Class 12 Physics, CBSE and UP Board, Electrostatic Potential and Capacitance, Additional Exercise,Chapter-2 Exercise 2.33,NCERT Solutions for Class 12 Physics
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 Vm⁻. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) 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?
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Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material, εr =3
Dielectric strength = 107 V/m
For safety, the field intensity never exceeds 10% of the dielectric strength.
Hence, electric field intensity, E = 10% of 107 = 106 V/m
Capacitance of the parallel plate capacitor, C = 50 pF = 50 x 10-12 F Distance between the plates is given by,
d =V/E
= 1000/106
=10–³m
Capacitance is given by the relation,
C=ε0 εr A/d
Where,
A = Area of each plate
ε0 = Permittivity of free space = 8.85×10-12 N⁻¹C²m⁻²
Therefore A =Cd/ε0 εr
= (50 x 10-12x 10–³)/( 8.85×10-12 x 3)
≈ 19 cm2
Hence, the area of each plate is about 19 cm2.