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NCERT Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance
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Electrostatic Potential and Capacitance
Chapter: 2
| Part – I |
EXERCISE
1. 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.
Ans:
Let the potential be zero at point P Where OP = 𝓍.
For 𝓍 < 0 (i.e., to the left of O), the potentials from the two charges cannot cancel each other out. Therefore, 𝓍 must be positive. If 𝓍 lies between O and A, then:
Another possibility is that 𝓍 could lie on the extension of OA, as shown in the figure below.
Thus, the electric potential is zero at points 10 cm and 40 cm away from the positive charge, on the side of the negative charge.
2. A regular hexagon of side 10 cm has a charge 5 C at each of its vertices. Calculate the potential at the centre of the hexagon.
Ans: V = Kq/ r
Given:
k = 8.99 × 109 Nm2C-2
r = α = 0.1 m
Vtotal = 6 × kq/r
= 6 × 449.5 109
= 2697 × 109 V
= 2.697 × 1012 V.
3. Two charges 2 C and –2 C are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
Ans: An equipotential surface is the plane on which total potential is zero. It is at the midpoint of the line AB and perpendicular to it. This is because the electric potential due to a point charge is inversely proportional to the distance from the charge.
(b) What is the direction of the electric field at every point on this surface?
Ans: The direction of the electric field at every point on this surface is normal to the plane in direction of AB. This is because the electric field lines are always perpendicular to equipotential surfaces.
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.
Ans: The electric field inside a conductor is zero. Electric fields inside a spherical conductor having charge distributed only.
(b) Just outside the sphere.
Ans: E = kQ / r2
= 9 × 109 Nm2 / C2 × (1.6× 10-7 C) / (0.12 )2
= 1.0 × 105 N/C.
(c) At a point 18 cm from the centre of the sphere?
Ans: At a point 18 cm from the centre of the sphere:
r = 18 cm = 0.18 m.
E = kQ / r2
= 9 × 109 Nm2 / C2 × (1.6× 10-7 C) / (0.18 )2
= 4.4 × 104 N/C.
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?
Ans: Given:
C0 = Σ0A / d
Distance between plates d1 = d/2
C = 2 × 6 × 8 = 96 F.
6. Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
Ans: Given:
C = 9pF = 9 × 10-12 F.
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Ans: Vi = Q / Ci
V = V1 + V2 + V3
Given that V1 = V2 = V3 and using the fact that Ci = C
Vi = V/3
Vi = 120V / 3
= 40 V.
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?
Ans: Total capacitance of the combination:
C = C1 + C2 + C3
= 2 pF + 3 pF + 4 pF
= 9 pF.
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Ans: Q = C × V
For C1 = 2 pF
Q1 = C1 × V = 2 pF × 100 V = 200 pC.
For C2 = 3 pF
Q2 = C2 × V = 3 pF × 100 V = 300 pC.
For C3 = 4 pF
Q3 = C3 × V = 4 pF × 100 V = 400 pC.
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?
Ans: C = Σ₀ A/d
Σ₀ is the permittivity of free space (8.85 x 1012 F/m)
A = 6 × 10-3 m2
d = 3 / 1000 m = 3 × 10-3 m
Now,
C = (8.85 × 10-12 F/m) × (6 × 10-3 m2) / (3 × 10-3 m)
= 1.77 × 10-11 F.
Q = CV
Q = (1.77 × 10-11 F) × (100 V)
= 1.77 × 10-9 C.
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.
Ans: Effect on Capacitance:
C1 = K × C0 = 6 × 17.7 pF
= 106.2 pF.
Q1 = C1 × V
= 106.2 × 10-12 F × 100 V.
= 10.62 × 10-9 C.
= 10.62 nC.
When the voltage supply remains connected, the potential difference between capacitor plates remains same.i.e, 100 V.
(b) After the supply was disconnected.
Ans: Effect on Capacitance:
C1 = 106.2 F due to the dielectric.
Q = 1.77 nC.
Effect on Voltage:
V1 = Q / C1
V1 = 1.77 × 10-9 C / 106.2 × 10-12 F
= 16.7 V.
10. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Ans: U = ½ CV2
Given:
C = 12 pF or 12 × 10-12 F
V = 50 V
U = ½ CV2
U = ½ × 12 × 10-12 F × (50 V)2
= ½ × 12 × 10-12 F × 2500 V2
= 1.5 × 10-8 J.
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?
Ans: Initially:
E1 = (1/2)CV2 = (1/2) × 600 × 10-12 × 2002 = 12 × 10-6 J
Ctotal = C1 + C2 = 600 + 600 = 1200 pF
Vnew = 200 / 2 = 100 V
The final energy is:
E2 = (1/2) Ctotal Vnew2
= (1/2) × 1200 × 10-12 × 1002 = 6 × 10-6 J
Energy lost = E1 – E2
= 12 × 10-6 – 6 × 10-6 = 6 × 10-6J.
