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

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

Dielectric constant of a material, `in_"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,

`"d" = "V"/"E"`

= `1000/10^6 = 10^-3  "m"`

Capacitance is given by relation 

C = `(in_0in_"r""A")/"d"`

Where,

A = Area of each plate

`in_0` = Permittivity of free space = `8.85 xx 10^-12  "N"^-1  "C"^2  "m"^-2`

∴ `"A" = ("Cd")/(in_0in_"r")`

= `(50 xx 10^-12 xx 10^-3)/(8.85 xx 10^-12 xx 3) ≈ 19  "cm"^2`

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