.66\times {{10}^{-27}}kg$

Substituting these values into the required ratio, we get,

$\frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}=\frac{9\times {{10}^{9}}\times {{\left( 1.6\times {{10}^{-19}} \right)}^{2}}}{6.67\times {{10}^{-11}}\times 9.1\times {{10}^{-3}}\times 1.67\times {{10}^{-22}}}$

$\Rightarrow \frac{k{{e}^{2}}}{G{{m}_{e}}{{m}_{p}}}\approx 2.3\times {{10}^{39}}$

We could infer that the given ratio is the ratio of electrical force to the gravitational force between a proton and an electron when the distance between them is kept constant.

4.

a) Explain the meaning of the statements ‘electric charge of a body is quantized’.

Ans: The given statement ‘Electric charge of a body is quantized’ means that only the integral number $(1,2,3,...,n)$ of electrons can be transferred from one body to another.

That is, charges cannot be transferred from one body to another in fraction.

b) Why can one ignore quantization of electric charge when dealing with macroscopic i.e., large scale charges?

Ans: On a macroscopic scale or large-scale, the number of charges is as large as the magnitude of an electric charge.

So, quantization is considered insignificant at a macroscopic scale for an electric charge and electric charges are considered continuous.

5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Ans: Rubbing two objects would produce charges that are equal in magnitude and opposite in nature on the two bodies.

This happens due to the reason that charges are created in pairs. This phenomenon is called charging by friction.

The net charge of the system however remains zero as the opposite charges equal in magnitude annihilate each other.

So, rubbing a glass rod with a silk cloth creates opposite charges of equal magnitude on both of them and this observation is found to be consistent with the law of conservation of charge.

6. Four point charges ${{q}_{A}}=2\mu C$, ${{q}_{B}}=-5\mu C$, ${{q}_{C}}=2\mu C$and ${{q}_{D}}=-5\mu C$ are located at the corners of a square ABCD with side 10cm. What is the force on the $1\mu C$ charge placed at the centre of this square?

Ans: Consider the square of side length $10cm$ given below with four charges at its corners and let O be its centre.

From the figure we find the diagonals to be,

$AC=BD=10\sqrt{2}cm$

$\Rightarrow AO=OC=DO=OB=5\sqrt{2}cm$

Now the repulsive force at O due to charge at A,

${{F}_{AO}}=k\frac{{{q}_{A}}{{q}_{O}}}{O{{A}^{2}}}=k\frac{\left( +2\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$…………………………………………… (1)

And the repulsive force at O due to charge at D,

${{F}_{DO}}=k\frac{{{q}_{D}}{{q}_{O}}}{O{{D}^{2}}}=k\frac{\left( +2\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$………………………………………….. (2)

And the attractive force at O due to charge at B,

${{F}_{BO}}=k\frac{{{q}_{B}}{{q}_{O}}}{O{{B}^{2}}}=k\frac{\left( -5\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$……………………………………………. (3)

And the attractive force at O due to charge at C,

${{F}_{CO}}=k\frac{{{q}_{C}}{{q}_{O}}}{O{{C}^{2}}}=k\frac{\left( -5\mu C \right)\left( 1\mu C \right)}{{{\left( 5\sqrt{2} \right)}^{2}}}$……………………………………………… (4)

We find that (1) and (2) are of same magnitude but they act in the opposite direction and hence they cancel out each other.

Similarly, (3) and (4) are of the same magnitude but in the opposite direction and hence they cancel out each other too.

Hence, the net force on charge at centre O is found to be zero.

7.

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?

Ans: An electrostatic field line is a continuous curve as the charge experiences a continuous force on being placed in an electric field.

As the charge doesn’t jump from one point to the other, field lines will not have sudden breaks.

b) Explain why two field lines never cross each other at any point?

Ans: If two field lines are seen to cross each other at a point, it would imply that the electric field intensity has two different directions at that point, as two different tangents (representing the direction of electric field intensity at that point) can be drawn at the point of intersection.

This is however impossible and thus, two field lines never cross each other.

8. Two point charges ${{q}_{A}}=3\mu C$and ${{q}_{B}}=-3\mu C$are located 20cm apart in vacuum.

a) What is the electric field at the midpoint O of the line AB joining the two charges?

Ans: The situation could be represented in the following figure. Let O be the midpoint of line AB.

We are given:

$AB=20cm$

$AO=OB=10cm$

Take E to be the electric field at point O, then,

The electric field at point O due to charge $+3\mu C$would be,

${{E}_{1}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( AO \right)}^{2}}}=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( 10\times {{10}^{-2}} \right)}^{2}}}N{{C}^{-1}}$along OB

The electric field at point O due to charge $-3\mu C$would be,

${{E}_{2}}=\left| \frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( OB \right)}^{2}}} \right|=\frac{3\times {{10}^{-6}}}{4\pi {{\varepsilon }_{0}}{{\left( 10\times {{10}^{-2}} \right)}^{2}}}N{{C}^{-1}}$along OB

The net electric field,

$\Rightarrow E={{E}_{1}}+{{E}_{2}}$

$\Rightarrow E=2\times \frac{9\times {{10}^{9}}\times 3\times {{10}^{-6}}}{{{\left( 10\times {{10}^{-2}} \right)}^{2}}}$

$\Rightarrow E=5.4\times {{10}^{6}}N{{C}^{-1}}$

Therefore, the electric field at mid-point O is $E=5.4\times {{10}^{6}}N{{C}^{-1}}$ along OB.

b) If a negative test charge of magnitude $1.5\times {{10}^{-19}}C$ is placed at this point, what is the force experienced by the test charge?

Ans: We have a test charge of magnitude $1.5\times {{10}^{-9}}C$ placed at mid-point O and we found the electric field at this point to be $E=5.4\times {{10}^{6}}N{{C}^{-1}}$.

So, the force experienced by the test charge would be F,

$\Rightarrow F=qE$

$\Rightarrow F=1.5\times {{10}^{-9}}\times 5.4\times {{10}^{6}}$

$\Rightarrow F=8.1\times {{10}^{-3}}N$

This force will be directed along OA since like charges repel and unlike charges attract.

9. A system has two charges ${{q}_{A}}=2.5\times {{10}^{-7}}C$and ${{q}_{B}}=-2.5\times {{10}^{-7}}C$located at points $A:\left( 0,0,-15cm \right)$ and $B:\left( 0,0,+15cm \right)$ respectively. What are the total charge and electric dipole moment of the system?

Ans: The figure given below represents the system mentioned in the question:

The charge at point A, ${{q}_{A}}=2.5\times {{10}^{-7}}C$

The charge at point B, ${{q}_{B}}=-2.5\times {{10}^{-7}}C$

Then, the net charge would be, $q={{q}_{A}}+{{q}_{B}}=2.5\times {{10}^{-7}}C-2.5\times {{10}^{-7}}C=0$

The distance between two charges at A and B would be,

$d=15+15=30cm$

$d=0.3m$

The electric dipole moment of the system could be given by,

$P=\mathop{q}_{A}\times d=\mathop{q}_{B}\times d$

$\Rightarrow P=2.5\times {{10}^{-7}}\times 0.3$

$\therefore P=7.5\times {{10}^{-8}}Cm$ along the $+z$axis.

Therefore, the electric dipole moment of the system is found to be $7.5\times {{10}^{-8}}Cm$ and it is directed along the positive $z$-axis.

10. An electric dipole with dipole moment $4\times {{10}^{-9}}Cm$ is aligned at $30{}^\circ$ with direction of a uniform electric field of magnitude $5\times {{10}^{4}}N{{C}^{-1}}$. Calculate the magnitude of the torque acting on the dipole.

Ans: We are given the following:

Electric dipole moment, $\overrightarrow{p}=4\times {{10}^{-9}}Cm$

Angle made by $\overrightarrow{p}$ with uniform electric field, $\theta =30{}^\circ$

Electric field, $\overrightarrow{E}=5\times {{10}^{4}}N{{C}^{-1}}$

Torque acting on the dipole is given by

$\tau =pE\sin \theta$

Substituting the given values we get,

$\Rightarrow \tau =4\times {{10}^{-9}}\times 5\times {{10}^{4}}\times \sin 30{}^\circ$

$\Rightarrow \tau =20\times {{10}^{-5}}\times \frac{1}{2}$

$\therefore \tau ={{10}^{-4}}Nm$

Thus, the magnitude of the torque acting on the dipole is found to be ${{10}^{-4}}Nm$.

11. A polythene piece rubbed with wool is found to have a negative charge of $3\times {{10}^{-7}}C$

a) Estimate the number of electrons transferred (from which to which?)

Ans: When polythene is rubbed against wool, a certain number of electrons get transferred from wool to polythene.

As a result of which wool becomes positively charged on losing electrons and polythene becomes negatively charged on gaining them.

We are given:

Charge on the polythene piece, $q=-3\times {{10}^{-7}}C$

Charge of an electron, $e=-1.6\times {{10}^{-19}}C$

Let n be the number of electrons transferred from wool to polythene, then, from the property of quantization we have,

$q=ne$

$\Rightarrow n=\frac{q}{e}$

Now, on substituting the given values, we get,

$\Rightarrow n=\frac{-3\times {{10}^{-7}}}{-1.6\times {{10}^{-19}}}$

$\therefore n=1.87\times {{10}^{12}}$

Therefore, the number of electrons transferred from wool to polythene would be$1.87\times {{10}^{12}}$.

b) Is there a transfer of mass from wool to polythene?

Ans: Yes, during the transfer of electrons from wool to polythene, along with charge, mass is transferred too.

Let $m$ be the mass being transferred in the given case and ${{m}_{e}}$ be the mass of the electron, then,

$m={{m}_{e}}\times n$

$\Rightarrow m=9.1\times {{10}^{-31}}\times 1.85\times {{10}^{12}}$

$\therefore m=1.706\times {{10}^{-18}}kg$

Thus, we found that a negligible amount of mass does get transferred from wool to polythene.

12.

a) Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times {{10}^{-7}}C$? The radii of A and B are negligible compared to the distance of separation.

Ans: We are given:

Charges on spheres $A$ and $B$ are equal,

${{q}_{A}}={{q}_{B}}=6.5\times {{10}^{-7}}C$

Distance between the centres of the spheres is given as,

$r=50cm=0.5m$

It is known that the force of repulsion between the two spheres would be given by Coulomb’s law as,

$F=\frac{{{q}_{A}}{{q}_{B}}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

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

Substituting the known values into the above expression, we get,

$F=\frac{9\times {{10}^{9}}\times {{(6.5\times {{10}^{-7}})}^{2}}}{{{(0.5)}^{2}}}=1.52\times {{10}^{-2}}N$

Thus, the mutual force of electrostatic repulsion between the two spheres is found to be$F=1.52\times {{10}^{-2}}N$.

b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Ans: It is told that the charges on both the spheres are doubled and the distance between the centres of the spheres is halved. That is,

${{q}_{A}}'={{q}_{B}}'=2\times 6.5\times {{10}^{-7}}=13\times {{10}^{-7}}C$

$r'=\frac{1}{2}(0.5)=0.25m$

Now, we could substitute these values in Coulomb’s law to get,

$F'=\frac{{{q}_{A}}'{{q}_{B}}'}{4\pi {{\varepsilon }_{0}}r{{'}^{2}}}$

$\Rightarrow F=\frac{9\times {{10}^{9}}\times {{(13\times {{10}^{-7}})}^{2}}}{{{(0.25)}^{2}}}$

$\Rightarrow F=0.243N$

The new mutual force of electrostatic repulsion between the two spheres is found to be $0.243N$.

13. Suppose the spheres $A$ and $B$ in question $12$ have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between $A$ and $B$?

Ans: We are given the following:

Distance between the spheres $A$ and $B$ is $r=0.5m$

The charge on each sphere initially is found to be ${{q}_{A}}={{q}_{B}}=6.5\times {{10}^{-7}}C$

Now, when an uncharged sphere $C$ is made to touch the sphere $A$, a certain amount of charge from $A$ will get transferred to the sphere $C$, making both $A$ and $C$ to have equal charges in them. So,

${{q}_{A}}'={{q}_{C}}=\frac{1}{2}(6.5\times {{10}^{-7}})=3.25\times {{10}^{-7}}C$

Now, when the sphere $C$ is made to touch the sphere $B$, there is a similar transfer of charge making both $C$ and $B$ to have equal charges in them. So,

${{q}_{C}}'={{q}_{B}}'=\frac{3.25\times {{10}^{-7}}+6.5\times {{10}^{-7}}}{2}=4.875\times {{10}^{-7}}C$

Thus, the new force of repulsion between the spheres $A$ and $B$ would now become,

$F'=\frac{{{q}_{A}}'{{q}_{B}}'}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$

$\Rightarrow F'=\frac{9\times {{10}^{9}}\times 3.25\times {{10}^{-7}}\times 4.875\times {{10}^{-7}}}{{{(0.5)}^{2}}}$

$\Rightarrow F'=5.703\times {{10}^{-3}}N$

14. Figure below shows tracks taken by three charged particles in a uniform electrostatic field. Give the signs of the three charges and also mention which particle has the highest charge to mass ratio?

Ans: From the known properties of charges, we know that the unlike charges attract and like charges repel each other.

So, the particles 1 and 2 that move towards the positively charged plate while repelling away from the negatively charged plate would be negatively charged and the particle 3 that moves towards the negatively charged plate while repelling away from the positively charged plate would be positively charged.

Now, we know that the charge to mass ratio (which is generally known as emf) is directly proportional to the displacement or the amount of deflection for a given velocity.

Since the deflection of particle 3 is found to be maximum among the three, it would have the highest charge to mass ratio.

15. Consider a uniform electric field $E=3\times {{10}^{3}}\hat{i}N/C$.

a) Find the flux of this field through a square of side $10cm$whose plane is parallel to the y-z plane.

Ans: We are given:

Electric field intensity, $\overrightarrow{E}=3\times {{10}^{3}}\hat{i}N/C$

Magnitude of electric field intensity, $\left| \overrightarrow{E} \right|=3\times {{10}^{3}}N/C$

Side of the square, $a=10cm=0.1m$

Area of the square, $A={{a}^{2}}=0.01{{m}^{2}}$

Since the plane of the square is parallel to the y-z plane, the normal to its plane would be directed in the x direction. So, angle between normal to the plane and the electric field would be, $\theta =0{}^\circ$

We know that the flux through a surface is given by the relation,

$\phi =\left| E \right|\left| A \right|\cos \theta$

Substituting the given values, we get,

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 0{}^\circ$

$\therefore \phi =30N{{m}^{2}}/C$

Thus, we found the net flux through the given surface to be $\phi =30N{{m}^{2}}/C$.

b) What would be the flux through the same square if the normal to its plane makes $60{}^\circ$ angle with the x-axis?

Ans: When the plane makes an angle of $60{}^\circ$ with the x-axis, the flux through the given surface would be,

$\phi =\left| E \right|\left| A \right|\cos \theta$

$\Rightarrow \phi =3\times {{10}^{3}}\times 0.01\times \cos 60{}^\circ$

$\Rightarrow \phi =30\times \frac{1}{2}$

$\Rightarrow \phi =15N{{m}^{2}}/C$

So, we found the flux in this case to be, $\phi =15N{{m}^{2}}/C$.

16. What is the net flux of the uniform electric field of exercise $1.15$ through a cube of side $20cm$ oriented so that its faces are parallel to the coordinate planes?

Ans: We are given that all the faces of the cube are parallel to the coordinate planes.

Clearly, the number of field lines entering the cube is equal to the number of field lines entering out of the cube. As a result, the net flux through the cube would be zero.

17. Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0\times {{10}^{3}}N{{m}^{2}}/C$

a) What is the net charge inside the box?

Ans: We are given that:

Net outward flux through surface of the box,

$\phi =8.0\times {{10}^{3}}N{{m}^{2}}/C$

For a body containing of net charge $q$, flux could be given by,

$\phi =\frac{q}{{{\varepsilon }_{0}}}$

Where, ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space

Therefore, the charge $q$ is given by

$q=\phi {{\varepsilon }_{0}}$

$\Rightarrow q=8.854\times {{10}^{-12}}\times 8.0\times {{10}^{3}}$

$\Rightarrow q=7.08\times {{10}^{-8}}$

$\Rightarrow q=0.07\mu C$

Therefore, the net charge inside the box is found to be $0.07\mu C$.

b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Ans: No, the net flux entering out through a body depends on the net charge contained within the body according to Gauss’s law.

So, if the net flux is given to be zero, then it can be inferred that the net charge inside the body is zero.

However, the net charge of the body being zero only implies that the body has equal amount of positive and negative charges and thus, we cannot conclude that there were no charges inside the box.

18. A point charge $+10\mu C$ is a distance $5cm$ directly above the centre of a square of side $10cm$, as shown in Figure below. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10cm$)

Ans: Consider the square as one face of a cube of edge length $10cm$ with a charge $q$ at its centre, according to Gauss's theorem for a cube, total electric flux is through all its six faces.

${{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}$

The electric flux through one face of the cube could be now given by,  $\phi =\frac{{{\phi }_{total}}}{6}$.

$\phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space

The net charge enclosed would be, $q=10\mu C=10\times {{10}^{-6}}C$

Substituting the values given in the question, we get,

$\phi =\frac{1}{6}\times \frac{10\times {{10}^{-6}}}{8.854\times {{10}^{-12}}}$

$\therefore \phi =1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$

Therefore, electric flux through the square is found to be $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

19. A point charge of $2.0\mu C$ is kept at the centre of a cubic Gaussian surface of edge length $9cm$. What is the net electric flux through this surface?

Ans: Let us consider one of the faces of the cubical Gaussian surface considered (square).

Since a cube has six such square faces in total, we could say that the flux through one surface would be one-sixth the total flux through the gaussian surface considered.

The net flux through the cubical Gaussian surface by Gauss’s law could be given by,

${{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}$

So, the electric flux through one face of the cube would be, $\phi =\frac{{{\phi }_{total}}}{6}$

$\Rightarrow \phi =\frac{1}{6}\frac{q}{{{\varepsilon }_{0}}}$……………………………….. (1)

But we have,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space

Charge enclosed, $q=10\mu C=10\times {{10}^{-6}}C$

Substituting the given values in (1) we get,

$\phi =\frac{1}{6}\times \frac{10\times {{10}^{-6}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow \phi =1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$

Therefore, electric flux through the square surface is $1.88\times {{10}^{5}}N{{m}^{2}}{{C}^{-1}}$.

20. A point charge causes an electric flux of $-1.0\times {{10}^{3}}N{{m}^{2}}/C$ to pass through a spherical Gaussian surface of $10cm$ radius centred on the charge.

a) If the radius of the Gaussian surface were doubled, how much flux could pass through the surface?

Ans: We are given:

Electric flux due to the given point charge, $\phi =-1.0\times {{10}^{3}}N{{m}^{2}}/C$

Radius of the Gaussian surface enclosing the point charge,$r=10.0cm$

Electric flux piercing out through a surface depends on the net charge enclosed by the surface according to Gauss’s law and is independent of the dimensions of the arbitrary surface assumed to enclose this charge.

Hence, if the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., $-{{10}^{3}}N{{m}^{2}}/C$.

b) What is the magnitude of the point charge?

Ans: Electric flux could be given by the relation,

${{\phi }_{total}}=\frac{q}{{{\varepsilon }_{0}}}$

Where,$q=$ net charge enclosed by the spherical surface

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space

$\Rightarrow q=\phi {{\varepsilon }_{0}}$

Substituting the given values,

$\Rightarrow q=-1.0\times {{10}^{3}}\times 8.854\times {{10}^{-12}}=-8.854\times {{10}^{-9}}C$

$\Rightarrow q=-8.854nC$

Thus, the value of the point charge is found to be $-8.854nC$.

21. A conducting sphere of radius $10cm$ has an unknown charge. If the electric field at a point $20cm$ from the centre of the sphere of magnitude $1.5\times {{10}^{3}}N/C$ is directed radially inward, what is the net charge on the sphere?

Ans: We have the relation for electric field intensity $E$ at a distance $\left( d \right)$ from the centre of a sphere containing net charge $q$ is given by,

$E=\frac{q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}$ ……………………………………………… (1)

Where,

Net charge, $q=1.5\times {{10}^{3}}N/C$

Distance from the centre, $d=20cm=0.2m$

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}=$ Permittivity of free space

$\frac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}$

From (1), the unknown charge would be,

$q=E\left( 4\pi {{\varepsilon }_{0}} \right){{d}^{2}}$

Substituting the given values we get,

$\Rightarrow q=\frac{1.5\times {{10}^{3}}\times {{\left( 0.2 \right)}^{2}}}{9\times {{10}^{9}}}=6.67\times {{10}^{-9}}C$

$\Rightarrow q=6.67nC$

Therefore, the net charge on the sphere is found to be$6.67nC$.

22. A uniformly charged conducting sphere of $2.4m$ diameter has a surface charge density of $80.0\mu C/{{m}^{2}}$.

a) Find the charge on the sphere.

Ans: Given that,

Diameter of the sphere, $d=2.4m$.

Radius of the sphere, $r=1.2m$.

Surface charge density,

$\sigma =80.0\mu C/{{m}^{2}}=80\times {{10}^{-6}}C/{{m}^{2}}$

Total charge on the surface of the sphere,

$Q=\text{Charge density }\times \text{ Surface area}$

$\Rightarrow \text{Q}=\sigma \times \text{4}\pi {{\text{r}}^{2}}=80\times {{10}^{-6}}\times 4\times 3.14\times {{\left( 1.2 \right)}^{2}}$

$\Rightarrow Q=1.447\times {{10}^{-3}}C$

Therefore, the charge on the sphere is found to be $1.447\times {{10}^{-3}}C$.

b) What is the total electric flux leaving the surface of the sphere?

Ans: Total electric flux $\left( {{\phi }_{total}} \right)$ leaving out the surface containing net charge $Q$ is given by Gauss’s law as,

${{\phi }_{total}}=\frac{Q}{{{\varepsilon }_{0}}}$…………………………………………………. (1)

Where, permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

We found the charge on the sphere to be,

$Q=1.447\times {{10}^{-3}}C$

Substituting these in (1), we get,

${{\phi }_{total}}=\frac{1.447\times {{10}^{-3}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow {{\phi }_{total}}=1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$

Therefore, the total electric flux leaving the surface of the sphere is found to be $1.63\times {{10}^{-8}}N{{C}^{-1}}{{m}^{2}}$.

23. An infinite line charge produces a field of magnitude $9\times {{10}^{4}}N/C$ at a distance of $2cm$. Calculate the linear charge density.

Ans: Electric field produced by the given infinite line charge at a distance $d$having linear charge density$\lambda$ could be given by the relation,

$E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}d}$

$\Rightarrow \lambda =2\pi {{\varepsilon }_{0}}Ed$…………………………………….. (1)

We are given:

$d=2cm=0.02m$

$E=9\times {{10}^{4}}N/C$

Permittivity of free space,

${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

Substituting these values in (1) we get,

$\Rightarrow \lambda =2\pi \left( 8.854\times {{10}^{-12}} \right)\left( 9\times {{10}^{4}} \right)\left( 0.02 \right)$

$\Rightarrow \lambda =10\times {{10}^{-8}}C/m$

Therefore, we found the linear charge density to be $10\times {{10}^{-8}}C/m$.

24. Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0\times {{10}^{-22}}C{{m}^{-2}}$. What is $E$ in the outer region of the first plate? What is $E$ in the outer region of the second plate? What is E between the plates?

Ans: The given nature of metal plates is represented in the figure below:

Here, A and B are two parallel plates kept close to each other. The outer region of plate A is denoted as $I$, outer region of plate B is denoted as $III$, and the region between the plates, A and B, is denoted as $II$.

It is given that:

Charge density of plate A, $\sigma =17.0\times {{10}^{-22}}C/{{m}^{2}}$

Charge density of plate B, $\sigma =-17.0\times {{10}^{-22}}C/{{m}^{2}}$

In the regions $I$and$III$, electric field E is zero. This is because the charge is not enclosed within the respective plates.

Now, the electric field $E$ in the region $II$ is given by

$E=\frac{|\sigma |}{{{\varepsilon }_{0}}}$

Where,

Permittivity of free space ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$

Clearly,

$E=\frac{17.0\times {{10}^{-22}}}{8.854\times {{10}^{-12}}}$

$\Rightarrow E=1.92\times {{10}^{-10}}N/C$

Thus, the electric field between the plates is $1.92\times {{10}^{-10}}N/C$.

25. An oil drop of $12$ excess electrons is held stationary under a constant electric field of $2.55\times {{10}^{4}}N{{C}^{-1}}$ in Millikan's oil drop experiment. The density of the oil is $1.26gm/c{{m}^{3}}$. Estimate the radius of the drop. $\left( g=9.81m{{s}^{-2}},e=1.60\times {{10}^{-19}}C \right)$.

Ans: It is given that:

The number of excess electrons on the oil drop, $n=12$

Electric field intensity, $E=2.55\times {{10}^{4}}N{{C}^{-1}}$

The density of oil, $\rho =1.26gm/c{{m}^{3}}=1.26\times {{10}^{3}}kg/{{m}^{3}}$

Acceleration due to gravity, $g=9.81m{{s}^{-2}}$

Charge on an electron $e=1.60\times {{10}^{-19}}C$

Radius of the oil drop $=r$

Here, the force (F) due to electric field E is equal to the weight of the oil drop (W).

Clearly,

$F=W$

$\Rightarrow Eq=mg$

$\Rightarrow Ene=\frac{4}{3}\pi {{r}^{2}}\rho \times g$

Where,

$q$ is the net charge on the oil drop $=ne$

$m$ is the mass of the oil drop $=\text{Volume of the oil drop}\times \text{Density of oil}$$=\frac{4}{3}\pi {{r}^{3}}\times p$

Therefore, radius of the oil drop can be calculated as

$r=\sqrt{\frac{3Ene}{4\pi \rho g}}$

$\Rightarrow r=\sqrt{\frac{3\times 2.55\times {{10}^{4}}\times 12\times 1.6\times {{10}^{-19}}}{4\times 3.14\times 1.26\times {{10}^{3}}\times 9.81}}$

$\Rightarrow r=\sqrt{946.09\times {{10}^{-21}}}$

$\Rightarrow r=9.72\times {{10}^{-10}}m$

Therefore, the radius of the oil drop is $9.72\times {{10}^{-10}}m$.

26. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?

a)

Ans: The field lines shown in (a) do not represent electrostatic field lines because field lines must be normal to the surface of the conductor which is a characterizing property of electric field lines.

b)

Ans: The lines shown in (b) do not represent electrostatic field lines because field lines cannot emerge from a negative charge and cannot terminate at a positive charge since the direction of the electric field is from positive to negative charge.

c)

Ans: The field lines shown in (c) do represent electrostatic field lines as they are directed outwards from positive charge in accordance with the property of electric field.

d)

Ans: The field lines shown in (d) do not represent electrostatic field lines because electric field lines should not intersect each other.

e)

Ans: The field lines shown in (e) do not represent electrostatic field lines because electric field lines do not form closed loops

27. In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of ${{10}^{5}}N{{C}^{-1}}$ per meter. What are the force and torque experienced by a system having a total dipole moment equal to ${{10}^{-7}}Cm$ in the negative z-direction?

Ans: We know that the dipole moment of the system, $P=q\times dl=-{{10}^{-7}}Cm$.

Also, the rate of increase of electric field per unit length is given as

$\frac{dE}{dl}={{10}^{5}}N{{C}^{-1}}$

Now, the force (F) experienced by the system is given by $F=qE$

$F=q\frac{dE}{dl}\times dl$

$\Rightarrow F=P\frac{dE}{dl}$

$\Rightarrow F=-{{10}^{-7}}\times {{10}^{5}}$

$\Rightarrow F=-{{10}^{-2}}N$

Clearly, the force is equal to $-{{10}^{-2}}N$ in the negative z-direction i.e., it is opposite to the direction of the electric field.

Thus, the angle between the electric field and dipole moment is equal to $180{}^\circ$.

Now, the torque is given by $\tau =PE\sin \theta$

$\tau =PE\sin 180{}^\circ =0$

Therefore, it can be concluded that the torque experienced by the system is zero.

28.

a) A conductor A with a cavity as shown in the Fig. 1.36(a) is given a charge $Q$. Show that the entire charge must appear on the outer surface of the conductor.

Ans: Firstly, let us consider a Gaussian surface that is lying within a conductor as a whole and enclosing the cavity. Clearly, the electric field intensity E inside the charged conductor is zero.

Now, let $q$ be the charge inside the conductor and ${{\varepsilon }_{0}}$, the permittivity of free space.

According to Gauss's law,

Flux is given by

$\phi =\overrightarrow{E}.ds=\frac{q}{{{\varepsilon }_{0}}}$

Here, $\phi =0$  as $E=0$ inside the conductor

Clearly,

$0=\frac{q}{8.854\times {{10}^{-12}}}$

$\Rightarrow q=0$

Therefore, the charge inside the conductor is zero.

And hence, the entire charge $Q$ appears on the outer surface of the conductor.

b) Another conductor B with charge $q$ is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is $Q+q$ [Fig. 1.36 (b)].

Ans: The outer surface of conductor A has a charge of $Q$.

It is given that another conductor B, having a charge $+q$ is kept inside conductor A and is insulated from conductor A.

Clearly, a charge of $-q$ will get induced in the inner surface of conductor A and a charge of $+q$ will get induced on the outer surface of conductor A.

Therefore, the total charge on the outer surface of conductor A amounts to $Q+q$.

c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Ans: A sensitive instrument can be shielded from a strong electrostatic field in its environment by enclosing it fully inside a metallic envelope.

Such a closed metallic body provides hindrance to electrostatic fields and thus can be used as a shield.

29. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left[ \frac{\sigma }{2{{\varepsilon }_{0}}} \right]\overset{\wedge }{\mathop{n}}\,$, where $\overset{\wedge }{\mathop{n}}\,$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.

Ans: Firstly, let us consider a conductor with a cavity or a hole as shown in the diagram below. It is known that the electric field inside the cavity is zero.

Let us assume E to be the electric field just outside the conductor, $q$ be the electric charge, $\sigma$ be the charge density, and ${{\varepsilon }_{0}}$, the permittivity of free space.

We know that charge $\left| q \right|=\sigma \times d$

Now, according to Gauss's law,

$\phi =E.ds=\frac{\left| q \right|}{{{\varepsilon }_{0}}}$

$\Rightarrow E.ds=\frac{\sigma \times d}{{{\varepsilon }_{0}}}$

$\Rightarrow E=\frac{\sigma }{{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$

where $\overset{\wedge }{\mathop{n}}\,$ is the unit vector in the outward normal direction.

Thus, the electric field just outside the conductor is $\frac{\sigma }{{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$.

Now, this field is actually a superposition of the field due to the cavity ${{E}_{1}}$ and the field due to the rest of the charged conductor ${{E}_{2}}$.

These electric fields are equal and opposite inside the conductor whereas equal in magnitude as well as direction outside the conductor.

Clearly,

${{E}_{1}}+{{E}_{2}}=E$

$\Rightarrow {{E}_{1}}={{E}_{2}}=\frac{E}{2}=\frac{\sigma }{2{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$

Therefore, the electric field in the hole is  $\frac{\sigma }{2{{\varepsilon }_{0}}}\overset{\wedge }{\mathop{n}}\,$.

Hence, proved.

30. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $\lambda$ without using Gauss's law. (Hint: Use Coulomb's law directly and evaluate the necessary integral)

Ans: Firstly, let us take a long thin wire XY as shown in the figure below. This wire is of uniform linear charge density $\lambda$.

Now, consider a point A at a perpendicular distance l from the mid-point O of the wire as shown in the figure below:

Consider E to be the electric field at point A due to the wire.

Also consider a small length element $dx$ on the wire section with $OZ=x$ as shown.

Let $q$ be the charge on this element.

Clearly, $q=\lambda dx$

Now, the electric field due to this small element can be given as

$dE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda dx}{{{\left( AZ \right)}^{2}}}$

However, $AZ=\sqrt{{{1}^{2}}+{{x}^{2}}}$

$\Rightarrow dE=\frac{\lambda dx}{4\pi {{\varepsilon }_{0}}\left( {{1}^{2}}+{{x}^{2}} \right)}$

Now, let us resolve the electric field into two rectangular components. Doing so, $dE\cos \theta$ is the perpendicular component and $dE\sin \theta$ is the parallel component.

When the whole wire is considered, the component $dE\sin \theta$ gets cancelled and only the perpendicular component $dE\cos \theta$  affects the point A.

Thus, the effective electric field at point A due to the element $dx$ can be written as

$d{{E}_{1}}=\frac{\lambda dx\cos \theta }{4\pi {{\varepsilon }_{0}}\left( {{l}^{2}}+{{x}^{2}} \right)}$    ....(1)

Now, in $\Delta AZO$, we have

$\tan \theta =\frac{x}{l}$

$x=l\tan \theta \text{ }......\text{(2)}$

On differentiating equation (2), we obtain

$dx=l{{\sec }^{2}}d\theta \text{ }......\text{(3)}$

From equation (2)

${{x}^{2}}+{{l}^{2}}={{l}^{2}}+{{l}^{2}}{{\tan }^{2}}\theta$

$\Rightarrow {{l}^{2}}\left( 1+{{\tan }^{2}}\theta \right)={{l}^{2}}{{\sec }^{2}}\theta$

$\Rightarrow {{x}^{2}}+{{l}^{2}}={{l}^{2}}{{\sec }^{2}}\theta \text{ }.....\text{(4)}$

Putting equations (3) and (4) in equation (1), we obtain

$d{{E}_{1}}=\frac{\lambda l{{\sec }^{2}}d\theta }{4\pi {{\varepsilon }_{0}}\left( {{l}^{2}}{{\sec }^{2}}\theta \right)}\cos \theta$

$\Rightarrow d{{E}_{1}}=\frac{\lambda \cos \theta d\theta }{4\pi {{\varepsilon }_{0}}l}\text{ }.....\text{(5)}$

Now, the wire is taken so long that it ends from $-\frac{\pi }{2}$ to $+\frac{\pi }{2}$.

Therefore, by integrating equation (5), we obtain the value of field ${{E}_{1}}$ as

$\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{d{{E}_{1}}}=\int\limits_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\frac{\lambda }{4\pi {{\varepsilon }_{0}}l}\cos \theta d\theta }$

$\Rightarrow {{E}_{1}}=\frac{\lambda }{4\pi {{\varepsilon }_{0}}l}\times \text{2}$

$\Rightarrow {{E}_{1}}=\frac{\lambda }{2\pi {{\varepsilon }_{0}}l}$

Thus, the electric field due to the long wire is derived to be equal to $\frac{\lambda }{2\pi {{\varepsilon }_{0}}l}$.

31. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called 'up quark (denoted by $u$)  of charge $\left( +\frac{1}{2} \right)e$ and the 'down' quark (denoted by $d$) of charge $-\left( \frac{1}{3} \right)e$ together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Ans: It is known that a proton has three quarks. Let us consider $n$ up quarks in a proton, each having a charge of $+\left( \frac{2}{3}e \right)$.

Now, the charge due to $n$ up quarks $=\left( \frac{2}{3}e \right)n$

The number of down quarks in a proton $=3-n$

Also, each down quark has a charge of $-\frac{1}{3}e$

Therefore, the charge due to $\left( 3-n \right)$ down quarks $=\left( -\frac{1}{3}e \right)\left( 3-n \right)$

We know that the total charge on a proton $=+e$

Therefore,

$e=\left( \frac{2}{3}e \right)n+\left( -\frac{1}{3}e \right)\left( 3-n \right)$

$\Rightarrow e=\left( \frac{2ne}{3} \right)-e+\frac{ne}{3}$

$\Rightarrow 2e=ne$

$\Rightarrow n=2$

Clearly, the number of up quarks in a proton, $n=2$

Thus, the number of down quarks in a proton $=3-n=3-2=1$

Therefore, a proton can be represented as $uud$.

A neutron is also said to have three quarks. Let us consider $n$ up quarks in a neutron, each having a charge of $+\left( \frac{2}{3}e \right)$ .

It is given that the charge on a neutron due to $n$ up quarks $=\left( +\frac{3}{2}e \right)n$

Also, the number of down quarks is $\left( 3-n \right)$, each having a charge of $=\left( -\frac{3}{2} \right)e$

Thus, the charge on a neutron due to $\left( 3-n \right)$ down quarks $=\left( -\frac{1}{3}e \right)\left( 3-n \right)$

Now, we know that the total charge on a neutron $=0$

Thus,

$0=\left( \frac{2}{3}e \right)n+\left( -\frac{1}{3}e \right)\left( 3-n \right)$

$\Rightarrow 0=\left( \frac{2ne}{3} \right)-e+\frac{ne}{3}$

$\Rightarrow e=ne$

$\Rightarrow n=1$

Clearly, the number of up quarks in a neutron, $n=1$

Thus, the number of down quarks in a neutron $=3-n=2$

Therefore, a neutron can be represented as $udd$.

33.

a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where $\mathbf{E}=\mathbf{0}$) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.

Ans: Firstly, let us assume that the small test charge placed at the null point of the given setup is in stable equilibrium.

By stable equilibrium, it means that even a slight displacement of the test charge in any direction will cause the charge to return to the null point as there will be strong restoring forces acting around it.

This further suggests that all the electric lines of force around the null point act inwards and towards the given null point.

But by Gauss law, we know that the net electric flux through a chargeless enclosing surface is equal to zero. This truth contradicts the assumption which we had started with. Therefore, it can be concluded that the equilibrium of the test charge is necessarily unstable.

b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed at a certain distance apart.

Ans: When we consider this configuration setup with two charges of the same magnitude and sign placed at a certain distance apart, the null point happens to be at the midpoint of the line joining these two charges.

As per the previous assumption, the test charge, when placed at this mid-point will experience strong restoring forces when it tries to displace itself.

But when the test charge tries to displace in a direction normal to the line joining the two charges, the test charge gets pulled off as there is no restoring force along the normal to the line considered.

Since stable equilibrium prioritizes restoring force in all directions, the assumption in this case also gets contradicted.

33. A particle of mass $m$ and charge $\left( -q \right)$ enters the region between the two charged plates initially moving along x- axis with speed $vx$ (like particle 1 in Fig 1.33). The length of plate is $L$ and a uniform electric field $E$ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is $\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$.

Compare this motion with motion of a projectile in the gravitational field discussed in section 4.10 of class XI textbook of Physics.

Ans: It is given that:

The charge on a particle of mass $m=-q$

Velocity of the particle $=vx$

Length of the plates $=L$

Magnitude of the uniform electric field between the plates $=E$

Mechanical force, $F=\text{ Mass }(m)\times \text{Acceleration }(a)$

Thus, acceleration, $a=\frac{F}{m}$

However, electric force, $F=qE$

Therefore, acceleration, $=\frac{qE}{m}$  .........(1)

Here, the time taken by the particle to cross the field of length $L$ is given by,

$t=\frac{\text{Length of the plate}}{\text{Velocity of the plate}}=\frac{L}{{{v}_{x}}}$  ......(2)

In the vertical direction, we know that the initial velocity, $u=0$

Now, according to the third equation of motion, vertical deflection $s$ of the particle can be derived as

$s=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow s=0+\frac{1}{2}\left( \frac{qE}{m} \right){{\left( \frac{L}{{{v}_{x}}} \right)}^{2}}$

$\Rightarrow s=\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$   .....(3)

Thus, the vertical deflection of the particle at the far edge of the plate is $\frac{qE{{L}^{2}}}{2m{{v}_{x}}^{2}}$.

In comparison, we can see that this is similar to the motion of horizontal projectiles under gravity.

34. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity ${{v}_{x}}=2.0\times {{10}^{6}}m{{s}^{-1}}$. If $E$between the plates separated by $0.5cm$ is $9.1\times {{10}^{2}}N/C$, where will the electron strike the upper plate? ($\left| e \right|=1.6\times {{10}^{-19}}C,{{m}_{e}}=9.1\times {{10}^{-31}}kg$ )

Ans: We are given the velocity of the particle, ${{v}_{x}}=2.0\times {{10}^{6}}m{{s}^{-1}}$.

Separation between the two plates, $d=0.5cm=0.005m$

Electric field between the two plates, $E=9.1\times {{10}^{2}}N/C$

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

mass of an electron, ${{m}_{e}}=9.1\times {{10}^{-31}}kg$

Let$s$be the deflection when the electron strikes the upper plate at the end of the plate $L$, then, we have the deflection given by,

$s=\frac{qE{{L}^{2}}}{2m{{v}_{x}}}$

$\Rightarrow L=\sqrt{\frac{2dm{{v}_{x}}}{qE}}$

Substituting the given values,

$\Rightarrow L=\sqrt{\frac{2\times 0.005\times 9.1\times {{10}^{-31}}\times {{\left( 2.0\times {{10}^{6}} \right)}^{2}}}{1.6\times {{10}^{-19}}\times 9.1\times {{10}^{2}}}}=\sqrt{0.025\times {{10}^{-2}}}=\sqrt{2.5\times {{10}^{-4}}}$

$\Rightarrow L=1.6\times {{10}^{-2}}=1.6cm$

Therefore, we found that the electron will strike the upper plate after travelling a distance of $1.6cm$.

NCERT Class 12 Physics Chapter 1 PDF

The NCERT Class 12 Physics Chapter 1 PDF deals with a wide range of concepts. The solutions consist of various objective type questions, including numerical problems, theoretical answers, short answer type questions as well as long-type questions. The thorough discussion of each question guides students to grasp all the concepts well. It helps them to ace in competitive exams like JEE as well as in their 12th board exams. For better guidance and assistance, download and refer to the PDF of NCERT Solution for 12 Physics Chapter 1 from Vedantu.

NCERT Solutions for Electric Charges and Fields Class 12: Question-Wise Explanations

NCERT Class 12 Physics Chapter 1 introduces students to one of the most important laws of Physics, Coulomb’s Law. Students will get to learn about insulators, conductors, and other materials in this chapter. Also, NCERT Solutions to chapter 1 broadly discusses the various sub-topics covered in this chapter. The PDF solution comprises a total of 34 questions, and their solutions will definitely guide all students to understand the applications of Coulomb’s law and other concepts.

Question 1: Physics Class 12 Chapter 1

The first question of the exercise helps students to identify the force between two charged spheres suspended at certain distances apart in the air.

Question 2:

The second solution gives students an idea about the electrostatic force on different charged spheres (due to one another) in air. Students will get to learn about the distance between two charged spheres and the force on the second sphere due to first.

Question 3:

The solution to this question provides an explanation of the ratio ke2 / G me mp is dimensionless and its significance.

Question 4:

Here, students get an elaborate explanation of the statement “electric charge of a body is quantized”. Also, students learn that the quantization of electric charges is of no use on a macroscopic scale.

Question 5:

This solution enlightens students about the phenomena of charging by friction.

Question 6:

Question 6 provides students with the answer of charge at the center of a 10 cm square with four charged points.

Question 7:

The solution to this question broadly discusses why electrostatic field line cannot have sudden breaks. Also, students get to know why two electric field lines never intersect at any point.

Question 8:

In this solution, students are provided with an answer to the electric field at the midpoint of two charged points suspended in the vacuum. Also, students learn about the force experienced by test charge at the midpoint.

Question 9:

Question 9 offers students with an explanation of the total charge and electric dipole moment of a system that has two charges.

Question 10:

The solution to this question elaborately discusses the magnitude of torque acting on the dipole.

Apart from these answers, class 12 Physics Chapter 1 NCERT solutions come with an extensive explanation of 24 other solutions. Some of the topics covered in the solutions are as follows.

• Coulomb’s law: applications.

• Calculation of linear charge density.

• Gaussian Surface: applications.

Important Questions on 12th Physics Chapter 1

1. ‘Electric charge of a body is quantised’ - Explain the meaning of this statement.

2. Give the reason behind one being able to ignore the quantisation of an electric charge while dealing with macroscopic or large scale charges?

3. When you rub a glass rod with a silk cloth, charges appear on both these materials. A similar phenomenon is also seen in the case of many other pairs of bodies. Explain the consistency of this observation with the law of conservation of charge.

4. Why is an electrostatic field line always a continuous curve, that is, a field line does not have any sudden breaks?

5. Provide the reason behind two field lines never crossing each other at any point.

6. Determine the net flux of the uniform electric field of Exercise 1.15 through a cube having a side equal to 20 cm, that is oriented in a manner such that its faces are parallel to the coordinate planes.

7. Consider the configuration of an arbitrary electrostatic field. Let a small test charge be placed at the null point, that is, E = 0, in this configuration. Prove that the equilibrium of this test charge is unstable.

8. If two charges of the same magnitude and sign are placed at a certain distance apart, then verify the result of this basic configuration.

Physics Class 12 Chapter 1: Marks Weightage

 Chapter Name Marks Weightage Electric Charges and Fields 15

The chapter carries a 15-point weightage. Students who are having difficulty grasping the issues discussed in this chapter can go to Vedantu's answers for assistance.

Advantages of NCERT Solutions for Class 12 Physics Chapter 1

Students who seek the NCERT solutions for Class 12 Physics Chapter 1 find it helpful for various reasons.

• The solutions are updated according to the latest NCERT guidelines.

• All the solutions to the theoretical and numerical questions are provided in a simple and logical manner.

• Each and every question is explained by highly experienced teachers.

All the solutions are designed to help students to develop a better understanding of the concepts of Electric Charges and Fields. Students who want to excel in this field of Physics can download and find assistance in the NCERT Solutions for Class 12 Physics Chapter 1 for free.

Why Students of Class 12 Opt for Vedantu?

• Experienced and qualified teachers offering expertise in respective subjects

• Interactive online classes with live engagement and instant doubt clearing

• Flexible timings for students with busy schedules or competitive exam preparations

• Reasonable and affordable pricing for accessibility

• Personalized doubt clearing sessions tailored to individual needs

• Customized study material catering to students' understanding and requirements

• Quality online education, expert guidance, and personalized attention for 12th class students

Conclusion

Vedantu's NCERT Solutions for Class 12 Physics Chapter 1 provide a comprehensive and reliable resource for students studying this subject. The solutions are well-structured, covering all the key topics and exercises from the chapter. With Vedantu's solutions, students can enhance their understanding of physics concepts and improve their problem-solving skills. The solutions are designed to be easily accessible and user-friendly, enabling students to grasp complex concepts with ease. Whether it's for exam preparation or concept clarification, Vedantu's NCERT Solutions for Class 12 Physics Chapter 1 serve as a valuable tool for students seeking academic success in physics.

FAQs on NCERT Solutions for Class 12 Physics Chapter 1 - Electric Charges And Fields

1. What are the Benefits of Class 12 Physics Chapter 1 NCERT Solutions?

NCERT Solutions of the chapter of Electric Charges and Fields class 12 come with an elaborate discussion of the topic and subtopics. Moreover, the solutions are explained in a simple manner for a better understanding of the concepts.

Students who want to ace in the CBSE 12th board examination as well as in competitive exams like JEE can seek the guidance of NCERT solutions. The solutions are created in light of the latest CBSE guidelines that help students to secure better marks. Students can completely rely on the solutions provided by Vedantu, as these are all solved by the subject matter expert of Physics.

2. How can I get the NCERT Class 12 Physics Chapter 1 PDF?

The NCERT solutions of Physics class 12 chapter 1 PDF can be downloaded from the Vedantu website and mobile application. Therefore, students who are facing problems in solving the very first chapter of class 12 Physics can seek guidance in the NCERT solutions. The way every solution is described has been considered helpful. Interestingly, students can download this PDF for absolutely free.

3. What are the Subtopics of the Chapter Electric Charges and Fields?

The first chapter of Physics Class 12 Electric Charges and Fields consists of several sub-topics under it, and these are as follows.

1. Introduction

2. Electric Charge

3. Conductors and Insulators

4. Charging by Induction

5. Basic properties of Electric Charge

6. Coulomb’s Law

7. Forces between multiple charges

8. Electric fields

9. Electric field lines

10. Electric flux

11. Electric Dipole

12. Dipole in a uniform external field

13. Continuous charge distribution

14. Gauss’s Law

15. Application of Gauss’s Law

Students who want to get an in-depth knowledge of these topics can consult the class 12 Physics chapter 1 NCERT solutions provided by Vedantu.

4. Are the NCERT Solutions for Class 12 Physics Chapter 1 sufficient for the exam preparation?

Yes, the NCERT Solutions for Class 12 Physics Chapter 1 is sufficient for the exam preparation. The solutions are well detailed and include every important point that will help students to score high. Each numerical problem is explained in a simple manner so that students can clear their basic concepts and learn the approach to solve questions in Physics. They can refer to Vedantu’s revision notes that contain all the important formulas and key points. The solution PDFs and other study materials such as important questions and revision notes can also be downloaded from the Vedantu app as well for free of cost.

5. How do you solve electric charges and fields?

Electric charges and fields are the first chapters of Class 12 Physics. Electric charges and fields are the basics to understand the concepts in further chapters. Solve every problem from your NCERT Physics book. Refer to Vedantu’s Solutions for Class 12 Physics to understand every concept. Every minute detail is included in this material, with an emphasis on strengthening your concept. You can download the Solutions for free from Vedantu’s website.

6. What is the first chapter of Class 12 Physics?

The first chapter of Class 12 Physics is Electric charges and Fields. It introduces the concepts of charges to the student and how it affects other materials. Students will learn about insulators, conductors, their working procedures, etc. Coulomb’s law and its applications are covered in detail. Vedantu’s NCERT Solution PDF for Class 12 Physics Chapter 1 consists of 34 questions to help students gain command over this topic. It is the best guide to understand the concepts.

7. What is electric charge and fields?

Electric Charge is the property of a material due to which it experiences a force in an electric, magnetic, or electromagnetic field. There are two different types of charges, positive charge and negative charge. The region around an electrically charged particle up to which it can exert an electrical force on another electrically charged particle is termed an electric field. It can cause either the force of attraction or repulsion, depending on the nature of the charges.

8. What are some tips to utilise NCERT books effectively for Class 12 exams?

Following are the basic and helpful tips to utilise NCERT books effectively for Class 12 exams:

• Make a timetable and stick to it. Your timetable should be unique to you and should be according to your needs and capabilities.

• Do not skip topics. Cover the entire syllabus. If you get stuck at any point, you should immediately get your doubts cleared. Refer to Vedantu’s Solutions for Class 12 to clear all your doubts and strengthen your concepts.

• Analyse the solutions after you have successfully solved a question. It will help to develop problem-solving capabilities.