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Class 12
Physics
Electrostatic Potential and Capacitance
a parallel plate capacitor is to be designed with...

Question:
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^-1. (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?

Answer:

Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, ∈_r=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 × 10^-12 F

Distance between the plates is given by,

Capacitance is given by the relation,

Where,

A = Area of each plate

∈₀ = Permittivity of free space = 8.85 x10^-12 N^-1C²m^-2

Hence, the area of each plate is about 19 cm².

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