Question

A capacitor of capacitance 15nF having dielectric slab of $\varepsilon_{r}$ = 2.5 dielectric strength 30 MV/m and potential difference = 30 volt. Calculate the area of plate

• A. 6.7 × 10^-4 m²
• B. 4.2 × 10^-4 m²
• C. 8.0 × 10^-4 m²
• D. 9.85 × 10^-4 m²

Answer 1

Answer: (A)

$C=\frac{A\varepsilon_{o}\varepsilon_{r}}{d}$
$\Rightarrow 15\times10^{-9}=\frac{A\times8.85\times10^{-12}\times2.5}{d}\quad...\left(i\right)$
Since,$\quad\quad E=\frac{V}{d}$
$\Rightarrow 30\times10^{6}=\frac{30}{d}$
$d=10^{-6} m$
from $\left(i\right)$
$15\times 10^{-9}=\frac{A\times 8.85\times 10^{-12}\times 2.5}{10^{-6}}$
$A=\frac{15\times10^{-9}\times10^{-6}}{8.85\times10^{-12}\times2.5}$
$=6.7\times10^{-4} m^{2}$