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 q1 =5 × 10–8 C

q2 = +3x10–8 C

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

Distance of point P from charge q1 = r

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

Therefore, Potential at point P is the sum of potentials caused by charges q1 and q2 respectively.

Physics Class 12 Electrostatic Potential Potential1

∴ V = (1/ 4πε0) x (q1/r) + (1/ 4πε0) x [q2/ (d -r)].......... (1)

Where ε0 = Permittivity of free space

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

0 = (1/ 4πε0) x (q1/r) + (1/ 4πε0) x [q2/ (d -r)]

=> (1/ 4πε0) x (q1/r) = - (1/ 4πε0) x [q2/ (d -r)]

=> (q1/r) = - (q2)/ (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)/ (4πε0)) x ((q1)/s)) + ((1)/ (4πε0)) x (q2)/ (s -d)).......... (2)

Where ε0 = Permittivity of free space

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

0 = ((1)/ (4πε0)) x ((q1)/s)) + ((1)/ (4πε0)) x (q2)/ (s -r))

=> (1/ 4πε0) x (q1/s) = - (1/ 4πε0) x [q2/ (s -r)]

=> (q1/s) = - [q2/ (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

Physics Class 12 Electrostatic Potential Hexagon

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)/4πε0)) x ((6 × q)/ (d))

Where, ε0 = Permittivity of free space and (1/4πε0)

= (9 × 109) Nm2C-2

∴ V = (9 × 109 × 6 × 5 × 10−6)/ (0.1)

= 2.7 × 106 V

Therefore, the potential at the centre of the hexagon is 2.7 × 106 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.

Physics Class 12 EquPotential Surface

Question 4.

A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C 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)/ (4πε0 r2)

Where, ε0 = Permittivity of free space

(4πε0 r2) = (9 x 109) Nm2 C-2

Therefore, E = ((1.6 × 10-7 x 9 x 109)/ (0.12)2)

= (105) Nm2 C-2

  • Electric field at a point 18 m from the centre of the sphere = E1

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

E1 = (q)/ (4πε0 dr2)

= ((9 x 109) x (1.6 x 10-7)/ (0.18)2)

= (4.4 x 104) N/C

Therefore, the electric field at a point 18 cm from the centre of the sphere is.4 x 104 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 C0 =8pF =8 x 10-12 F

d2 = (d1)/ (2)

Using formula, C = (Kε0A)/ (d)

Therefore, Ck = (Kε0A)/ (d/2)

= [(2 Kε0A)/ (d)]

By dividing,

(Ck / C) = [(2 Kε0A)/ (d)] x (d/ε0A)

=2K

Or CK =2kC0

= (2 x 6 x 8 10-12)

=96 x 10-12

CK =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.

∴ V1 = (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: C1 = 2 pF, C2 = 3 pF and C3 = 4 pF

For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,

Therefore Ceq = (C1 + C2 + C3) = (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 = VC = 100 × 2 = 200 pC = 2 × 10–10 C

For C = 3 pF, charge = VC = 100 × 3 = 300 pC = 3 × 10–10 C

For C = 4 pF, charge = VC = 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 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?

Answer:

Given:

Area of each plate A= 6 × 10–3 m2

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

Using, C = (Aε0)/ (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 10-9C ≈1.8 x 10-9C

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 10-11F

New capacitance, C1 = kC =6 x 1.771 x 10-11 F= 106pF

New charge, q1 = (C1V) = (106 x100) pC =1.06 x 10-8C

Potential across the plates remains 100V.

  • Dielectric constant, k =6

Initial capacitance, C=1.771 x 10-11F

New capacitance, C1 = 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,

V1 = (q/C1)

= (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) CV2

= (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,

E1 = (1/2) CV2

=12× (600 × 10−12) × (200)2J

= 1.2 × 10−5J

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:

E2 = (1/2) C’V2

= (1/2) × (300 × (200)2) J

= 0.6 × 10−5J

E2=6 × 10−6J

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

 

 

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