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 1

given, The parallel plate capacitor is designed with a voltage rating, V = 1kV= 1000 V.

dielectric constant, K = 3

capacitance of capacitor , C = 50pF = 50 × 10^-12 F

dielectric strength = 10^7 V/m

given, E = 10% of dielectric strength.

= 10 × 10^7/100 = 10^6 V/m

we know, $\boxed{\bf{V=E.d}}$

so, d = V/E = 10³/10^6 = 10^-3 m

hence, seperation between plates , d = 10^-3 m

using formula, $\boxed{\bf{C=\frac{K\epsilon_0A}{d}}}$

or, $A=\frac{C.d}{K\epsilon_0}$

= (50 × 10^-12 × 10^-3)/(3 × 8.85 × 10^-12)

= 1.9 × 10^-3 m²

= 19 × 10^-4 m²

[ we know, 1cm² = 10^-4 m² ]

= 19 cm²

hence, minimum area of the plates , A = 19 cm²

Answer 2

Answer:

given, The parallel plate capacitor is designed with a voltage rating, V = 1kV= 1000 V.

dielectric constant,

K = 3 capacitance of capacitor ,

C = 50pF = 50 × 10^-12 F

dielectric strength = 10^7 V/m given,

E = 10% of dielectric strength.= 10 × 10^7/100 = 10^6 V/m

we know, so, d = V/E = 10³/10^6 = 10^-3 m hence,

separation between plates , d = 10^-3 m

using formula = (50 × 10^-12 × 10^-3)/(3 × 8.85 × 10^-12) = 1.9 × 10^-3 m²

= 19 × 10^-4 m² [ we know, 1cm² = 10^-4 m² ]

= 19 cm² hence,

minimum area of the plates ,

A = 19 cm²

Explanation: