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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 33
Total: 37
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 V m–1. For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Solution:  
V = 1 × 103 V , K = 3 , C = 50 × 10−12 F
Given E = 10% of dielectric strength
or E =
10
100
 × 4107 V m−1 = 106Vm−1
As V = E.d , so d =
V
E
 =
13
106
 = 10−3 m
and C =
kε0A
d
or A =
C.d
kε0
 =
50×10−12×10−3
3×8.85×10−12
 or A = 1.9 × 10−3m2 = 19 cm2
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