Q. 334.3( 6 Votes )

# 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 10^{7} 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 difference of a parallel plate capacitor(V) = 1kV = 1000V

Dielectric constant of a material(ϵ_{r}) = 3

Dielectric strength = 10^{7}V/m

Electric field intensity(E) = 10%of 10^{7}

⇒ E = 10^{6}V/m

(since, the field intensity never exceeds 10% of the dielectric strength)

Capacitance of the parallel plate capacitor(C) = 50pF = 50 × 10^{-12}F

Distance between the plates (d),

⇒

⇒ d = 10^{-3}m

∵ Capacitance =

where, ϵ_{0} = Permittivity of free space = 8.85 × 10^{-12}N^{-1} m^{-2} C^{2}

A = area of plates

∴

⇒ A = 50 × 10^{-12}F × 10^{-3}m/8.85 × 10^{-12}N^{-1} m^{-2} C^{2} × 3

⇒ A = 1.88 × 10^{-3}m^{2}

∴ Minimum area of the plates required = 1.88 × 10^{-3}m^{2}

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