Class 12 - Physics - Electrostatic Potential

**Question1.**

Two charges 5 × 10^{–8} C and –3 × 10^{–8} 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:

Case 1:

Let q_{1} =5 × 10^{–8} C

q_{2} = +3x10^{–8} C

Distance between the two charges, d =16cm =0.16m

Distance of point P from charge q_{1} = r

The electric potential (V) at point P will be 0

Therefore, Potential at point P is the sum of potentials caused by charges q_{1} and q_{2} respectively.

∴ V = (1/ 4πε_{0}) x (q_{1}/r) + (1/ 4πε_{0}) x [q_{2}/ (d -r)].......... (1)

Where ε_{0 }= Permittivity of free space

For V=0, equation (1) changes to:

0 = (1/ 4πε_{0}) x (q_{1}/r) + (1/ 4πε_{0}) x [q_{2}/ (d -r)]

=> (1/ 4πε_{0}) x (q_{1}/r) = - (1/ 4πε_{0}) x [q_{2}/ (d -r)]

=> (q_{1}/r) = - (q_{2})/ (d -r)

=> (5 x 10^{-8})/(r) = - (-3 x 10^{-8})/ (0.16-r)

=> 5 (0.16-r) = 3r

=> 0.8 = 8r

=> r =0.1m

=10cm

This shows, the potential is 0 at a distance of 10cm from the positive charge between the charges.

Consider a point P on the outside the system of two charges at a distance of s from the negative charge, where potential is 0.

Then potential is given as:

V = ((1)/ (4πε_{0})) x ((q_{1})/s)) + ((1)/ (4πε_{0})) x (q_{2})/ (s -d)).......... (2)

Where ε_{0 }= Permittivity of free space

For V=0, equation (2) changes to:

0 = ((1)/ (4πε_{0})) x ((q_{1})/s)) + ((1)/ (4πε_{0})) x (q_{2})/ (s -r))

=> (1/ 4πε_{0}) x (q_{1}/s) = - (1/ 4πε_{0}) x [q_{2}/ (s -r)]

=> (q_{1}/s) = - [q_{2}/ (s -r)]

=> (5 x 10^{-8})/(s) = = - (-3 x 10^{-8})/ (s - 0.16)

=> 5(s - 0.16) = 3s

=> 0.8 = 2s

=> s= 0.4m

Or s=40cm

This shows, the potential is 0 at a distance of 40cm from the positive charge between the charges outside the system of charges.

**Question2.**

A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer:

Consider the figure which shows six equal amount of charges at the vertices of the regular hexagon = q

Where,

Charge, q = 5 μC = 5 × 10^{−6} C

Side of the hexagon, l = AB = BC = CD = DE = EF = FA = 10 cm

Distance of each vertex from centre O, d = 10 cm

Electric potential at point O,

V = ((1)/4πε_{0})) x ((6 × q)/ (d))

Where, ε_{0} = Permittivity of free space and (1/4πε_{0})

= (9 × 10^{9}) Nm^{2}C^{-2}

∴ V = (9 × 10^{9} × 6 × 5 × 10^{−6})/ (0.1)

= 2.7 × 10^{6} V

Therefore, the potential at the centre of the hexagon is 2.7 × 10^{6} V.

**Question 3.**

Two charges 2 μC and –2 μC 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:

- An equipotential surface is the plane on which total potential is 0 everywhere. This plane is normal to the line AB.
- The plane is located at the mid -point of line AB because the magnitude of charges is the same.
- The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.

**Question 4.**

A spherical conductor of radius 12 cm has a charge of 1.6 × 10^{–7}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:

Given:

- Radius of the spherical conductor, r = 12 cm = 0.12 m

Charge is uniformly distributed over the conductor, q = 1.6 × 10^{-7} 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.

- Electric field E just outside the conductor is given by the relation,

E = (q)/ (4πε_{0} r^{2})

Where, ε_{0 }= Permittivity of free space

(4πε_{0} r^{2}) = (9 x 10^{9}) Nm^{2} C^{-2}

Therefore, E = ((1.6 × 10^{-7} x 9 x 10^{9})/ (0.12)^{2})

= (10^{5}) Nm^{2} C^{-2}

- Electric field at a point 18 m from the centre of the sphere = E
_{1}

Distance of the point from the centre, d = 18 cm = 0.18 m

E_{1} = (q)/ (4πε_{0} dr^{2})

= ((9 x 10^{9}) x (1.6 x 10^{-7})/ (0.18)^{2})

= (4.4 x 10^{4}) N/C

Therefore, the electric field at a point 18 cm from the centre of the sphere is.4 x 10^{4 }N/C

**Question 5.**

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^{–12} 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?

Answer:

Given:

Capacitor C_{0} =8pF =8 x 10^{-12} F

d_{2} = (d_{1})/ (2)

Using formula, C = (Kε_{0}A)/ (d)

Therefore, C_{k} = (Kε_{0}A)/ (d/2)

= [(2 Kε_{0}A)/ (d)]

By dividing,

(C_{k} / C) = [(2 Kε_{0}A)/ (d)] x (d/ε_{0}A)

=2K

Or C_{K} =2kC_{0}

= (2 x 6 x 8 10^{-12})

=96 x 10^{-12}

C_{K} =96pF

**Question 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:

- Capacitance of each of 3 capacitors, C =9pF

Equivalent capacitance of the 3 capacitors is:-

(1/C’) = (1/C) + (1/C) + (1/C)

= (3/C)

= (3/9)

= (1/3)

=> (1/C’) = (1/3)

=> C’ = 3pF

Therefore, total capacitance of the combination is 3 pF.

- Supply voltage, V = 100 V

Potential difference (V1) across each capacitor is equal to one-third of the supply voltage.

∴ V_{1} = (V/3)

= (120/3)

= 40 V

Therefore, the potential difference across each capacitor is 40 V.

**Question 7.**

Three capacitors of capacitances 2 pF, 3 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:

Given:

Capacitances of the given capacitors: C_{1} = 2 pF, C_{2} = 3 pF and C_{3} = 4 pF

For the parallel combination of the capacitors, equivalent capacitor is given by C_{eq} the algebraic sum,

Therefore C_{eq} = (C_{1} + C_{2} + C_{3}) = (2 + 3 + 4) = 9 pF

Therefore, total capacitance of the combination is 9 pF.

(b) Supply voltage, V = 100 V

The voltage through all the three capacitors is same = V = 100 V

Charge on a capacitor of capacitance C and potential difference V is given by the relation, q = VC

For C = 2 pF, charge = V_{C} = 100 × 2 = 200 pC = 2 × 10^{–10} C

For C = 3 pF, charge = V_{C} = 100 × 3 = 300 pC = 3 × 10^{–10} C

For C = 4 pF, charge = V_{C }= 100 × 4 = 400 pC = 4 × 10^{–10} C

**Question 8.**

In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^{–3} 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:

Given:

Area of each plate A= 6 × 10^{–3} m^{2}

Distance between each plates d= 3mm = 3 x 10^{-3} m

Using, C = (Aε_{0})/ (d)

= (6 × 10^{–3} x 8.85 x 10^{-12})/ (3 x 10^{-3}) = 1.77 x 10^{-11 }F

Also, q = (CV)

= (1.77 x 10^{-11 }F x 100) =1.77 x 10^{-9}C ≈1.8 x 10^{-9}C

**Question 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:

Given:

- Dielectric constant of the mica sheet, k =6

If voltage supply remained connected, voltage between two plates will be constant.

Supply voltage, V = 100V

Initial capacitance, C=1.771 x 10^{-11}F

New capacitance, C_{1} = kC =6 x 1.771 x 10^{-11} F= 106pF

New charge, q_{1} = (C_{1}V) = (106 x100) pC =1.06 x 10^{-8}C

Potential across the plates remains 100V.

- Dielectric constant, k =6

Initial capacitance, C=1.771 x 10^{-11}F

New capacitance, C_{1} = kC = (6 x 1.771 x 10^{-11}) F= 106pF

If supply voltage is removed, then there will be constant amount of charge in the plates which is (1.771 x 10^{-11}) F

Potential across the plates is given by,

V_{1} = (q/C_{1})

= (1.771 x 10^{-11})/ (106 x10^{-12})

=16.7V

**Question 10.**

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

Answer:

Given:

Capacitor of the capacitance C = 12pF = 12 x 10^{-12} F

Potential difference, V =50V

Electrostatic energy stored in the capacitor is given as:

E = (1/2) CV^{2}

= (1/2) x 12 x 10^{-12} x (50)^{2} J

=1.5 x 10^{-8} J

The electrostatic energy stored in the capacitor is 1.5 x 10^{-8} J

**Question 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?

Answer:

Given:

Capacitance of the capacitor, C = 600 pF

Potential difference, V = 200 V

Electrostatic energy stored in the capacitor is given by,

E_{1} = (1/2) CV^{2}

=12× (600 × 10^{−12}) × (200)^{2}J

= 1.2 × 10^{−5}J

If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it,

then equivalent capacitance (C') of the combination is given by,

(1/C’) = (1/C) + (1/C)

= (1/600) + (1/600)

= (2/600)

= (1/300)

Therefore, C’ = 300pF

New electrostatic energy can be calculated as follows:

E_{2} = (1/2) C’V^{2}

= (1/2) × (300 × (200)^{2}) J

= 0.6 × 10^{−5}J

E_{2}=6 × 10^{−6}J

Therefore, the electrostatic energy lost in the process is 6 × 10^{−6} J.

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