$\mathrm{Q}_{2}=\frac{\sum Q}{\mathrm{R}_{1}+\mathrm{R}_{2}} \times \mathrm{R}_{2}$
$\sigma_{1}=\frac{\mathrm{Q}_{1}}{4 \pi \mathrm{R}_{1}^{2}}=\frac{\sum Q}{\mathrm{R}_{1}+\mathrm{R}_{2}} \times \frac{\mathrm{R}_{1}}{4 \pi \mathrm{R}_{1}^{2}} \propto \frac{1}{\mathrm{R}_{1}}$
$\sigma_{2}=\frac{\mathrm{Q}_{2}}{4 \pi \mathrm{R}_{2}^{2}}=\frac{\sum Q}{\mathrm{R}_{1}+\mathrm{R}_{2}} \times \frac{\mathrm{R}_{2}}{4 \pi \mathrm{R}_{2}^{2}} \propto \frac{1}{\mathrm{R}_{2}}$
$\frac{\sigma_{1}}{\sigma_{2}}=\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}$
($1$) The value of $R$ is. . . . meter.
($2$) The value of $b$ is. . . . . .meter.

$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled ${E_1}$
$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is ${E_2}.$
Then ${E_1}/{E_2}$ value is

Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).
