NCERT Class 12 Physics — Electrostatic Potential and Capacitance, Chapter 2 Exercises. All 11 questions solved.

The exercises at the end of Chapter 2 fall into three groups. Questions 2.1 to 2.4 are about electrostatic potential — where it is zero, how the potentials of several charges add, and what the field does on an equipotential surface and around a charged conductor. Questions 2.5 to 2.9 are about capacitance: the parallel plate capacitor, the effect of a dielectric, and capacitors joined in series and in parallel. Questions 2.10 and 2.11 are about the energy a capacitor stores. The arithmetic runs on three results:

• Potential of a point charge — $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$, with $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\,\text{C}^{-2}$. Potential is a scalar, so the potentials of several charges simply add, sign included.
• Parallel plate capacitor — $C = \dfrac{K\varepsilon_0 A}{d}$, where $K$ is the dielectric constant of the material filling the gap ($K = 1$ for air) and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$.
• Energy stored in a capacitor — $U = \tfrac12 CV^2 = \dfrac{Q^2}{2C} = \tfrac12 QV$.

Every distance in centimetres or millimetres, and every capacitance in picofarads ($1\ \text{pF} = 10^{-12}\ \text{F}$), is converted to SI units before it goes into a formula.

Key insight. Two habits settle almost every question here. First, potential is a number, not an arrow: add the potentials of the charges with their signs and ignore direction completely (2.1, 2.2, 2.3). Second, with a capacitor, always ask what is being held fixed. While a battery is connected the voltage cannot change, so the charge adjusts; once the capacitor is isolated the charge cannot change, so the voltage adjusts (2.9, 2.11). Most wrong answers in this exercise come from holding the wrong quantity constant.

Question 2.1

Two charges $5\times10^{-8}\ \text{C}$ and $-3\times10^{-8}\ \text{C}$ are located $16\ \text{cm}$ apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution. Put the positive charge at the origin and the negative charge at $x = 16\ \text{cm}$, and look for a point $P$ a distance $x$ from the positive charge where the two potentials cancel. Because potential falls off as $1/r$, the cancelling point must be nearer the smaller charge, $3\times10^{-8}\ \text{C}$, so that its smaller size is made up for by its smaller distance. That leaves two places to look: between the charges, and on the far side of the negative charge.

Between the charges, $P$ is $x$ from the positive charge and $(0.16 – x)$ from the negative one:

$$\frac{1}{4\pi\varepsilon_0}\left[\frac{5\times10^{-8}}{x} – \frac{3\times10^{-8}}{0.16 – x}\right] = 0 \quad\Rightarrow\quad 5(0.16 – x) = 3x \quad\Rightarrow\quad x = 0.10\ \text{m}$$

Beyond the negative charge, $P$ is $x$ from the positive charge and $(x – 0.16)$ from the negative one:

$$\frac{5}{x} = \frac{3}{x – 0.16} \quad\Rightarrow\quad 5x – 0.80 = 3x \quad\Rightarrow\quad x = 0.40\ \text{m}$$

On the other side of the positive charge there is no such point, because every point there is nearer the larger charge, whose positive potential always wins.

Question 2.2

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

Solution. A regular hexagon is made of six equilateral triangles meeting at the centre, so every vertex is exactly one side length, $r = 0.10\ \text{m}$, from the centre. Each charge therefore contributes the same potential, and because potential is a scalar the six contributions just add:

$$V = 6\times\frac{1}{4\pi\varepsilon_0}\frac{q}{r} = 6\times\frac{(9\times10^9)(5\times10^{-6})}{0.10} = 6\times4.5\times10^5 = 2.7\times10^6\ \text{V}$$

The electric field at the centre is zero, since the six equal fields cancel in opposite pairs. The potential is not, because there is nothing to cancel: each term is positive.

Question 2.3

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

Question 2.3 (a)

Solution. Take any point on the plane that passes through the midpoint of $AB$ and is perpendicular to it. Such a point is the same distance $r$ from $A$ as from $B$, so the potentials of the two charges are $+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ and $-\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$, and they cancel. The potential is zero at every point of that plane, which makes it an equipotential surface.

Question 2.3 (b)

Solution. The electric field is always perpendicular to an equipotential surface, because no work is done in moving a charge along the surface, so the field can have no component along it. The field also points from higher to lower potential: the $A$ side, near the positive charge, is at positive potential and the $B$ side is at negative potential. So on the plane the field is along the normal, pointing from $A$ towards $B$. Adding the two fields directly gives the same thing: at any point of the plane the components perpendicular to $AB$ cancel by symmetry, and the components along $AB$ both point towards $B$.

Question 2.4

A spherical conductor of radius $12\ \text{cm}$ has a charge of $1.6\times10^{-7}\ \text{C}$ distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point $18\ \text{cm}$ from the centre of the sphere?

Question 2.4 (a)

Solution. In electrostatics the field inside a conductor is zero: if it were not, the free charges inside would move, and the charges come to rest only when the field inside has vanished. Gauss’s law says the same for this sphere, since a spherical surface drawn inside it encloses no charge.

Question 2.4 (b)

Solution. Outside a uniformly charged sphere the field is the same as if all the charge sat at the centre. Just outside the surface, $r = R = 0.12\ \text{m}$:

$$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{R^2} = \frac{(9\times10^9)(1.6\times10^{-7})}{(0.12)^2} = \frac{1440}{0.0144} = 1.0\times10^5\ \text{N C}^{-1}$$

The charge is positive, so the field points radially outwards.

Question 2.4 (c)

Solution. The same formula applies at $r = 0.18\ \text{m}$, measured from the centre:

$$E = \frac{(9\times10^9)(1.6\times10^{-7})}{(0.18)^2} = \frac{1440}{0.0324} = 4.4\times10^4\ \text{N C}^{-1}$$

Question 2.5

A parallel plate capacitor with air between the plates has a capacitance of $8\ \text{pF}$ ($1\ \text{pF} = 10^{-12}\ \text{F}$). 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 $6$?

Solution. From $C = K\varepsilon_0A/d$, the capacitance is proportional to $K$ and inversely proportional to $d$, while the plate area is unchanged. So there is no need to know $A$ or $d$; the two changes simply scale the original value. Halving $d$ doubles $C$, and filling the gap with a dielectric of $K = 6$ multiplies it by $6$:

$$C’ = 8\ \text{pF}\times2\times6 = 96\ \text{pF}$$

Question 2.6

Three capacitors each of capacitance $9\ \text{pF}$ 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 $120\ \text{V}$ supply?

Question 2.6 (a)

Solution. In series the reciprocals add, because each capacitor carries the same charge and the voltages across them add:

$$\frac{1}{C} = \frac19 + \frac19 + \frac19 = \frac13 \quad\Rightarrow\quad C = 3\ \text{pF}$$

Question 2.6 (b)

Solution. The charge drawn from the supply is $Q = CV = (3\times10^{-12})(120) = 3.6\times10^{-10}\ \text{C}$, and in a series chain every capacitor carries this same charge. Each therefore has a potential difference

$$V_1 = \frac{Q}{C_1} = \frac{3.6\times10^{-10}}{9\times10^{-12}} = 40\ \text{V}$$

This is just the $120\ \text{V}$ shared equally, as it must be for three identical capacitors.

Question 2.7

Three capacitors of capacitances $2\ \text{pF}$, $3\ \text{pF}$ and $4\ \text{pF}$ 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 $100\ \text{V}$ supply.

Question 2.7 (a)

Solution. In parallel every capacitor has the same voltage and the charges add, so the capacitances add directly:

$$C = 2 + 3 + 4 = 9\ \text{pF}$$

Question 2.7 (b)

Solution. Each capacitor is connected straight across the supply, so each has $V = 100\ \text{V}$, and $Q = CV$ for each one separately:

$$Q_1 = (2\times10^{-12})(100) = 2\times10^{-10}\ \text{C},\quad Q_2 = 3\times10^{-10}\ \text{C},\quad Q_3 = 4\times10^{-10}\ \text{C}$$

Question 2.8

In a parallel plate capacitor with air between the plates, each plate has an area of $6\times10^{-3}\ \text{m}^2$ and the distance between the plates is $3\ \text{mm}$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100\ \text{V}$ supply, what is the charge on each plate of the capacitor?

Solution. With air between the plates $K = 1$, and $d = 3\ \text{mm} = 3\times10^{-3}\ \text{m}$:

$$C = \frac{\varepsilon_0 A}{d} = \frac{(8.85\times10^{-12})(6\times10^{-3})}{3\times10^{-3}} = 1.77\times10^{-11}\ \text{F} = 17.7\ \text{pF}$$

The charge follows from $Q = CV$. One plate carries $+Q$ and the other $-Q$:

$$Q = (1.77\times10^{-11})(100) = 1.77\times10^{-9}\ \text{C}$$

To two significant figures these are $18\ \text{pF}$ and $1.8\times10^{-9}\ \text{C}$, which is how the book’s answer key prints them.

Question 2.9

Explain what would happen if in the capacitor given in Exercise 2.8, a $3\ \text{mm}$ 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.

Solution. The sheet is $3\ \text{mm}$ thick, the whole gap, so it fills the space between the plates and the capacitance becomes $K$ times its air value in both cases:

$$C’ = 6\times17.7\ \text{pF} = 106\ \text{pF}$$

What happens next depends on what is held fixed.

Question 2.9 (a)

Solution. While the supply is connected it holds the potential difference at $100\ \text{V}$. The capacitance has gone up six times, so the capacitor must now hold six times the charge, and the supply pushes the extra charge on:

$$Q’ = C’V = (1.06\times10^{-10})(100) = 1.06\times10^{-8}\ \text{C}$$

Question 2.9 (b)

Solution. Once the supply is disconnected, the charge on the plates has nowhere to go, so it stays at $1.77\times10^{-9}\ \text{C}$. The capacitance is still six times larger, so the voltage must fall to a sixth. Physically, the mica is polarised and the charges induced on its faces partly cancel the field between the plates:

$$V’ = \frac{Q}{C’} = \frac{1.77\times10^{-9}}{1.06\times10^{-10}} = \frac{100}{6} = 16.7\ \text{V}$$

Note on the book’s answer. The key gives $C = 108\ \text{pF}$ and $Q = 1.08\times10^{-8}\ \text{C}$ in (a) because it multiplies the rounded $18\ \text{pF}$ of Question 2.8 by $6$; unrounded, it is $106\ \text{pF}$ and $1.06\times10^{-8}\ \text{C}$. Its $16.6\ \text{V}$ in (b) is $100/6 = 16.67\ \text{V}$ cut short rather than rounded. These are rounding differences, not errors.

Question 2.10

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

Solution. Capacitance and voltage are both given, so $U = \tfrac12CV^2$ is the form to use:

$$U = \tfrac12(12\times10^{-12})(50)^2 = \tfrac12(12\times10^{-12})(2500) = 1.5\times10^{-8}\ \text{J}$$

Question 2.11

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

Solution. Before. The first capacitor holds

$$Q = CV = (600\times10^{-12})(200) = 1.2\times10^{-7}\ \text{C},\qquad U_1 = \tfrac12CV^2 = \tfrac12(600\times10^{-12})(200)^2 = 1.2\times10^{-5}\ \text{J}$$

After. Once it is disconnected from the supply the charge is fixed, and connecting the second capacitor only lets that charge spread over both. The two are now in parallel, with total capacitance $1200\ \text{pF}$ and a common voltage

$$V’ = \frac{Q}{C_1 + C_2} = \frac{1.2\times10^{-7}}{1200\times10^{-12}} = 100\ \text{V}$$

so the stored energy is

$$U_2 = \tfrac12(1200\times10^{-12})(100)^2 = 6\times10^{-6}\ \text{J}$$

Loss. $U_1 – U_2 = 1.2\times10^{-5} – 6\times10^{-6} = 6\times10^{-6}\ \text{J}$. Exactly half the energy has gone. Charge is conserved but energy is not: while the charge rushes from one capacitor to the other, the current heats the connecting wires and some energy is radiated.

Common mistakes

• Question 2.1: looking for the zero only between the charges. Potential falls off as $1/r$ on both sides of the negative charge, so there is a second zero beyond it, at $40\ \text{cm}$ from the positive charge. Students who stop at the first root miss it; others search on the wrong side, next to the larger charge, where no zero can exist.
• Question 2.2: adding the potentials as vectors and getting zero. The field at the centre of the hexagon is zero, and it is tempting to carry that over. Potential is a scalar with no direction, so six equal positive potentials add to $2.7\times10^6\ \text{V}$.
• Question 2.3(b): saying the field is zero on the plane because the potential is. Zero potential does not mean zero field. The field depends on how the potential changes from place to place, and across the plane it changes fast; the field is along $AB$, strongest at the midpoint.
• Question 2.4(a): using $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ inside the sphere. The charge on a conductor sits on its surface and the field inside is zero. It is the potential inside that is not zero — it is constant, equal to its value at the surface.
• Question 2.6(b): putting $120\ \text{V}$ across each capacitor. That is the rule for parallel connections. In series the charge is common and the voltages add up to the supply voltage, so each gets $40\ \text{V}$.
• Question 2.8: leaving $d$ in millimetres. $3\ \text{mm}$ must go in as $3\times10^{-3}\ \text{m}$; left as $3$, it makes the capacitance a thousand times too small.
• Question 2.9: giving the same answer for (a) and (b). The mica multiplies $C$ by $6$ in both cases, but with the supply connected $V$ stays at $100\ \text{V}$ and $Q$ rises sixfold, while with it disconnected $Q$ stays put and $V$ falls to a sixth.
• Question 2.11: assuming no energy is lost. Charge is conserved when the capacitors are joined, so it is natural to assume energy is as well. It is not: halving the voltage while doubling the capacitance halves $\tfrac12CV^2$, and the missing half is dissipated as heat in the wires.

Practise next

• Chapter 3, Current Electricity — potential difference now drives a steady current through a conductor instead of charging a capacitor.
• Chapter 1, Electric Charges and Fields — the fields and Gauss’s law results this chapter turns into potentials and energies.