# NCERT Solution : Electrostatic Potential & Capacitance

Q:8. In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10–3 m2 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 ?

Sol. A = 6 × 10–3 m2

d = 3 × 10–3 m , V =100 volt

$\displaystyle C_0 = \frac{\epsilon_0 A}{d}$

$\displaystyle C_0 = \frac{8.85\times 10^{-12}\times 6\times 10^{-3}}{3\times 10^{-3}}$

On Solving , Co = 17.7 × 10–12 F

Co = 17.7 pF

and q = CoV

q = 17.7 × 10–12 × 100

q = 1.77 × 10–9 C

Q: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.

Sol: Given, C0 = 17.7 pF, CK = KC0 = 6C0

(a) While the supply remained connected; voltage will remain constant

qK = CK V = 6 C0 V = 6 × 17.7 × 10-10 C = 1.06 × 10-8C

Charge on capacitor becomes six times.

(b) When supply is disconnected; Charge will remain constant

q = CK VK

VK= q/Ck = q/KC0 = 16.7 V

Potential difference between plates becomes one-sixth.

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

Sol. C = 12 × 10–12 F , V = 50 V

Electrostatic energy , $\displaystyle U = \frac{1}{2}CV^2$

$\displaystyle U = \frac{1}{2}(12\times 10^{-12})(50)^2$

U = 1.5 × 10–8 joule

Q:11. A 600pF capacitor is charged by a 200V 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 ?

Sol. C1 = C2 = 600 pF

V1 = 200 V , V2 = 0

$\displaystyle \Delta U = \frac{C_1 C_2}{2(C_1 + C_2)} (V_1 – V_2)^2$

Loss of Energy = C1 C2(V1 -V2)2 /2(C1+C2)

By putting the given values we get ,

Loss of energy ΔU = 6 × 10–6 J

Q: 12. A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 × 10-9 C from a point P(0, 0, 3cm) to a point Q(0, 4cm, 0) via a point R (0, 6cm, 9cm).

Sol:
Let $\displaystyle \vec{r_1} \, and \, \vec{r_2}$ are the position vectors for points P and Q

r1 =3 cm =0.03 m

r2 = 4 cm = 0.04 m

Position of point R does not affect the result.

Also q = 8 mC = 8 × 10-3 C.

$\displaystyle V = \frac{q}{4\pi \epsilon_0 r}$

For Point P , $\displaystyle V_1 = (9\times 10^9 ) \frac{8\times 10^{-3}}{0.03}$

V1 = 24 × 108

For Point Q , $\displaystyle V_2 = (9\times 10^9 ) \frac{8\times 10^{-3}}{0.04}$

V2 = 18 × 108

Potential Difference = ( V2 – V1)

= 18 × 108 – 24 × 108

= -6 × 108 volt

Work done = charged moved × potential difference,

W = (-2 × 10-9) (- 6 × 108)

= 1.2 J

Q:13. 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.

Sol:
Geometrically, distance of centre of the cube from each vertex = √3b/2

Potential at centre due to charge at one vertex $\displaystyle = \frac{q}{4\pi\epsilon_0 (\sqrt 3 b/2)} = \frac{q}{2\sqrt3\pi\epsilon_0 b}$

Potential at centre due to charge at eight vertices,

$\displaystyle V = \frac{8q}{2\sqrt3\pi\epsilon_0 b}= \frac{4q}{\sqrt 3 \pi \epsilon_0 b}$

Electric field = zero.

It is due to symmetry of charge about the point.

Q:14. Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid point.

Sol: (a) Potential at C due to charge at A

$\displaystyle V_1 = \frac{1}{4 \pi \epsilon_0} \frac{1.5 \times 10^{-6}}{0.15}$

= 9 × 109 × 10-5

= 9 × 104 volt

Potential at C due to charge at B

$\displaystyle V_2 = \frac{1}{4 \pi \epsilon_0} \frac{2.5 \times 10^{-6}}{0.15}$

= 15 × 104 volt

Total potential at C is

V = V1 + V2

= 9 × 104 + 15 × 104

= 24 × 104 volt

= 2.4 × 105 volt

Electric field at C due to charge at A

$\displaystyle E_1 = \frac{1}{4 \pi \epsilon_0} \frac{1.5 \times 10^{-6}}{(0.15)^2}$

= 6 × 105 N/C (Along CB)

Electric field at C due to charge at B

$\displaystyle E_2 = \frac{1}{4 \pi \epsilon_0} \frac{2.5 \times 10^{-6}}{(0.15)^2}$

= 10 × 105 N/C (Along CA)

Resultant intensity of electric field at C

E = E2 – E1

= 10 × 105 – 6 × 105

= 4 × 105 (Along CA)

(b) Potential at D due to charge at A

$\displaystyle V_{DA} = \frac{1}{4 \pi \epsilon_0} \frac{1.5 \times 10^{-6}}{0.18}$

= 7.5 × 104 volt

Potential at D due to charge at B

$\displaystyle V_{DB} = \frac{1}{4 \pi \epsilon_0} \frac{2.5 \times 10^{-6}}{0.18}$

= 12.5 × 104 volt

Total potential at D is

V = VDA + VDB

= 7.5 × 104 + 12.5 × 104

= 20 × 104 volt

Q:15. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell ?
(b) Is the electric field inside a cavity with no charge zero, even if the shell is not spherical, but has any irregular shape? Explain.

Sol:
(a) Surface charge density on the inner and outer shell. Taking a Gaussian surface of radius r > r1 but r < r2. Since the Gaussian surface is inside the conductor, therefore electric field is everywhere zero. $\displaystyle \oint \vec{E}.\vec{ds} = \frac{q_{enclosed}}{\epsilon_0}$

$\displaystyle \oint \vec{E}.\vec{ds} = \frac{q + q’}{\epsilon_0}$ ; where q’ be the charge on the inner shell surface.

$\displaystyle \oint \vec{E}.\vec{ds} = 0$ (Since E = 0)

$\displaystyle \frac{q + q’}{\epsilon_0} = 0$

q’ = -q

The conducting shell has no net charge, yet its inner shell has -q surface charge. Because the net charge on the shell is zero and no charge can be internal to the conductor, there must be +q charge on the outer surface of the conductor, other than +Q.

Total charge on outer surface of the shell = Q + q

Surface charge density of inner surface $\displaystyle \sigma_1 = \frac{-q }{4 \pi r_1^2 }$

Surface charge density of outer surface $\displaystyle \sigma_2 = \frac{Q+q }{4 \pi r_2^2 }$

(b) By Gauss’s law, the net charge on the inner surface enclosing the cavity (not having any charge) must be zero. For a cavity of arbitrary shape, this is not enough to claim that the electric field inside must be zero. The cavity may have positive and negative charges with total charge zero. To dispose of this possibility, take a closed loop, part of which is inside the cavity along a field line and the rest gives a net work done by the field in carrying a test charge over a closed loop. We know this is impossible for an electrostatic field. Hence there are no field lines inside the cavity i.e., no field, and no charge on the inner surface of the conductor, whatever be its shape.

Q:16.(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged
surface to another given by $\displaystyle (\vec{E_2}-\vec{E_1}).\hat{n} = \frac{\sigma}{\epsilon_0}$ where $latex \displaystyle \hat{n}$ is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of n ̂ is from side 1 to side 2.)
Hence show that just outside a conductor, the electric field is $\displaystyle \frac{\sigma}{\epsilon_0}\hat{n}$.

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

Sol: The normal component of electric field intensity due to thin infinite plane sheet of charge

$\displaystyle \vec{E_1} = – \frac{\sigma}{2 \epsilon_0}\hat{n}$ (on left side )

$\displaystyle \vec{E_2} = \frac{\sigma}{2 \epsilon_0}\hat{n}$ (on right side )

Discontinuity from one side of a charged surface to another given by

$\displaystyle (\vec{E_2}-\vec{E_1}) = \frac{\sigma}{2 \epsilon_0}\hat{n} + \frac{\sigma}{2 \epsilon_0}\hat{n}$

$\displaystyle (\vec{E_2}-\vec{E_1}) = \frac{\sigma}{\epsilon_0}\hat{n}$

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

Since , inside a closed conductor $\displaystyle \vec{E_1} = 0$

$\displaystyle \vec{E_2} = \frac{\sigma}{\epsilon_0}\hat{n}$

For (b), use the fact that work done by electrostatic field on a closed loop is zero.

Q:17. A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Sol: The charge +q spreads uniformly on the outer surface of A and q uniformly spreads on the inner surface of B. An electric field E ⃗ is produced between the two shells which will be directed radially outwards as shown in figure. Let us consider a coaxial cylindrical Gaussian surface of radius r. The electric flux through this cylindrical Gaussian surface is given by

$\displaystyle \phi_E = \int \vec{E}.\vec{ds} = \int E ds cos0$

$\displaystyle \phi_E = E \int ds = E (2 \pi r l)$

The flux through the end faces of the Gaussian cylinder is zero because E is parallel to them. Hence φE is the flux through whole of the Gaussian surface.

Applying Gauss’s law,

$\displaystyle E(2\pi r l) = \frac{q}{\epsilon_0}$

$\displaystyle E = \frac{q}{\epsilon_0 (2\pi r l)}$

$\displaystyle E = \frac{\lambda}{2 \pi \epsilon_0 r}$

Q:18. In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å.
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation ?

Sol: (a) Given, q1 = +1.6 × 10-19C, q2 = -1.6 × 10-19 C, r = 0.53 Å = 0.53 × 10-10 m

By formula, potential energy,

$\displaystyle U = \frac{1}{4 \pi \epsilon_0}.\frac{q_1 q_2}{r}$

$\displaystyle U = 9 \times 10^9 \times \frac{1.6 \times 10^{-19}\times (-1.6 \times 10^{-19})}{0.53 \times 10^{-10}}$

U = -27.17 eV

(b) Kinetic energy of electron = +27.17/2 = 13.585 eV.

Total energy of electron =-27.17+13.585 =-13.585 eV.

When electron is free, energy becomes zero.

Then Work done = increase in energy of electron =0-(-13.585) = 13.835 eV

(c) Potential energy,

$\displaystyle U = \frac{q_1 q_2}{4 \pi \epsilon_0}(\frac{1}{r_1} – \frac{1}{r_2})$

$\displaystyle U = 9 \times 10^9 \times (1.6 \times 10^{-19}) (-1.6 \times 10^{-19}) [\frac{1}{0.53 \times 10^{-10}}- \frac{1}{1.06 \times 10^{-10}}]$

U = -13.58 eV

Kinetic energy of electron = + 13.585 eV

Q:19. If one of the two electrons of a H2 molecule is removed, we get a hydrogen molecular ion H2+. In the ground state of an H2+ , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Sol:

By formula, potential energy of the system

U = P.E. of first proton and electron system + P.E. of second proton and electron system + P.E. of Applying values, we get

$U = 9 \times 10^9 \times\frac{(1.6\times 10^{-19})(-1.6\times 10^{-19})}{1\times 10^{-10}} + 9 \times 10^9 \times\frac{(1.6\times 10^{-19})(-1.6\times 10^{-19})}{1\times 10^{-10}} + 9 \times 10^9 \times\frac{(1.6\times 10^{-19})(1.6\times 10^{-19})}{1.5\times 10^{-10}}$

U = -30.78 × 10-19 J

U = -19.2 eV (zero of potential energy is taken to be at infinite)

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