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# A parallel plate capacitor is to be designed with a voltage rating 1 kV,

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?

Solution:

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

Dielectric constant of a material, =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 10= 106 V/m

Capacitance of the parallel plate capacitor, C = 50 pF = 50 × 10−12 F

Distance between the plates is given by,

$d=\frac{V}{E}$

$=\frac{1000}{10^{6}}=10^{-3} \mathrm{~m}$

Capacitance is given by the relation,

$C=\frac{\in_{0} \in, A}{d}$

Where,

A = Area of each plate

$€_{0}=$ Permittivity of free space $=8.85 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^{2} \mathrm{~m}^{-2}$

$\therefore A=\frac{C d}{\epsilon_{0} \in_{r}}$

$=\frac{50 \times 10^{-12} \times 10^{-3}}{8.85 \times 10^{-12} \times 3} \approx 19 \mathrm{~cm}^{2}$

Hence, the area of each plate is about $19 \mathrm{~cm}^{2}$.