Question

A capacitor of capacitance $9nF$ having dielectric slab of $\epsilon r$ = $2.4$ dielectric strength $20MV/m^{-1}$ and P.D. = $20V$ calculate area of plates.

• A. $2.1\times 10^{-4}{m}^2$
• B. $4.2\times 10^{-4}{m}^2$
• C. $1.4\times 10^{-4}{m}^2$
• D. $2.4\times 10^{-4}{m}^2$

Answer 1

Answer: (B)

C = 9nF, $\epsilon r$ = 2.4, V = 20 volt
Dielectric strength = 20MV/m
Let separation between plants = d
$E=\frac{v}{d}$
$20\times 10^6=\frac{20}{d}$
$d=10^{-6}m$
Now,
$C=\frac{{\in }_{o}A{\in }_r}{d}$
$9\times 10^{-9}=\frac{8.85\times 10^{-12}\times A\times 2.4}{10^{-6}}$
$A=\frac{9\times 10^{-15}}{8.85\times 2.4\times 10^{-12}}$
$A=0.42\times 10^{-3}$
$A=4.2\times 10^{-4}{m}^2$