Two circular metal plates each of radius ' $r$ ' are kept parallel to each other distance 'd' apart. The…
Two circular metal plates each of radius ' $r$ ' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is ' $\mathrm{C}_1$ '. If the radius of each of the plates is increased to $\sqrt{2}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes ' $\mathrm{C}_2$ '. The ratio of the capacitances $\mathrm{C}_1: \mathrm{C}_2$. is
$1: 1$
$1: 2$
$1: 4$
$4: 1$
Solution
For the $1^{\text {st }}$ capacitor,
$\mathrm{C}_1=\frac{\varepsilon_0 \mathrm{~A}_1}{\mathrm{~d}}=\frac{\varepsilon_0 \pi \mathrm{r}^2}{\mathrm{~d}}$ For the $2^{\text {nd }}$ capacitor,
$\begin{aligned}
& \mathrm{C}_2=\frac{\varepsilon_0 \mathrm{~A}_2}{\mathrm{~d}}=\frac{\varepsilon_0 \pi 2 \mathrm{r}^2}{\frac{d}{2}}=\frac{\varepsilon_0 4 \pi \mathrm{r}^2}{\mathrm{~d}}=4 \mathrm{C}_1 \\
\therefore \quad & \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\mathrm{C}_1}{4 \mathrm{C}_1}=\frac{1}{4}
\end{aligned}$