Class 12 Physics
CBSE and UP Board
Electromagnetic Waves
Chapter-8 Exercise 8.1
NCERT Solutions for Class 12 Physics Chapter 8 Question 1
Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoffs first rule (junction rule) valid at each plate of the capacitor? Explain.
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Radius of each circular plate, r = 12 cm = 0.12 m
Distance between the plates, d = 5 cm = 0.05 m
Charging current, I = 0.15 A
Permittivity of free space, ε0 = 8.85 x 10⁻12 C2 N⁻1 m-2
Ans (a).
Capacitance between the two plates is given by the relation,
C = ε0 A/d
Where, A = Area of each plate = πr2
C= ε0 πr2/d = [8.85 x 10⁻12 x π x (0.12)2 ]/0.05
= 8.0032 x 10⁻12 F = 80.032 pF
Charge on each plate, q = CV
Where,
V = Potential difference across the plates
Differentiation on both sides with respect to time (t) gives:
dq/dt = C dV/dt
But dq/dt = current (I)
Therefore ,dV/dt = I /C
=> 0.15/(8.0032 x 10⁻12) = 1.87 x 10⁹ V/s
Therefore, the change in potential difference between the plates is 1.87 x 109 V/s.
Ans (b).
The displacement current across the plates is the same as the conduction current.
Hence, the displacement current, id is 0.15 A.
Ans (c).
Yes
Kirchhoffs first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.