Question

A capacitor of capacitance $9\,nF$ having dielectric slab of $\varepsilon{r}$ = $2.4$ dielectric strength $20\, MV/m^{-1}$ and P.D. = $20\, V$ 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,$\quad$$\quad$ $\varepsilon{r}$ = 2.4, V = 20 volt
Dielectric strength = 20MV/m
Let separation between plants = d
$E=\frac{v}{d}$
$20\times10^{6}=\frac{20}{d}$
$d=10^{-6}m$
Now,
$C=\frac{\in_{o}A\in_{r}}{d}$
$9\times10^{-9}= \frac{8.85\times10^{-12}\times A\times2.4}{10^{-6}}$
$A=\frac{9\times10^{-15}}{8.85\times2.4\times10^{-12}}$
$A=0.42\times10^{-3}$
$A=4.2\times10^{-4 }m^{2}$