Question

A parallel plate capacitor is constructed using a dielectric material with dielectric constant of 3 and whose dielectric strength is 2x10^8 V/m. the desired capacitance is 0.250 mF, and the capacitors must withstand a maximum potential difference of 4 kV. Find the minimum area of the capacitor plates.

Answer 1

maximum.separation between plates of capacitor be d

Than

​d= ∆V/E ; E = dielectric strength

Hence :

d=4*10³ / 2*10⁸ =2*10^-5 m

Now capacitance

$\dpi{120} \large C=\frac {k \epsilon _{o}A}{d}$

$\dpi{120} \large 0.250*10^{-3}=\frac {3*8.85*10^{-12}*A}{2*10^{-5}}$

A= 188m²