NCERT Exercise Q & A

NCERT Exercise Problems

Q2.1: Two charges \(5 \times 10^{-8} \mathrm{C}\) and \(-3 \times 10^{-8} \mathrm{C}\) are located 16 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.

Answer: There are two charges,
\(
\begin{aligned}
& q_1=5 \times 10^{-8} C \\
& q_2=-3 \times 10^{-8} C
\end{aligned}
\)
Distance between the two charges, \({d}=16 \mathrm{~cm}=0.16 \mathrm{~m}\)
consider a point \(p\) on the line joining the two charges, as shown in the figure.

\(r=\) Distance of point P from charge \(q_1\)
Let the electric potential \((V)\) at point \(P\) be zero.
Potential at point \(P\) is the sum of potentials caused by charges \(q_1\) and \(q_2\) respectively.
\(
\therefore V=\frac{q_1}{4 \pi \epsilon_0 r}+\frac{q_2}{4 \pi \epsilon_0(d-r)} \dots(i)
\)
Where,
\(\epsilon_0=\) Permittivity of free space
For \(V=0\), equation (i) reduces to
\(
\begin{aligned}
& \frac{q_1}{4 \pi \epsilon_0 r}=-\frac{q_2}{4 \pi \epsilon_0(d-r)} \\
& \frac{q_1}{r}=\frac{-q_2}{d-r} \\
& \frac{5 \times 10^{-8}}{r}=-\frac{\left(-3 \times 10^{-8}\right)}{(0.16-r)} \\
& \frac{0.16}{r}-1=\frac{3}{5} \\
& \frac{0.16}{r}=\frac{8}{5} \\
& \therefore r=0.1 \mathrm{~m}=10 \mathrm{~cm}
\end{aligned}
\)
Therefore, the potential is zero at a distance of 10 cm from the positive charge between the charges.
Suppose point \(P\) is outside the system of two charges at a distance \(s\) from the negative charge, where potential is zero, as shown in the following figure.

For this arrangement, potential is given by,
\(
V=\frac{q_1}{4 \pi \epsilon_0 s}+\frac{q_2}{4 \pi \epsilon_0(s-d)} \dots(ii)
\)
For \(V=0\), equation (ii) reduces to
\(
\begin{aligned}
& \frac{q_1}{4 \pi \epsilon_0 s}=-\frac{q_2}{4 \pi \epsilon_0(s-d)} \\
& \frac{q_1}{s}=\frac{-q_2}{s-d} \\
& \frac{5 \times 10^{-8}}{s}=-\frac{\left(-3 \times 10^{-8}\right)}{(s-0.16)} \\
& 1-\frac{0.16}{s}=\frac{3}{5} \\
& \frac{0.16}{s}=\frac{2}{5} \\
& \therefore s=0.4 \mathrm{~m}=40 \mathrm{~cm}
\end{aligned}
\)
Therefore, the potential is zero at a distance of 40 cm from the positive charge outside the system of charges.

Q2.2: A regular hexagon of side 10 cm has a charge \(5 \mu \mathrm{C}\) at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer: The given figure shows six equal amount of charges, \(q\), at the vertices of a regular hexagon.

Where,
Charge, \(q=5 \mu \mathrm{C}=5 \times 10^{-6} \mathrm{C}\)
Side of the hexagon, \(l=\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DE}=\mathrm{EF}=\mathrm{FA}=10 \mathrm{~cm}\)
Distance of each vertex from centre \(0, d=10 \mathrm{~cm}\)
Electric potential at point O ,
\(
V=\frac{6 \times q}{4 \pi \epsilon_0 d}
\)
Where,
\(
\begin{aligned}
& \epsilon_0=\text { Permittivity of free space } \\
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{C}^{-2} \mathrm{~m}^{-2} \\
& \begin{aligned}
& \therefore V=\frac{6 \times 9 \times 10^9 \times 5 \times 10^{-6}}{0.1} \\
& \quad=2.7 \times 10^6 \mathrm{~V}
\end{aligned}
\end{aligned}
\)
Therefore, the potential at the centre of the hexagon is \(2.7 \times 10^6 \mathrm{~V}\).

Q2.3: Two charges \(2 \mu \mathrm{C}\) and \(-2 \mu \mathrm{C}\) are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?

Answer: (a) The situation is represented in the given figure.

An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.
(b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.

Q2.4: A spherical conductor of radius 12 cm has a charge of \(1.6 \times 10^{-7} \mathrm{C}\) distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?

Answer: (a) Radius of the spherical conductor, \(r=12 \mathrm{~cm}=0.12 \mathrm{~m}\)
Charge is uniformly distributed over the conductor, \(q=1.6 \times 10^{-7} \mathrm{C}\) Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.

(b) Electric field \(E\) just outside the conductor is given by the relation,
\(
E=\frac{q}{4 \pi \epsilon_0 r^2}
\)
Where,
\(
\begin{aligned}
& \epsilon_0=\text { Permittivity of free space } \\
& \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} \\
& \begin{aligned}
& \therefore E=\frac{1.6 \times 10^{-7} \times 9 \times 10^{-9}}{(0.12)^2} \\
& \quad=10^5 \mathrm{~N} \mathrm{C}^{-1}
\end{aligned}
\end{aligned}
\)

(c) Electric field at a point 18 m from the centre of the sphere \(=E_1\)
Distance of the point from the centre, \(d=18 \mathrm{~cm}=0.18 \mathrm{~m}\)
\(
\begin{aligned}
E_1 & =\frac{q}{4 \pi \epsilon_0 d^2} \\
& =\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{\left(18 \times 10^{-2}\right)^2} \\
& =4.4 \times 10^4 \mathrm{~N} / \mathrm{C}
\end{aligned}
\)
Therefore, the electric field at a point 18 cm from the centre of the sphere is
\(
4.4 \times 10^4 \mathrm{~N} / \mathrm{C}
\)

Q2.5: A parallel plate capacitor with air between the plates has a capacitance of \(8 \mathrm{pF}\left(1 \mathrm{pF}=10^{-12} \mathrm{~F}\right)\). 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 ?

Answer: Capacitance between the parallel plates of the capacitor, \(\mathrm{C}=8 \mathrm{pF}\)
Initially, distance between the parallel plates was \(d\) and it was filled with air. Dielectric constant of air, \(k=1\)
Capacitance, \(C\), is given by the formula,
\(
\begin{aligned}
C & =\frac{k \epsilon_0 A}{d} \\
& =\frac{\epsilon_0 A}{d} \dots(i)
\end{aligned}
\)
Where,
\(A=\) Area of each plate
\(\epsilon_0=\) Permittivity of free space
If distance between the plates is reduced to half, then new distance, \(d^{\prime}=\frac{d}{2}\)
Dielectric constant of the substance filled in between the plates, \(k=6\)
Hence, capacitance of the capacitor becomes
\(
C^{\prime}=\frac{k^{\prime} \epsilon_0 A}{d}=\frac{6 \epsilon_0 A}{\frac{d}{2}} \dots(ii)
\)
Taking ratios of equations (i) and (ii), we obtain
\(
\begin{aligned}
C^{\prime} & =2 \times 6 C \\
& =12 C \\
& =12 \times 8=96 \mathrm{pF}
\end{aligned}
\)
Therefore, the capacitance between the plates is 96 pF.

Q2.6: Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Answer: (a) Capacitance of each of the three capacitors, \(C=9 \mathrm{pF}\)
Equivalent capacitance \(\left(C^{\prime}\right)\) of the combination of the capacitors is given by the relation,
\(
\begin{aligned}
\frac{1}{C^{\prime}} & =\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \\
& =\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3} \\
\therefore C^{\prime} & =3 \mu \mathrm{~F}
\end{aligned}
\)
Therefore, total capacitance of the combination is \(3 \mu \mathrm{~F}\).

(b) Supply voltage, \(V=100 \mathrm{~V}\)
Potential difference \((V)\) across each capacitor is equal to one-third of the supply voltage.
\(
\therefore V^{\prime}=\frac{V}{3}=\frac{120}{3}=40 \mathrm{~V}
\)
Therefore, the potential difference across each capacitor is 40 V.

Q2.7: Three capacitors of capacitances \(2 \mathrm{pF}, 3 \mathrm{pF}\) and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Answer: (a) Capacitances of the given capacitors are
\(
\begin{aligned}
& C_1=2 \mathrm{pF} \\
& C_2=3 \mathrm{pF} \\
& C_3=4 \mathrm{pF}
\end{aligned}
\)
For the parallel combination of the capacitors, equivalent capacitor \(C^{\prime}\) is given by the algebraic sum,
\(
C^{\prime}=2+3+4=9 \mathrm{pF}
\)
Therefore, total capacitance of the combination is 9 pF.

(b) Supply voltage, \(V=100 \mathrm{~V}\)
The voltage through all the three capacitors is same \(=V=100 \mathrm{~V}\)
Charge on a capacitor of capacitance \(C\) and potential difference \(V\) is given by the relation,
\(
q=V C \ldots(i)
\)
For \(\mathrm{C}=2 \mathrm{pF}\),
\(
\text { Charge }=V C=100 \times 2=200 \mathrm{pC}=2 \times 10^{-10} \mathrm{C}
\)
For \(\mathrm{C}=3 \mathrm{pF}\),
\(
\text { Charge }=V C=100 \times 3=300 \mathrm{pC}=3 \times 10^{-10} \mathrm{C}
\)
For \(\mathrm{C}=4 \mathrm{pF}\),
\(
\text { Charge }=V C=100 \times 4=200 \mathrm{pC}=4 \times 10^{-10} \mathrm{C}
\)

Q2.8: In a parallel plate capacitor with air between the plates, each plate has an area of \(6 \times 10^{-3} \mathrm{~m}^2\) and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Answer: Area of each plate of the parallel plate capacitor, \(A=6 \times 10^{-3} \mathrm{~m}^2\)
Distance between the plates, \(d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}\)
Supply voltage, \(V=100 \mathrm{~V}\)
Capacitance \(C\) of a parallel plate capacitor is given by,
\(
C=\frac{\in_0 A}{d}
\)
Where,
\(
\begin{aligned}
& \in_0=\text { Permittivity of free space } \\
& =8.854 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \mathrm{C}^{-2} \\
& \begin{aligned}
\therefore C & =\frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} \\
& =17.71 \times 10^{-12} \mathrm{~F} \\
& =17.71 \mathrm{pF}
\end{aligned}
\end{aligned}
\)
Potential \(V\) is related with the charge \(q\) and capacitance \(C\) as
\(
\begin{aligned}
& V=\frac{q}{C} \\
& \begin{aligned}
\therefore q & =V C \\
& =100 \times 17.71 \times 10^{-12} \\
& =1.771 \times 10^{-9} \mathrm{C}
\end{aligned}
\end{aligned}
\)
Therefore, capacitance of the capacitor is 17.71 pF and charge on each plate is \(1.771 \times\) \(10^{-9} \mathrm{C}\)

Q2.9: Explain what would happen if in the capacitor given in Exercise 2.8 , a 3 mm thick mica sheet (of dielectric constant \(=6\) ) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.

Answer: (a) Dielectric constant of the mica sheet, \(k=6\)
Initial capacitance, \(C=1.771 \times 10^{-11} \mathrm{~F}\)
New capacitance, \(C^{\prime}=k C=6 \times 1.771 \times 10^{-11}=106 \mathrm{pF}\)
Supply voltage, \(V=100 \mathrm{~V}\)
New charge, \(q^{\prime}=C^{\prime} V=6 \times 1.771 \times 10^{-9}=1.06 \times 10^{-8} \mathrm{C}\)
Potential across the plates remains 100 V.

(b) Dielectric constant, \(k=6\)
Initial capacitance, \(C=1.771 \times 10^{-11} \mathrm{~F}\)
New capacitance, \(C^{\prime}=k C=6 \times 1.771 \times 10^{-11}=106 \mathrm{pF}\)
If supply voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge \(=1.771 \times 10^{-9} \mathrm{C}\)
Potential across the plates is given by,
\(
\begin{aligned}
\therefore V^{\prime} & =\frac{q}{C^{\prime}} \\
& =\frac{1.771 \times 10^{-9}}{106 \times 10^{-12}} \\
& =16.7 \mathrm{~V}
\end{aligned}
\)

Q2.10: A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?

Answer: Capacitor of the capacitance, \(C=12 \mathrm{pF}=12 \times 10^{-12} \mathrm{~F}\)
Potential difference, \(V=50 \mathrm{~V}\)
Electrostatic energy stored in the capacitor is given by the relation,
\(
\begin{aligned}
E & =\frac{1}{2} C V^2 \\
& =\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2 \\
& =1.5 \times 10^{-8} \mathrm{~J}
\end{aligned}
\)
Therefore, the electrostatic energy stored in the capacitor is \(1.5 \times 10^{-8} \mathrm{~J}\).

Q2.11: A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Answer: Capacitance of the capacitor, \(C=600 \mathrm{pF}\)
Potential difference. \(V=200 \mathrm{~V}\)
Electrostatic energy stored in the capacitor is given by,
\(
\begin{aligned}
E & =\frac{1}{2} C V^2 \\
& =\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^2 \\
& =1.2 \times 10^{-5} \mathrm{~J}
\end{aligned}
\)
If the supply is disconnected from the capacitor and another capacitor of capacitance \(C=\) 600 pF is connected to it, then equivalent capacitance \(\left(C^{\prime}\right)\) of the combination is given by,
\(
\begin{aligned}
& \begin{aligned}
\frac{1}{C^{\prime}} & =\frac{1}{C}+\frac{1}{C} \\
\quad & =\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300} \\
\therefore C^{\prime} & =300 \mathrm{pF}
\end{aligned}
\end{aligned}
\)
New electrostatic energy can be calculated as
\(
\begin{aligned}
E^{\prime} & =\frac{1}{2} \times C^{\prime} \times V^2 \\
& =\frac{1}{2} \times 300 \times(200)^2 \\
& =0.6 \times 10^{-5} \mathrm{~J}
\end{aligned}
\)
\(
\begin{aligned}
\text { Loss in electrostatic energy } & =E-E^{\prime} \\
& =1.2 \times 10^{-5}-0.6 \times 10^{-5} \\
& =0.6 \times 10^{-5} \\
& =6 \times 10^{-6} \mathrm{~J}
\end{aligned}
\)
Therefore, the electrostatic energy lost in the process is \(6 \times 10^{-6} \mathrm{~J}\).

NCERT Exemplar Problems

Very Short Answer (VSA)

Q2.14: Consider two conducting spheres of radii \(R_1\) and \(R_2\) with \(R_1>R_2\). If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Answer: Since, the two spheres are at the same potential, therefore
\(
\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{R_1}=\frac{1}{4 \pi \varepsilon_0} \frac{q_2}{R_2} \Rightarrow \frac{R_1}{\varepsilon_0} \frac{q}{4 \pi R_1^2}=\frac{R_2}{\varepsilon_0} \frac{q_2}{4 \pi R_2^2}
\)
or \(\quad \sigma_1 R_1=\sigma_2 R_2 \Rightarrow \frac{\sigma_1}{\sigma_2}=\frac{R_2}{R_1}\)
As \(R_2>R_1\), this imply that \(\sigma_1>\sigma_2\).
The charge density of the smaller sphere is more than that of the larger one.

Q2.15: Do free electrons travel to region of higher potential or lower potential?

Answer: The force on a charge particle in electric field \(F=q E\)
The free electrons (negative charge) experience electrostatic force in a direction opposite to the direction of electric field.
The direction of electric field is always from higher potential to lower. Hence direction of travel of electrons is from lower potential to region of higher potential.

Q2.16: Can there be a potential difference between two adjacent conductors carrying the same charge?

Answer: Yes, if the sizes are different.
Explanation: We define capacitance of a conductor \(\mathrm{C}=\mathrm{Q} / \mathrm{V}\) is the charge of conductor and \(V\) is the potential of the conductor. For given charge potential \(V \propto 1 / C\). The capacity of conductor depends on its geometry, so two adjacent conductors carrying the same charge of different dimensions may have different potentials.

Q2.17: Can the potential function have a maximum or minimum in free space?

Answer: No, the potential function does not have a maximum or minimum in free space, it is because the absence of atmosphere around conductor prevents the phenomenon of electric discharge or potential leakage.

Q2.18: A test charge \(q\) is made to move in the electric field of a point charge \(Q\) along two different closed paths (Fig. below). First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases?

Answer: Work done will be zero in both the cases.
Explanation: The electrostatic field is conservative, and in this field work done by electric force on the charge in a closed loop is zero. In this question both are closed paths, hence the work done in both the cases will be zero.

Short Answer (SA)

Q2.19: Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Answer: Let us assume that in a closed equipotential surface with no charge the potential is changing from position to position. Let the potential just inside the surface is different to that of the surface causing in a potential gradient ( \(\mathrm{dV} / \mathrm{dr}\) )It means \(\mathrm{E} \neq 0\) electric field comes into existence, which is given by as \(\mathrm{E}=-\mathrm{dV} / \mathrm{dr}\). It means there will be field lines pointing inwards or outwards from the surface. These lines cannot be again on the surface, as the surface is equipotential. It is possible only when the other end of the field lines are originated from the charges inside. This contradicts the original assumption. Hence, the entire volume inside must be equipotential.

Q2.20: A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

Answer: The capacitance of the parallel plate capacitor, filled with dielectric medium of dielectric constant
\(K\) is given by \(C=K \epsilon_0 A / d\)
The capacitance of the parallel plate capacitor decreases with the removal of dielectric medium as for air or vacuum \(K=1\) and for dielectric \(K>1\).
If we disconnect the battery from capacitor, then the charge stored will remain the same due to conservation of charge.
The energy stored in an isolated charge capacitor \(U=q^2 / 2 C\) as \(q\) is constant, energy stored \(U \propto 1 / C\). As \(C\) decreases with the removal of dielectric medium, therefore energy stored increases.
The potential difference across the plates of the capacitor is given by \(V=q / C\)
Since \(q\) is constant and \(C\) decreases which in turn increases \(V\) and therefore \(E\) increases as \(E=V / d\). Important point:

Q2.21: Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Answer: The electric field \(\mathrm{E}=\mathrm{dV} / \mathrm{dr}\) suggests that electric potential decreases along the direction of electric field. Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path. Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Q2.22: Calculate potential energy of a point charge \(-q\) placed along the axis due to a charge \(+Q\) uniformly distributed along a ring of radius \(R\). Sketch P.E. as a function of axial distance \(z\) from the centre of the ring. Looking at graph, can you see what would happen if \(-q\) is displaced slightly from the centre of the ring (along the axis)?

Answer:The potential energy \((U)\) of a point charge q placed at-potential \(V, ~U=q V\) In our case a negative charged particle is placed at the axis of a ring having charge Q . Let the ring has radius a,the electric potential at an axial distance \(z\) from the centre of the ring is
\(
V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\sqrt{z^2+a^2}}
\)
Hence potential energy of a point charge \(-q\) is
\(
U=q V=(-q)\left[\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\sqrt{z^2+a^2}}\right]
\)
\(
\Rightarrow U=-\frac{1}{4 \pi \varepsilon_0} \frac{Q q}{\sqrt{z^2+a^2}}=\frac{1}{4 \pi \varepsilon_0 a} \frac{-Q q}{\sqrt{1+\left(\frac{z}{a}\right)^2}}
\)
\(
\begin{aligned}
& \text { At } z=0, U=-\frac{1}{4 \pi \varepsilon_0} \frac{Q q}{a} \\
& \text { At } z \rightarrow \infty, U \rightarrow 0
\end{aligned}
\)

The variation of potential energy with \(z\) is shown in the figure.
The charge \(-q\) displaced would perform oscillations. Nothing can be concluded just by looking at the graph.

Q2.23: Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius \(R\).

Answer:

Let us take point \(P\) to be at a distance \(z\) from the centre of the ring, as shown in the figure. The charge element \(d q\) is at a distance \(z\) from the point \(P\). Therefore, \(V\) can be written as
\(
V=\frac{1}{4 \pi \varepsilon_0} \int \frac{d q}{r}=\frac{1}{4 \pi \varepsilon_0} \int \frac{d q}{\sqrt{z^2+a^2}}
\)
Since each element \(d q\) is at the same distance from point \(P\), so we have net potential
\(
V=\frac{1}{4 \pi \varepsilon_0} \frac{1}{\sqrt{z^2+a^2}} \int d q=\frac{1}{4 \pi \varepsilon_0} \frac{1}{\sqrt{z^2+a^2}}[Q]
\)
The net electric potential \(V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{\sqrt{z^2+a^2}}\)

Long Answer (LA)

Q2.24: Find the equation of the equipotentials for an infinite cylinder of radius \(r_0\), carrying charge of linear density \(\lambda\).

Answer: We know the integral relation between electric field gives potential difference between two points. Let the field lines must be radically outward. Draw a cylindrical Gaussian surface of radius \(r\) and length \(l\).
\(
V(r)-V\left(r_0\right)=-\int_{r_0}^r \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{r}} \dots(i)
\)
The electric field due to line charge need to be obtained in order to find the potential at distance \(r\) from the line charge. For this we need to apply Gauss’ theorem.
Let the field lines must be radically outward. Draw a cylindrical Gaussian surface of radius \(r\) and length \(l\).

\(
\oint \overrightarrow{\mathbf{E}} \cdot \mathbf{d} \overrightarrow{\mathbf{S}}=\frac{1}{\varepsilon_0}(\lambda l)
\)
\(
\text { or } \quad E_r 2 \pi r l=\frac{1}{\varepsilon_0} \lambda l \Rightarrow E_r=\frac{\lambda}{2 \pi \varepsilon_0 r} \dots(ii)
\)
Hence, if \(r_0\) is the radius of the cylindrical wire, then from (i) and (ii)
\(
V(r)-V\left(r_0\right)=-\int_{r_0}^r \frac{\lambda}{2 \pi \varepsilon_0 r} \cdot d r=-\frac{\lambda}{2 \pi \varepsilon_0} \int_{r_0}^r \frac{d r}{r}
\)
\(
V(r)-V\left(r_0\right)=-\frac{\lambda}{2 \pi \varepsilon_0}[\ln ]_{r_0}^r
\)
\(
\Rightarrow \quad V(r)-V\left(r_0\right)=\frac{\lambda}{2 \pi \varepsilon_0} \ln \frac{r_0}{r}=-\frac{\lambda}{2 \pi \varepsilon_0} \ln \frac{r}{r_0}
\)
\(
\begin{aligned}
&\text { For a given } V \text {, }\\
&\begin{aligned}
\cdot \ln \frac{r}{r_0} & =-\frac{2 \pi \varepsilon_0}{\lambda}\left[V(r)-V\left(r_0\right)\right] \\
\Rightarrow \quad r & =r_0 e^{\left.-\frac{2 \pi \varepsilon_0}{\lambda} V(r)-V\left(r_0\right)\right]}
\end{aligned}
\end{aligned}
\)
The equipotential surfaces are cylinders of radius \(r=r_0 e^{\frac{2 \pi x_0}{\lambda}\left[V(r)-V\left(r_0\right)\right]}\)

Q2.25: Two point charges of magnitude \(+q\) and \(-q\) are placed at \((-d / 2,0,0)\) and \((d / 2,0,0)\), respectively. Find the equation of the equipoential surface where the potential is zero.

Answer: Let the required plane lies at a distance \(x\) from the origin as shown in figure.

The potential at the point \(P\) due to charges is given by
\(
V_P=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_1}+\frac{1}{4 \pi \varepsilon_0} \frac{(-q)}{r_2}
\)
If net electric potential at this point is zero, then
\(
\begin{array}{ll}
& 0=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_1}-\frac{1}{4 \pi \varepsilon_0} \frac{q}{r_2} \Rightarrow \frac{1}{r_1}=\frac{1}{r_2} \text { or } r_1=r_2 \\
& r_1=\sqrt{(x+d / 2)^2+h^2} \text { and } r_2=\sqrt{(x-d / 2)^2+h^2} \\
\text { or } & (x-d / 2)^2+h^2=(x+d / 2)^2+h^2 \\
\Rightarrow & x^2-d x+d^2 / 4=x^2+d x+d^2 / 4 \\
\text { or } & 2 d x=0 \Rightarrow x=0
\end{array}
\)
The equation of the required plane is \(x=0\), i.e., \(y-z\) plane.

Q2.26: A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage \((U)\) as \(\varepsilon=\alpha U\) where \(\alpha=2 \mathrm{~V}^{-1}\).A similar capacitor with no dielectric is charged to \(\mathrm{U}_0=78 \mathrm{~V}\). It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Answer: Let the final voltage be \(U\) : If \(C\) is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is
\(
Q_1=C U
\)
The capacitor with the dielectric has a capacitance \(\varepsilon C\). Hence the charge on the capacitor is
\(
Q_2=\varepsilon U=\alpha C U^2
\)
The initial charge on the capacitor that was charged is
\(
Q_0=\mathrm{CU}_0
\)
From the conservation of charges,
\(
\begin{aligned}
& Q_0=Q_1+Q_2 \\
& \text { Or, } \mathrm{CU}_0=\mathrm{CU}+\alpha \mathrm{CU}^2 \\
& \Rightarrow \alpha U^2+U-u_0=0 \\
& \therefore U=\frac{-1 \pm \sqrt{1+4 \alpha U_0}}{2 \alpha}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& =\frac{-1 \pm \sqrt{1+624}}{4} \\
& =\frac{-1 \pm \sqrt{625}}{4} \text { volts }
\end{aligned}\\
&\text { As } \mathrm{U} \text { is positive }\\
&U=\frac{\sqrt{625}-1}{4}=\frac{24}{4}=6 \mathrm{~V}
\end{aligned}
\)

Q2.27: A capacitor is made of two circular plates of radius \(R\) each, separated by a distance \(d \ll R\). The capacitor is connected to a constant voltage. A thin conducting disc of radius \(r \ll R\) and thickness \(t \ll r\) is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the dise is \(m\).

Answer: When the disc is in touch with the bottom plate, the entire plate is a equipotential. A change \(q^{\prime}\) is transferred to the disc.
The electric fleld on the disc is
\(
=\frac{V}{d}
\)
\(
\therefore q=-\varepsilon_0 \frac{V}{d} \pi r^2
\)
The force acting on the disc is
\(
-\frac{V}{d} \times q^{\prime}=\varepsilon_0 \frac{V^2}{d^2} \pi r^2
\)
If the disc is to be lifted, then
\(
\begin{aligned}
& \varepsilon_0 \frac{V^2}{d^2} \pi r^2=m g \\
& \Rightarrow V=\sqrt{\frac{m g d^2}{\pi \varepsilon_0 r^2}}
\end{aligned}
\)

Q2.28: (a) In a quark model of elementary particles, a neutron is made of one up quarks [charge \((2 / 3) e\) ] and two down quarks [charges \(-(1 / 3) e]\). Assume that they have a triangle configuration with side length of the order of \(10^{-15} \mathrm{~m}\). Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV .
(b) Repeat above exercise for a proton which is made of two up and one down quark.

Answer: This system is made up of three charges. The potential energy of the system is equal to the algebraic sum of the PE of each pair. So,
(a) \(q_d=\frac{1}{3} e\) charge on down quark
\(q_u=+\frac{2}{3} e\) charge on up quark

\(
\begin{aligned}
&\text { Potential energy } \mathrm{U}=\frac{k q_1 q_2}{r}\\
&\begin{aligned}
& k=\frac{1}{4 \pi \varepsilon_0} \\
& \mathrm{U}=\frac{k q_1 q_2}{r}+\frac{k q_1 q_3}{r}+\frac{k q_2 q_3}{r} \\
& \therefore \mathrm{U}_{\mathrm{n}}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(-q_d\right)\left(-q_d\right)}{r}+\frac{\left(-q_d\right) q_u}{4 \pi \varepsilon_0}+\frac{q_u\left(-q_d\right)}{4 \pi \varepsilon_0 r}
\end{aligned}
\end{aligned}
\)
\(
=\frac{q_d}{4 \pi \varepsilon_0 r}\left[q_d-2 q_u\right]
\)
\(
=\frac{9 \times 10^9 \times \frac{1}{3} e}{10^{-15}}\left[\frac{1}{3} e-2 \cdot \frac{2}{3} e\right]
\)
\(
=-7.68 \times 10^{-14} \mathrm{~J}
\)
\(
=-4.8 \times 10^5 \mathrm{eV}=-0.48 \mathrm{MeV}
\)
So, charges inside neutron \(\left[1 \mathrm{q}_{\mathrm{u}}\right.\) and \(\left.2 \mathrm{q}_d\right]\) are attracted by energy of 0.48 MeV .
Energy released by a neutron when converted into energy is 939 MeV .
\(
\therefore \text { Required ratio }=\frac{1-0.481 \mathrm{~MeV}}{939 \mathrm{~MeV}}=0.0005111=5.11 \times 10^{-4}
\)
(b) P.E. of proton consists of 2 up and 1 down quark

\(
\begin{aligned}
& \mathrm{r}=10^{-15} \mathrm{~m} \\
& \mathrm{q}_{\mathrm{d}}=-\frac{1}{3} e, q_u=\frac{2}{3} e \\
& \mathrm{U}_{\mathrm{p}}=\frac{1}{4 \pi \varepsilon_0} \frac{q_u \times q_u}{r}+\frac{q_u\left(-q_d\right)}{4 \pi \varepsilon_0 r}+\frac{q_u\left(-q_d\right)}{4 \pi \varepsilon_0 r} \\
& =\frac{q}{4 \pi \varepsilon_0 r}\left[q_u-q_d-q_d\right] \\
& =\frac{q}{4 \pi \varepsilon_0 r}\left[q_u-2 q_d\right] \\
& =\frac{9 \times 10^9}{10^{-13}} \frac{2}{3} e\left[\frac{2}{3} e-2 \frac{1}{3} e\right]=0
\end{aligned}
\)

Q2.29: Two metal spheres, one of radius \(R\) and the other of radius \(2 R\), both have same surface charge density \(\sigma\). They are brought in contact and separated. What will be new surface charge densities on them?

Answer: The charges on metal spheres before contact, are \(Q_1=\sigma \cdot 4 \pi R^2\)
and \(Q_2=\sigma \cdot 4 \pi(2 R)^2=4\left(\sigma \cdot 4 \pi R^2\right)=4 Q_1\)
Let the charges on the metal spheres, after coming in contact becomes \(Q_1^{\prime}\) and \(Q_2^{\prime}\).
Applying law of conservation of charges,
\(
Q_1^{\prime}+Q_2^{\prime}=Q_1+Q_2=5 Q_1=5\left(\sigma \cdot 4 \pi R^2\right) \dots(i)
\)
When metal spheres come in contact, they acquire equal potentials. Therefore, we have
\(
\frac{1}{4 \pi \varepsilon_0} \frac{Q_1^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{Q_2^{\prime}}{2 R} \Rightarrow Q_1^{\prime}=\frac{Q_2^{\prime}}{2} \dots(ii)
\)
On solving (i) and (ii), we get
\(
\begin{aligned}
& \therefore \quad Q_1^{\prime}=\frac{5}{3}\left(\sigma \cdot 4 \pi R^2\right) \text { and } Q_2^{\prime}=\frac{10}{3}\left(\sigma \cdot 4 \pi R^2\right) \\
& \therefore \quad \sigma_1=\frac{5 \sigma}{3} \quad \text { and } \sigma_2=\frac{5}{6} \sigma
\end{aligned}
\)

Q2.30: In the circuit shown in Fig. 2.7, initially \(\mathrm{K}_1\) is closed and \(\mathrm{K}_2\) is open. What are the charges on each capacitors. Then \(K_1\) was opened and \(K_2\) was closed (order is important), What will be the charge on each capacitor now? \([C=1 \mu \mathrm{~F}]\)

Answer: In the circuit, when initially \(\mathrm{K}_1\) is closed and \(\mathrm{K}_2\) is open, the capacitors \(\mathrm{C}_1\) and \(\mathrm{C}_2\) connected in series with battery acquire equal charge.

In fig, when \(K_1\) is closed and \(K_2\) is open, \(C_3\) is out of contact.
\(
\therefore Q_3=0
\)
\(C_1, C_2\) are in series (Capacitors connected in series must have the same charge) with the battery \(E=9 \mathrm{~V}\),
\(
\begin{aligned}
& \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{6}+\frac{1}{3}=\frac{1+2}{6}=\frac{3}{6}=\frac{1}{2} \\
& C_s=2 \mu F \\
& \therefore Q_1=E \cdot C_s=9 \times 2=18 \mu C, \\
& Q_2=E \cdot C_s=9 \times 2=18 \mu C
\end{aligned}
\)

Later \(K_1\), is opened and \(K_2\) is closed, charge on \(C_1\) remains uncharged, i.e.,
\(
Q_1^{\prime}=Q_1=18 \mu C
\)
The charged capacitor \(\mathrm{C}_2\) now connects in parallel with uncharged capacitor \(\mathrm{C}_3\), considering common potential of parallel combination as \(V^{\prime}\).
\(
\begin{aligned}
&\text { Then, } C_2 V^{\prime}+C_3 V^{\prime}=Q_2\\
&\begin{aligned}
& \Rightarrow \quad V^{\prime}=\frac{Q_2}{C_2+C_3}=\frac{18}{3 C+3 C}=3 \mathrm{~V} \\
& \text { Hence } \quad Q_2^{\prime}=3 C V^{\prime}=9 \mu \mathrm{C} \\
& \text { Also } \quad Q_3^{\prime}=3 C V^{\prime}=9 \mu \mathrm{C} \\
& \text { and } \quad Q_1^{\prime}=18 \mu \mathrm{C}
\end{aligned}
\end{aligned}
\)

Note: Charge on \(C_2\) is now shared between \(C_2\) and \(C_3 \cdot\) As \(C_2=C_3\), therefore,
\(
Q_2^{\prime}=Q_3^{\prime}=\frac{Q_2}{2}=\frac{18}{2}=9 \mu \mathrm{C}
\)

Q2.31: Calculate potential on the axis of a disc of radius \(R\) due to a charge \(Q\) uniformly distributed on its surface.

Answer: 

Let us consider a point \(P\) on the axis of the disc at a distance \(x\) from the centre of the disk and take the plane of the disk to be perpendicular to the \(x\)-axis. Let the disc is divided into a number of charged rings as shown in figure.
The electric potential of each ring, of radius \(r\) and width \(d r\), have charge \(d q\) is given by
\(
d q=\sigma d A=\sigma 2 \pi r d r
\)
and potential is given by
\(
d V=\frac{1}{4 \pi \varepsilon_0} \frac{d q}{\sqrt{r^2+x^2}}=\frac{1}{4 \pi \varepsilon_0} \frac{(\sigma 2 \pi r d r)}{\sqrt{r^2+x^2}}
\)
The total electric potential at \(P\), is given by
\(
V=\frac{\sigma}{4 \varepsilon_0} \int_0^R \frac{2 r d r}{\sqrt{r^2+x^2}}=\frac{\sigma}{4 \varepsilon_0}\left[\frac{\sqrt{r^2+x^2}}{1 / 2}\right]_0^R
\)
\(
\Rightarrow \quad V=\frac{\sigma}{2 \varepsilon_0}\left[\sqrt{R^2+x^2}-x\right]=\frac{Q}{2 \pi \varepsilon_0 R^2}\left[\sqrt{R^2+x^2}-x\right]
\)

Q2.32: Two charges \(q_1\) and \(q_2\) are placed at \((0,0, d)\) and \((0,0,-d)\) respectively. Find locus of points where the potential a zero.

Answer:

Let us take a point on the requited plane as \((x, y, z)\). The two charges lies on \(z\)-axis at a separation of \(2 d\). The potential at the point \(P\) due to two charges is given by
\(
\frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}+\frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}}=0
\)
\(
\therefore \cdot \frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}=\frac{-q_2}{\sqrt{x^2+y^2+(z+d)^2}}
\)
On squaring and simplifying, we get
\(
x^2+y^2+z^2+\left[\frac{\left(q_1 / q_2\right)^2+1}{\left(q_1 / q_2\right)^2-1}\right](2 z d)+d^2=0
\)
The standard equation of sphere is
\(
x^2+y^2+z^2+2 u x+2 u y+2 w z+g=0
\)
with centre \((-u,-v,-w)\) and radius
\(
\sqrt{u^2+v^2+w^2-g}
\)
Hence centre of sphere will be
\(
\left(0,0,-2d\left[\frac{q_1^2+q_2^2}{q_1^2-q_2^2}\right]\right)
\)
And radius is
\(
r=\sqrt{\left(d\left[\frac{q_1^2+q_2^2}{q_1^2-q_2^2}\right]\right)^2-d^2}=\frac{2 q_1 q_2 d}{q_1^2-q_2^2}
\)

Note : if \(q_1=-q_2 \Rightarrow\) Then \(\mathrm{z}=0\), which is a plane through mid-point.

Q2.33: Two charges \(-q\) each are separated by distance \(2 d\). A third charge \(+q\) is kept at mid point \(O\). Find potential energy of \(+q\) as a function of small distance \(x\) from O due to \(-q\) charges. Sketch P.E. v/s \(x\) and convince yourself that the charge at \(O\) is in an unstable equilibrium.

Answer: In fig, two charges \(-q\) each are shown at A and B , where \(A B=2 d\). A charge \(+q\) is kept at mid point \(O\) of \(A B, O P=x\) is small displacement of \(+q\) charge.

Therefore, potential energy of \(+q\) charge at P due to the two charges \(-q\) each is
\(
U=\frac{1}{4 \pi \epsilon_0}\left[-\frac{q^2}{d+x}-\frac{q^2}{d-x}\right]=\frac{q^2}{4 \pi \epsilon_0} \frac{2 d}{d^2-x^2} \dots(i)
\)
So, \(U\) is the P.E. as a function of \(x\) which is shown below.

\(
\frac{\mathrm{dU}}{\mathrm{dx}}=\frac{-q^2 \cdot 2 d}{4 \pi \epsilon_0} \cdot \frac{2 x}{\left(d^2-x^2\right)^2} \dots(ii)
\)
\(
\text { At } x=0 \text {, From (i) } U_0=\frac{-2 q^2}{4 \pi \epsilon_0 d} \text {, and from (ii), } \frac{d U}{d x}=0
\)
\(\therefore x=0\) is an equilibrium point.
Now
\(
\frac{\mathrm{d}^2 \mathrm{U}}{\mathrm{dx}^2}=\left(\frac{-2 d q^2}{4 \pi \epsilon_0}\right)\left[\frac{2}{\left(d^2-x^2\right)^2}-\frac{8 x^2}{\left(d^2-x^2\right)^3}\right]
\)
\(
=\left(\frac{-2 d q^2}{4 \pi \epsilon_0}\right) \frac{1}{\left(d^2-x^2\right)^3}\left[2\left(d^2-x^2\right)^2-8 x^2\right]
\)
At \(x=0\) \(\frac{d^2 \mathrm{U}}{\mathrm{dx}^2}=\left(\frac{-2 d q^2}{4 \pi \epsilon_0}\right)\left(\frac{1}{d^6}\right)\left(2 d^2\right)\), which is \(<0\).
Hence, the equilibrium of charge \(+q\) at \(\mathrm{O}\) is unstable.