NCERT Exemplar Solutions Class 12 Physics Electrostatic Potential and Capacitance — Very Short, Short and Long Answer Questions Q14–33

NCERT Exemplar Problems, Class 12 Physics — Electrostatic Potential and Capacitance, Very Short Answer Questions 14 to 18, Short Answer Questions 19 to 23 and Long Answer Questions 24 to 33. All 20 questions solved. The book numbers them 2.14 to 2.33.

Most of this page is built from three ideas: potential is the work per unit charge and so is a scalar that adds, a conductor in equilibrium is an equipotential with no field inside, and potential energy tells you about stability, since a minimum means oscillation and a maximum means the charge runs away. The capacitor questions add the rule that charge is conserved when capacitors are joined.

Key insight. Write the potential or potential energy first, then read off everything else from it. Fields come from the slope of $V$, work from the difference of $V$, and stability from the curvature of $U$. In Questions 22 and 33 the sign of $U”$ at the centre settles the answer at once.

Very Short Answer Questions

Question 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.

Solution. The potential of a charged sphere is $V = \dfrac{Q}{4\pi\varepsilon_0 R}$. With $Q = \sigma\,4\pi R^2$ this becomes $V = \dfrac{\sigma R}{\varepsilon_0}$. The potentials are equal, so $\sigma R$ is the same for both spheres and $\sigma \propto 1/R$. The smaller sphere therefore has the greater surface charge density. This is why charge crowds on sharp points.

Question 15

Do free electrons travel to region of higher potential or lower potential?

Solution. The electric field points from high potential to low potential, and the force on a positive charge is along the field. An electron is negative, so the force on it is opposite to the field. It is therefore pushed towards the region of higher potential, which lowers its potential energy $-eV$.

Question 16

Can there be a potential difference between two adjacent conductors carrying the same charge?

Solution. Yes. The potential of a conductor depends on its charge and on its capacitance, which is fixed by its size, shape and surroundings: $V = Q/C$. Two conductors with equal $Q$ but different $C$ (a large and a small sphere, say, or conductors of different shapes) have different potentials.

Question 17

Can the potential function have a maximum or minimum in free space?

Solution. No. In a region free of charge the potential obeys Laplace’s equation $\nabla^2 V = 0$, and a function satisfying it cannot have a local maximum or minimum inside the region; its extreme values occur only on the boundary. Physically, a maximum would need field lines pointing away from that point in every direction, and field lines can only begin on charges.

Question 18

A test charge $q$ is made to move in the electric field of a point charge $Q$ along two different closed paths (Fig. 2.6). 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?

Solution. The figure shows the field lines of $Q$ spreading out radially. The first loop is made of two radial segments and two circular arcs between them, enclosing an angle $\theta$. The second loop is a rectangle of sides $a$ and $b$ in the same field.

The electrostatic field of a point charge is conservative: the work done is $q\,[V(\text{final}) – V(\text{initial})]$, which depends only on the end points. Along the arcs the displacement is perpendicular to $\vec E$, so no work is done there, and the work on the two radial parts cancels. For any closed path the end points coincide, so the work is zero. The shape and area of the loop do not matter.

Short Answer Questions

Question 19

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

Solution. Let the closed surface be at potential $V_0$ and suppose, for contradiction, that the potential inside is not constant. Then there is a gradient inside, and so an electric field. Field lines must begin and end on charges, or run off to infinity. Inside the surface there is no charge, so a line cannot start or end there. It cannot leave through the surface either, because the surface is an equipotential, and field lines cross an equipotential at right angles, never along it. A field line also cannot form a closed loop, because the field is conservative. So no field line can exist inside, which contradicts our assumption. Hence $\vec E = 0$ inside and the potential equals $V_0$ throughout the enclosed volume.

(Equivalently, Laplace’s equation inside, with the constant value $V_0$ on the boundary, has only the constant solution.)

Question 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.

Solution. The battery is disconnected first, so the isolated plates keep their charge: $Q$ is constant.

• Capacitance $C = K\varepsilon_0 A/d$ falls, because $K$ drops to $1$.
• Charge $Q$ stays the same.
• Voltage $V = Q/C$ rises, since $Q$ is fixed and $C$ has fallen.
• Field $E = V/d$ rises, for the same $d$.
• Energy $U = \dfrac{Q^2}{2C}$ rises as $C$ falls. (The extra energy is the work done in pulling the dielectric out against the attraction of the plates.)

Question 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.

Solution. Take the charged conductor to be positive at potential $V_1 > 0$; the argument is the same for a negative one. Potential falls steadily along a field line, so it falls from $V_1$ down to $0$ at infinity.

The uncharged conductor is an equipotential at some potential $V_2$. Its net charge is zero, so it has induced charge of both signs, and lines of the charged body’s field end on its negative part and start again from its positive part. Lines that end on it come from the charged body, and along any line the potential falls, so $V_2 < V_1$. Lines that start from it must end either on a negative charge or at infinity, and there is no other conductor, so they run to infinity; along them the potential falls from $V_2$ to $0$, so $V_2 > 0$. Hence $0 < V_2 < V_1$.

Question 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)?

Solution. Every element of the ring is at the same distance $\sqrt{R^2+z^2}$ from the point on the axis, so the potential there is simply the total charge over that distance (the result of Question 23 below). The potential energy of $-q$ is

$$U(z) = -q\,V(z) = -\frac{qQ}{4\pi\varepsilon_0\sqrt{R^2+z^2}}$$

Sketch. At $z = 0$, $U = -\dfrac{qQ}{4\pi\varepsilon_0 R}$, the most negative value. As $|z|$ increases $U$ rises symmetrically on both sides and tends to zero from below as $z \to \pm\infty$. The curve is an inverted bell, symmetric about $z = 0$, with its lowest point at the centre.

Stability. The centre is a minimum of $U$, so it is a stable equilibrium. If $-q$ is displaced slightly along the axis, the force $-\dfrac{dU}{dz}$ points back towards the centre, and the charge oscillates about the centre of the ring (simple harmonic motion for small displacements).

Question 23

Calculate potential on the axis of a ring due to charge $Q$ uniformly distributed along the ring of radius $R$.

Solution. Divide the ring into small elements $dq$. A point on the axis at distance $z$ from the centre is at distance $r = \sqrt{R^2+z^2}$ from every element. Potential is a scalar, so the contributions add without any direction bookkeeping:

$$V = \int \frac{dq}{4\pi\varepsilon_0 r} = \frac{1}{4\pi\varepsilon_0\sqrt{R^2+z^2}}\int dq = \frac{Q}{4\pi\varepsilon_0\sqrt{R^2+z^2}}$$

Long Answer Questions

Question 24

Find the equation of the equipotentials for an infinite cylinder of radius $r_0$, carrying charge of linear density $\lambda$.

Solution. By Gauss’s law with a coaxial cylinder of radius $r > r_0$ and length $l$, $E\,(2\pi r l) = \lambda l/\varepsilon_0$, so

$$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

directed radially outward. The potential difference between radii $r_0$ and $r$ is

$$V(r) – V(r_0) = -\int_{r_0}^{r}\frac{\lambda}{2\pi\varepsilon_0 r’}\,dr’ = \frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r_0}{r}$$

Solving for $r$:

$$r = r_0\,e^{-2\pi\varepsilon_0\,[V(r)-V(r_0)]/\lambda}$$

For a fixed value of $V$ the radius $r$ is fixed, so the equipotentials are coaxial cylinders around the charged cylinder.

Question 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 equipotential surface where the potential is zero.

Solution. For a point $(x,y,z)$ the distances to the two charges are $\sqrt{(x+d/2)^2+y^2+z^2}$ and $\sqrt{(x-d/2)^2+y^2+z^2}$. The potential is zero where

$$\frac{q}{4\pi\varepsilon_0}\left[\frac{1}{\sqrt{(x+d/2)^2+y^2+z^2}} – \frac{1}{\sqrt{(x-d/2)^2+y^2+z^2}}\right] = 0$$

The two distances must be equal, so $(x+d/2)^2 = (x-d/2)^2$, which gives $2xd = 0$, that is $x = 0$. This is the perpendicular bisector plane of the two charges. (The printed key’s figure labels the left charge $2q$, a typo; the question says $+q$.)

Question 26

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage ($U$) as $\varepsilon = \alpha U$ where $\alpha = 2\ \text{V}^{-1}$. A similar capacitor with no dielectric is charged to $U_0 = 78$ V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Solution. Let $C$ be the capacitance of the empty capacitor. With the dielectric, the capacitance at voltage $U$ is $\varepsilon C = \alpha U C$, so the charge on it is $\alpha C U^2$. The empty capacitor starts with charge $CU_0$. When they are joined in parallel they reach a common voltage $U$, and charge is conserved:

$$CU_0 = CU + \alpha C U^2 \;\Rightarrow\; 78 = U + 2U^2 \;\Rightarrow\; 2U^2 + U – 78 = 0$$

$$U = \frac{-1 \pm \sqrt{1 + 624}}{4} = \frac{-1 \pm 25}{4}$$

The positive root is $U = 6$ V (the other root is negative and has no meaning).

Question 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 disc is $m$.

Solution. The disc rests on the bottom plate, so it is part of that plate and at the same potential. The field between the plates is $E = V/d$. Just above the disc the surface charge density is $\sigma = \varepsilon_0 E = \varepsilon_0 V/d$, so the charge on the disc is

$$q’ = \sigma\,\pi r^2 = \frac{\varepsilon_0 V\pi r^2}{d}$$

What field acts on this charge? Not the full field $V/d$. The charge sits on the surface of a conductor, where the field is $\sigma/\varepsilon_0$ just outside and $0$ inside. The field the charge itself produces is half of each, so the other charges produce the average of the two, $\dfrac{\sigma}{2\varepsilon_0}$. This gives the familiar result that a surface charge feels $\dfrac{\sigma^2}{2\varepsilon_0}$ per unit area. The upward force on the disc is

$$F = q’\cdot\frac{E}{2} = \frac{\varepsilon_0 V^2\pi r^2}{2d^2}$$

The disc just lifts when $F = mg$:

$$\frac{\varepsilon_0 V^2\pi r^2}{2d^2} = mg \;\Rightarrow\; V_{\min} = \sqrt{\frac{2mgd^2}{\pi\varepsilon_0 r^2}} = \frac{d}{r}\sqrt{\frac{2mg}{\pi\varepsilon_0}}$$

Printed key error. The key gives $\sqrt{mgd^2/(\pi\varepsilon_0 r^2)}$, which is smaller by a factor $\sqrt2$. It multiplies the charge $q’$ by the full field $V/d$, which counts the disc’s own field acting on itself. A charge does not exert a net force on itself, so only the field of the rest of the system, half of $V/d$, should be used.

Question 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 $10^{-15}$ 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.

Solution. The three quarks sit at the corners of an equilateral triangle of side $r = 10^{-15}$ m. The potential energy is the sum over the three pairs, each $\dfrac{q_iq_j}{4\pi\varepsilon_0 r}$.

(a) Neutron (up, down, down). In units of $\dfrac{e^2}{4\pi\varepsilon_0 r}$:

$$\left(\tfrac23\right)\left(-\tfrac13\right)\times 2 + \left(-\tfrac13\right)\left(-\tfrac13\right) = -\tfrac49 + \tfrac19 = -\tfrac13$$

$$U = -\frac13\cdot\frac{e^2}{4\pi\varepsilon_0 r} = -\frac13\cdot\frac{9\times10^9\,(1.6\times10^{-19})^2}{10^{-15}} \approx -7.68\times10^{-14}\ \text{J}$$

Dividing by $1.6\times10^{-13}$ J per MeV, $U \approx -0.48$ MeV. Compared with $939$ MeV the ratio is about $5.1\times10^{-4}$, so the electrostatic energy is a tiny fraction (about $0.05\%$) of the rest energy.

(b) Proton (up, up, down). In the same units:

$$\left(\tfrac23\right)\left(\tfrac23\right) + \left(\tfrac23\right)\left(-\tfrac13\right)\times 2 = \tfrac49 – \tfrac49 = 0$$

so the electrostatic potential energy of the proton in this model is zero. (The printed key does not give part (b).)

Question 29

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

Solution. The initial charges are $Q_1 = 4\pi R^2\sigma$ and $Q_2 = 4\pi(2R)^2\sigma = 16\pi R^2\sigma$, a total of $20\pi\sigma R^2$ which is conserved. In contact the two spheres become one conductor and reach the same potential, so after separation

$$\frac{Q_1′}{R} = \frac{Q_2′}{2R} \;\Rightarrow\; Q_2′ = 2Q_1’$$

Then $3Q_1′ = 20\pi\sigma R^2$, so $Q_1′ = \dfrac{20}{3}\pi\sigma R^2$ and $Q_2′ = \dfrac{40}{3}\pi\sigma R^2$. The new densities are

$$\sigma_1 = \frac{Q_1′}{4\pi R^2} = \frac53\sigma, \qquad \sigma_2 = \frac{Q_2′}{4\pi(2R)^2} = \frac56\sigma$$

The smaller sphere ends up with the larger density, as in Question 14.

Question 30

In the circuit shown in Fig. 2.7, initially $K_1$ is closed and $K_2$ 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? [$C = 1\ \mu$F]

Solution. The figure shows a $9$ V battery in series with $C_1 = 6C$ and, through $K_1$, $C_2 = 3C$ in a parallel branch; $K_2$ connects $C_3 = 3C$ in parallel with $C_2$.

Initially ($K_1$ closed, $K_2$ open) $C_1$ and $C_2$ are in series across $9$ V. Series capacitors carry the same charge:

$$C_{eq} = \frac{6\times3}{6+3} = 2\ \mu\text{F}, \qquad Q = 2\times9 = 18\ \mu\text{C}$$

So $Q_1 = Q_2 = 18\ \mu$C, with $V_1 = 3$ V and $V_2 = 6$ V (check: $3\times6 = 18$). $C_3$ is not connected, so $Q_3 = 0$.

Finally $K_1$ is opened first, which isolates $C_1$ with its $18\ \mu$C. Then $K_2$ is closed, joining $C_2$ (holding $18\ \mu$C) to the uncharged $C_3$. The $18\ \mu$C is shared in parallel, with a common voltage

$$V = \frac{18}{3+3} = 3\ \text{V}, \qquad Q_2′ = Q_3′ = 3\times3 = 9\ \mu\text{C}$$

Printed key error. The key finds $V_1 = 3$ V and $V_2 = 6$ V correctly but then writes $Q_2 = 9\ \mu$C initially and $4.5\ \mu$C each finally. Since $C_1$ and $C_2$ are in series they must carry equal charge, and $Q_2 = C_2V_2 = 3\times6 = 18\ \mu$C, which then splits into $9\ \mu$C and $9\ \mu$C.

Question 31

Calculate potential on the axis of a disc of radius $R$ due to a charge $Q$ uniformly distributed on its surface.

Solution. The surface charge density is $\sigma = \dfrac{Q}{\pi R^2}$. Treat the disc as a set of thin rings. A ring of radius $s$ and width $ds$ carries $dq = \sigma\,2\pi s\,ds$, and by Question 23 it contributes at a point a distance $z$ along the axis

$$dV = \frac{\sigma\,2\pi s\,ds}{4\pi\varepsilon_0\sqrt{s^2+z^2}} = \frac{\sigma}{2\varepsilon_0}\frac{s\,ds}{\sqrt{s^2+z^2}}$$

Integrating from $s = 0$ to $R$, with $\displaystyle\int\frac{s\,ds}{\sqrt{s^2+z^2}} = \sqrt{s^2+z^2}$:

$$V(z) = \frac{\sigma}{2\varepsilon_0}\left[\sqrt{R^2+z^2} – |z|\right] = \frac{Q}{2\pi\varepsilon_0R^2}\left[\sqrt{R^2+z^2} – z\right]\quad (z>0)$$

For large $z$, the bracket tends to $\dfrac{R^2}{2z}$, so $V \to \dfrac{Q}{4\pi\varepsilon_0 z}$, the point-charge result, as it should.

Question 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 is zero.

Solution. The potential at $(x,y,z)$ is zero when

$$\frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}} + \frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}} = 0$$

This can only happen if $q_1$ and $q_2$ have opposite signs. Moving the second term across and squaring:

$$q_1^2\left[x^2+y^2+(z+d)^2\right] = q_2^2\left[x^2+y^2+(z-d)^2\right]$$

$$(q_1^2-q_2^2)(x^2+y^2+z^2+d^2) + 2zd\,(q_1^2+q_2^2) = 0$$

• If $|q_1| \ne |q_2|$: dividing through,

$$x^2+y^2+z^2 + 2zd\,\frac{q_1^2+q_2^2}{q_1^2-q_2^2} + d^2 = 0$$

This is a sphere with centre $\left(0,\,0,\,-d\,\dfrac{q_1^2+q_2^2}{q_1^2-q_2^2}\right)$ and radius $\dfrac{2d\,|q_1q_2|}{|q_1^2-q_2^2|}$ (completing the square gives radius$^2 = d^2\dfrac{(q_1^2+q_2^2)^2-(q_1^2-q_2^2)^2}{(q_1^2-q_2^2)^2}$). – If $q_1 = -q_2$: the term $q_1^2-q_2^2$ vanishes and the condition reduces to $z = 0$, the plane midway between the charges.

(The printed key’s centre carries a stray factor $2$ in front of $d$; the centre is at $-d\,\dfrac{q_1^2+q_2^2}{q_1^2-q_2^2}$, as derived above.)

Question 33

Two charges $-q$ each separated by distance $2d$. 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.

Solution. Displace $+q$ by $x$ along the line joining the fixed charges. Its distances from the two $-q$ charges are $d-x$ and $d+x$, so

$$U(x) = -\frac{q^2}{4\pi\varepsilon_0}\left(\frac{1}{d-x}+\frac{1}{d+x}\right) = -\frac{2q^2d}{4\pi\varepsilon_0\,(d^2-x^2)}$$

At $x = 0$, $U(0) = -\dfrac{2q^2}{4\pi\varepsilon_0 d}$. As $|x|$ grows the denominator $d^2-x^2$ shrinks, so $U$ becomes more negative, falling without limit as $x \to \pm d$.

Sketch. An inverted U: symmetric about $x=0$, with its highest point at $x = 0$ and the curve dropping on both sides.

Stability. $U'(0) = 0$, so $O$ is an equilibrium. For small $x$, $U \approx U(0) – \dfrac{2q^2}{4\pi\varepsilon_0 d^3}x^2$, so $U”(0) < 0$ and $O$ is a maximum of the potential energy. A small push moves the charge down the slope, towards one of the negative charges, and the force $-dU/dx$ then points away from O. The equilibrium is unstable.

(The sketch label in the printed key drops the minus sign on $U(0)$.)

Common mistakes

• Using the full field $V/d$ on a surface charge, in Question 27. A charge on a conductor’s surface feels only the average of the field outside and inside, which is $\sigma/2\varepsilon_0$. Using the full field overcounts by $2$ and makes $V_{\min}$ too small by $\sqrt2$; this is how the printed key goes wrong.
• Treating series capacitors as sharing charge, in Question 30. Capacitors in series carry the same charge; it is the voltages that divide. Only after $K_2$ closes, with $C_2$ and $C_3$ in parallel, does charge divide, and then in proportion to capacitance.
• Reasoning with fields instead of $Q$, in Question 20. The battery is disconnected before the dielectric comes out, so $Q$ is fixed and $V = Q/C$ rises. If the battery stayed connected it would be $V$ that is fixed.
• Missing the sign of the potential energy in Questions 22 and 33. Both involve opposite charges, so $U$ is negative with the extreme value at the centre. In Question 22 that is a minimum (stable); in Question 33 the same shape is flipped and it is a maximum (unstable).
• Forgetting that equal potential means equal $\sigma R$, in Questions 14 and 29. Smaller radius means larger surface charge density at the same potential.
• Adding potentials as vectors in Questions 23 and 31. Potential is a scalar; every ring or element at the same distance contributes the same way, with no cancellation.
• Dropping the $|q_1|\ne|q_2|$ condition in Question 32. The locus is a sphere only when the magnitudes differ; for $q_1 = -q_2$ the quadratic term vanishes and a plane is left.

Practise next

• Exemplar Multiple Choice Questions — the thirteen questions on potential, equipotentials and capacitors, including the printed-key error in Question 11 and the misprint in Question 6.
• NCERT textbook Chapter 2 — Electrostatic Potential and Capacitance for the end-of-chapter problems on potential, capacitors and dielectrics.
• Exemplar Electric Charges and Fields — the previous unit, whose fields and Gauss’s law are the starting point for everything here.
• Exemplar Current Electricity — the next unit, which puts these potentials to work in circuits.