- Home
- Standard 12
- Physics
A capacitor of capacitance $15 \,nF$ having dielectric slab of $\varepsilon_{r}=2.5$ dielectric strength $30 \,MV / m$ and potential difference $=30\; volt$ then the area of plate is ....... $ \times 10^{-4}\; m ^{2}$
$6.7$
$4.2$
$8.0$
$9.85$
Solution
The expression for capacitance is given by,
$C=\frac{A \varepsilon_{0} \varepsilon_{r}}{d}$ $……(I)$
Substitute $15 \times 10^{-9}$ for $C, 8.85 \times 10^{-12}$ for $\varepsilon_{0}$ and $2.5$ for $\varepsilon,$ in equation $(I).$
$15 \times 10^{-9}=\frac{A\left(8.85 \times 10^{-12}\right)(2.5)}{d} \quad \ldots \ldots$ $(II)$
The expression for electric field in given by,
$E=\frac{V}{d}$
$30 \times 10^{6}=\frac{30}{d}$
$d=10^{-6} m$
Substitute $10^{-6} m$ for $d$ in equation $( II )$.
$15 \times 10^{-9}=\frac{A\left(8.85 \times 10^{-12}\right)(2.5)}{10^{-6}}$
$A=\frac{15 \times 10^{-9} \times 10^{-6}}{8.85 \times 10^{-12} \times 2.5}$
$=6.7 \times 10^{-4} m ^{2}$
