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

Q. Two charges 2 µC and – 2 µC are placed at points A and B 6 cm apart.

(i) Identify an equipotential surface of the system.

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

Ans. The situation is represented in the given figure.

(i) 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. (ii) The direction of the electric field at every point on this surface is normal to the plane  in the direction of AB.

Q. Describe schematically the equipotential surfaces corresponding to :

(i) a constant electric field in the z-direction,

(ii) a field that uniformly increases in magnitude but remains in a constant. (say z) direction.

(iii) a single positive charge at the origin and

(iv) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Ans. (i) Equidistant planes parallel to the X-Y plane are the equipotential surfaces.

(ii) Planes parallel to the X-Y plane are the equipotential surfaces with the exception that when the planes get closer, the field increases.

(iii) Concentric spheres centered at the origin are equipotential surfaces.

(iv) A periodically varying shape near the given grid is the equipotential surface. This shape gradually reaches the shape of planes parallel to the grid at a larger distance.

Q. Fig. (a) and (b) show the field lines of a positive and negative point charge respectively. Ans. (i) As we know,

$$\text{v}∝\frac{1}{2}$$

VP > VQ

[As P is closer w.r.t. Q to positive charge]

VA < VB

i.e., VA is more negative w.r.t. VB as A is more closer to negative charge w.r.t. B.

Hence, VP – VQ = positive

and VB – VA = positive

(ii) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Thus, sign of potential energy difference of a small negative charge between Q and P is positive.

Similar to this potential energy at A > potential energy at B, thus difference is positive.

(iii) In moving a small positive charge from Q to P. Work has to be done by an external agency against the electric field. Thus, work done by electric field is negative.

(iv) In moving a small negative charge from B to A work has to be done by the external agency. Hence, it is positive.

(v) Due to force of repulsion on negative charge, velocity decreases. Hence, the kinetic energy decreases in going from B to A.

Q. Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of The system of charges form an electric quadrupole.

It can be considered that the system of the electric quadrupole has three charges.

Charge + q placed at point X
Charge – 2q placed at point Y
Charge + q placed at point Z
XY = YZ = a
YP = r

PX = r + a
PZ = r – a

Electrostatic potential caused by the system of three charges at point P is given by,

$$\text{V}=\frac{1}{3\pi\epsilon_0}\bigg[\frac{q}{XP}-\frac{2q}{XP}+\frac{q}{ZP}\bigg]\\=\frac{1}{4\pi\epsilon_0}\bigg[\frac{q}{r+a}-\frac{2q}{r}+\frac{q}{r-a}\bigg]\\=\frac{q}{4\pi\epsilon_0}\bigg[\frac{r(r-a)-2(r+a)(r-a)+r(r+a)}{r(r+a)(r-a)}\bigg]\\=\frac{q}{4\pi\epsilon_0}\bigg[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r(r^2-a)}\bigg]\\=\frac{q}{4\pi\epsilon_0}\bigg[\frac{2a^2}{r(r^2-a^2)}\bigg]\\=\frac{2qa^2}{4\pi\epsilon_0r^3\bigg(1-\frac{a^2}{r^2}\bigg)}\\\text{Since}\space\frac{r}{a}\text{\textgreater}\text{\textgreater}1,\\\qquad\therefore\frac{a}{r}\text{\textless}\text{\textless}1\\\qquad\frac{a^2}{r^2}\text{is taken as negligible.}\\\therefore \text{V}=\frac{2qa^2}{4\pi\epsilon_0r^3}\\\text{It can be inferred that potential,}\\\text{v}∝\frac{1}{r^3}\\\text{However, it is known that for a dipole,}\\\text{V}∝\frac{1}{r^3}\\\text{And, for a monopole,}\\\text{V}∝\frac{1}{5}$$

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

Ans. Length of the side of a cube = b Charge at each of its vertices = q
A cube of side b is shown in the following figure.
d = Diagonal of one of the six faces of the cube $$d^2 =\sqrt{b^2+b^2}\\=\sqrt{2b^2}\\\text{d}=b\sqrt{2}\\\text{l = Length of the diagonal of thecube}\\l=\sqrt{d^2+b^2}\\=\sqrt{(\sqrt{2}b)^2+b^2}\\\sqrt{2b^2+b^2}=\sqrt{3b^2}\\\text{l}=b\sqrt{3}\\r=\frac{l}{2}=\frac{b\sqrt{3}}{2}$$

This is the distance between the centre of the cube and one of the eight vertices. The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.

$$\text{V}=\frac{8q}{4\pi\epsilon_0r}\\=\frac{8q}{4\pi\epsilon_0 ×\frac{b\sqrt{3}}{2}}\\=\frac{4q}{\sqrt{3}\pi\epsilon_0b}$$

Therefore, the potential at the centre of the cube is

$$\space\frac{4q}{\sqrt{3}\pi\epsilon_0b}$$

The electric field at the centre of the cube, due to the eight charges, gets cancelled. This is because the charge are distributed symmetrically with respect to the centre of the cube. Hence, the electric field is zero at the centre.

Q. Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$$\text(E_2-E_1).\hat{n}=\frac{\sigma}{\epsilon_0}$$

$$\text{where}\space \hat{n}\space\text{is a unit vector normal to the surface at a point and}$$

$$\sigma\text{(The direction of}\space \hat{n}\text{is from side 1 to side 2).}$$

Hence show that just outside a conductor, the electric field is

$$\sigma\space\frac{\hat{n}}{\epsilon_0}.$$

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

[Hint : For (i) Use Gauss’ law. For (ii) use the fact that work done by electrostatic field on a closed loop is zero.]

Ans. (i) Electric field on one side of a charged body is E1 and electric field on the other side of the same body is E2. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$$\vec{E}_1=-\frac{\sigma}{2\epsilon_0}\hat{n}\space\space…(i)\\\text{where}\\\hat{n}=\text{Unit vector normal to the surface at a point}\\\sigma = \text{Surface charge density at that point Electric field due to the other surface of the charged body,}\\\vec{E}_2=\frac{\sigma}{2\epsilon_0}\hat{n}\space\space…(ii)\\\text{Electric field at any point due to the two surfaces,}\\\vec{E_2}-\vec{E_1}=\frac{\sigma}{2\epsilon_0}\hat{n}-\bigg(\frac{-\sigma}{2\epsilon_0}\hat{n}\bigg)=\frac{\sigma}{\epsilon_0}\hat{n}\\(\vec{E_2}-\vec{E_1})\hat{n}=\frac{\sigma}{\epsilon_0}\space…(iii)\\\text{Since inside a closed conductor,}\\\vec{E_1}=0\\\therefore\vec{E}=\vec{E}_2=-\frac{\sigma}{2\epsilon_0}\hat{n} \text{Therefore, the electric field just outside the conductor is}\space\frac{\sigma}{\epsilon_0}\hat{n}.$$

(ii) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of
electrostatic field is continuous from one side of charged surface to the other.

Q. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2.

Ans. Consider F be the force applied to separate the plates of a parallel plate capacitor by a distance of Δx. Hence, work done by the force to do so

= F Δ x

(W) = Force × Displacement

As a result, the potential energy of the capacitor increases by an amount given as uAΔx.
Where, u = Energy density A = Area of each plate
d = Distance between the plates
V = Potential difference across the platesThe work done will be equal to the increase in the potential energy i.e.,
FΔx = uAΔx

$$\text{F}=u\text{A}=\bigg(\frac{1}{2}\epsilon_0\space E^{2}\bigg)\text{A}\\\text{Electric field intensity is given by,}\\\text{E}=\frac{V}{d}\\\qquad\text{F}=\frac{1}{2}\epsilon_0\bigg(\frac{V}{d}\bigg)\text{EA}\\=\frac{1}{2}\bigg(\epsilon_0\text{A}\frac{V}{d}\bigg)E\\\text{However, capacitance,}\\\text{C}=\frac{\epsilon_0A}{d}\\\therefore\text{F}=\frac{1}{2}\text{QE}\\\text{The physical origin of the factor,}\space\frac{1}{2}\text{in the force formula lies in the fact that just outside the conductor, field is E and inside it is zero.}\\$$

$$\text{Thus, it is the average value, E/2, of the field that contributes to the force.}$$

Two tiny spheres carrying charges 1.5μC and 2.5μC are located 30 cm apart. Find the potential and electric field:

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

(ii) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid-point.

Two charges placed at points A and B are shown in the given figure. O is the mid-point of the line joining the two charges. Magnitude of charge located at A,
q1 = 1.5 μC Magnitude of charge located at B,
q2 = 2.5 μC Distance between the two charges,
d = 30 cm= 0.3 m
(i) Let V1 and E1 are the electric potential and electric field respectively at O.
V1 = Potential due to charge at A + Potential due to charge at B

$$\text{V}_1=\frac{q_1}{4\pi\epsilon_0\bigg(\frac{d}{2}\bigg)}+\frac{q_1}{4\pi\epsilon_0\bigg(\frac{d}{2}\bigg)}\\=\frac{1}{4\pi\epsilon_0\bigg(\frac{d}{2}\bigg)}[q_1+q_2]\\\text{Where},\epsilon_0=\text{Permittivity of free space}\\\frac{1}{4\pi\epsilon_0}=9×10^{9}\text{NC}^{2}m^{-2}\\V_1=\frac{9×10^{9}×10^{-6}}{\bigg(\frac{0.30}{2}\bigg)}[2.5+1.5]\\=2.4×10^{5}\text{V}\\\text{E}_1=\text{Electric field due to q}_2 – \text{Electric field due to q}_1$$

=\frac{q_2}{4\pi\epsilon_0\bigg(\frac{d}{2}\bigg)}-\frac{q_1}{4\pi\epsilon_0\bigg(\frac{d}{2}\bigg)^{2}}\\=\frac{9×10^{9}×10^{-6}}{\bigg(\frac{0.30}{2}\bigg)^{2}}[2.5-1.5]\\=4×10^{5}\space\text{Vm}^{-1}\\\text{Therefore, the potential at mid-point is} 2.4 × 10^5 \text{V and the electric field at mid-point is}\space4 × 10^{5} \text{Vm}^{\normalsize– 1}.\\\text{The field is directed from the larger charge to the smaller charge.}\\\text{(ii) Consider a point Z such that normal distance OZ = 10 cm = 0.1 m, as shown in the following figure.} \text{Let},\text{V}_2\space\text{and}\space\text{E}_2\text{are the electric potential and electric field respectively at Z.}\\\text{It can be observed from one figure that distance,}\\BZ=AZ\\=\sqrt{(0.10)^{2}(0.15)^{2}}\\=0.18\space\text{m}\\\text{V}_2=\text{Electric potential due to A + Electric Potential due to B}\\=\frac{q_1}{4\pi\epsilon_0(\text{AZ})}+\frac{q_2}{4\pi\epsilon_0(\text{BZ})}\\=\frac{9×10^{9}×10^{-6}}{0.18}[1.5+2.5]\\=2×10^{5}\text{V}\\\text{Electric field due to}\space q_1\text{at}\space Z,\\\text{E}_B=\frac{q_2}{4\pi\epsilon_0(\text{BZ})^{2}}\\

$$=\frac{9×10^{9}×2.5×10^{-6}}{(0.18)^{2}}\\=0.69×10^{6}\text{V\space m}^{-1}\\\text{The resultant field intensity at Z,}\\\text{E}=\sqrt{\text{E}^{2}_{A}+\text{E}^{2}_{B}+2\text{E}_\text{A}\text{E}_{B}\text{cos}\space2\theta}\\\text{cos}\space\theta=\frac{0.10}{0.18}=\frac{5}{9}=0.5556\\\theta=\text{cos}^{-1}(0.5556)=56.25\degree\\\therefore\space2\theta=112.5\degree\\\text{cos}\space2\theta=-0.38\\\text{E}=\sqrt{(0.416×10^{6})^{2}+(0.69×10^{6})^{2}+2×0.416×0.69×10^{12}×(-0.38)}\\=6.6×10^{5}\text{V}\space m^{-1}\\\text{Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is}\space\\2.0 × 10^{5} \text{V and electric field is 6.6 × 105 V m– 1.}$$

Q. Obtain the equivalent capacitance of the network in Fig. For a 300 V supply, determine the charge and voltage across each capacitor. Ans. Capacitance of capacitor C1 is 100 pF.
Capacitance of capacitor C2 is 200 pF.
Capacitance of capacitor C3 is 200 pF.
Capacitance of capacitor C4 is 100 pF.
Supply potential, V = 300 V
Capacitors C2 and C3 are connected in series. Let their equivalent capacitance be C′

$$\therefore\space\frac{1}{C^{‘}}=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}\\\text{C}^{‘}=100\text{pF}\\\text{Capacitors C}_1 \text{and C}^{′} \text{are in parallel. Let their equivalent capacitance be C}^{′′}.\\ \therefore\space\text{C}^{”}=\text{C}^{‘}_1+\text{C}_1\\= 100 + 100 = 200 \text{pF}\\ \text{C}^{”}\text{and}\space\text{C}_4\text{are connected in series. Let their equivalent capacitance be C.}\\ \therefore\space\frac{1}{C^{‘}}=\frac{1}{C^{”}}+\frac{1}{C_4}=\frac{1}{200}+\frac{1}{100}\\=\frac{2+1}{200}\\ \text{C}=\frac{200}{3}\text{pF}\\ \text{Hence, the equivalent capacitance of the circuit is}\space\frac{200}{3}\text{pF.}$$

$$\text{Potential difference across C′′ = V′′}\\\text{Potential difference across C}_4 = \text{V}_4\\\therefore\space\text{V}^{”}+\text{V}_4=V=300V\\\text{Charge on C}_4\text{ is given by},\\Q_4=CV\\\text{(As C}_4\space\text{and}\space\text{C}^{”}\text{are in series then they have same charge = Q}_4)\\=\frac{200}{3}×10^{-12}×300\\= 2 × 10^{\normalsize– 8}\text{C}\\\therefore V_4=\frac{Q_4}{C_4}=\frac{2×10^{-8}}{100×10^{-12}}=200V\\\therefore\text{Voltage across C}_1\space\text{is given by,}\\V_1=V-V_4=300-200=100V\\\text{Hence, potential difference, V}_1\text{across C}_1\text{is 100 V.}\\\text{Charge on C}_1 \text{is given by,}$$

$$\text{Q}_1=\text{C}_1\text{V}_1\\=100×10^{-12}×100=10^{-8}\text{C}\\\text{V}_2=\text{V}_3=50V\\\text{Therefore, charge on C}_2 \space\text{is given by,}\\\text{Q}_2={C}_2{V}_2\\= 200 × 10^{\normalsize– 12} × 50 = 10^{\normalsize– 8} C\\\text{Hence, the equivalent capacitance of the given circuit is}\space\frac{200}{3}\space{\text{pF}}\space\text{with}.Q_1 = 10^{– 8} C, V_1 = 100 V\\Q_2 = 10^{– 8} C, V_2 = 50 V\\Q_3 = 10^{– 8} C, V_3 = 50 V\\Q_4 = 2 × 10^{\normalsize– 8} C, V_4 = 200 V$$

Q. A 4 µ F capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µ F capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

Ans. Capacitance of a charge capacitor,
C1 = 4µF = 4 × 10– 6 F
Supply voltage, V1 = 200 V
Electrostatic energy stored in capacitor is given by,

$$\text{E}=\frac{1}{2}\text{CV}^{2}\\\text{Thus, energy stored in C}_1,\\\text{E}_1=\frac{1}{2}×\text{CV}^{2}\\\text{E}_1=\frac{1}{2}×4×10^{-6}×(200)^{2}\\=8×10^{-2}\text{J}$$ When C2 is connected to the circuit, the potential
acquired by it is V2.

As per conservation of charge, initial charge on capacitor C1 is equal to the final charge on capacitors, C1 and C2.

$$\therefore\space V_2(\text{C}_1+\text{C}_2)=\text{C}_1\text{V}_1\\\text{V}_2×(4+2)×10^{-6}=4×10^{-6}×200\\V_2=\frac{400}{3}V\\\text{Electrostatic energy for the combination of two capacitors is given by,}\\\text{E}_2=\frac{1}{2}(C_1+C_2)=C_1V_1\\V_2×(4+2)×10^{-6}=4×10^{-6}×200\\V_2=\frac{400}{3}V\\\text{Electrostatic energy for the combination of two capacitors is given by,}\\\text{E}_2=\frac{1}{2}(C_1+C_2)V_2^{2}\\=\frac{1}{2}(2+4)×10^{-6}×\bigg(\frac{400}{3}\bigg)^{2}$$

$$\approx 5.33 × 10^{\normalsize– 2} J\\\text{Hence, amount of electrostatic energy lost by capacitor C}_1,\\=E_1-E_2\\=0.08 – 0.0533 = 0.0267\\=2.67 × 10^{\normalsize– 2} J$$

Q. (i) A comb run through one’s dry hair attracts small bits of paper. Why ? What happens if the hair is wet or if it is a rainy day ? (Remember, a paper does not conduct electricity)
(ii) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary?
(iii) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why ?

Ans. (i) This is because the comb gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction. If the hair is wet, or if it is rainy day, friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper.

(ii) To enable them to conduct charge (produced by friction) to the ground as too much of static electricity accumulated may result in spark and result in fire.

(iii) It is for conduction of electrostatic charge (developed by friction) to ground so that spark will not occur.

Q. An electrical technician requires a capacitance of 2 mF in a circuit across a potential difference of 1 kV. A large number of 1 mF capacitors are available to him each of which can withstand a potential difference of not more than 400 V.
Suggest a possible arrangement that requires the minimum number of capacitors.

Ans. (i) The electric field at P due to charge + 10 μC is given by

C = 2 µF

Potential difference,

V = 1 kV = 1000 V

Capacitance of each capacitor,

C1 = 1 µF

Each capacitor can withstand a potential difference,

V1 = 400 V

Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V.

Hence, number of capacitors in each row is given as

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

$$\text{Hence, there are three capacitors in each row.}\\\text{Capacitance of each row =}\frac{1}{1+1+1}=\frac{1}{3}\mu F \text{Hence, there are three capacitors in each row.}\\\text{Capacitance of each row =}\frac{1}{1+1+1}=\frac{1}{3}\mu F\\\text{Let there are m rows, each having three capacitors, which are connected in parallel.}\\ \text{Hence, equivalent capacitance of the circuit is given as}\\\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+............\text{m\space terms}\\=\frac{m}{3}\\\text{However, capacitance of the circuit is given as 2} \mu F.\\\therefore\space\frac{m}{3}=2\\\text{m}=6\\\text{Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e.,} \\\text{18 capacitors are required for the given arrangement.}$$