(a) For sphere with radius a, Potential $=\mathrm{V}$
(since, both the charged spheres are connected by a wire)



For sphere with radius $\mathit{b}$,
Potential $=\mathit{V}$
(since both the charged spheres are connected by a wire)



$=\frac{{\mathit{Q}}_{\mathit{b}}}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{b}}^2}$
......(1)
$\frac{{\mathit{Q}}_{\mathit{a}}}{{\mathit{Q}}_{\mathit{b}}}=\frac{{\mathit{C}}_{\mathit{a}}\mathit{V}}{{\mathit{C}}_{\mathit{b}}\mathit{V}}$

.......(2)
(since, ${\mathit{C}}_{\mathit{d}}∕{\mathit{C}}_{\mathit{b}}=\mathit{a}∕\mathit{b}$ )
Putting in equation (1)
$\frac{{\mathit{E}}_{\mathit{a}}}{{\mathit{E}}_{\mathit{b}}}=\left(\frac{\mathit{a}}{\mathit{b}}\right)\times \left(\frac{{\mathit{b}}^2}{{\mathit{a}}^2}\right)$
$\therefore \frac{{\mathit{E}}_{\mathit{a}}}{{\mathit{E}}_{\mathit{b}}}=\frac{\mathit{b}}{\mathit{a}}$
OR
(b) (i) Initially,
Charge on A capacitor $=\mathrm{CV}$
Charge on $\mathrm{B}$ capacitor $=0$
After connecting A with B,

$=\frac{\mathit{Q}+0}{\mathit{C}+2\mathit{C}}=\frac{\mathit{Q}}{3\mathit{C}}$
Final charge on A capacitor after redistribution
$=\mathit{C}{\mathit{V}}^{\prime }=\frac{\mathit{C}\mathit{Q}}{3\mathit{C}}=\frac{\mathit{Q}}{3}$
Final charge on B capacitor after redistribution
$=2\mathit{C}{\mathit{V}}^{\prime }=\frac{2\mathit{C}\mathit{Q}}{3\mathit{C}}=\frac{2\mathit{Q}}{3}$
So, the ratio of charges $=\frac{\mathit{Q}∕3}{2\mathit{Q}∕3}=1:2$
(ii) Initially energy stored in $\mathrm{A}=\frac{1}{2}{\mathrm{CV}}^2$
Finally energy stored in $\mathrm{A}=\frac{1}{2}\mathrm{C}{\mathit{V}}^2$
$=\frac{1}{2}\times \mathit{C}\times {\left(\frac{\mathit{Q}}{3\mathit{C}}\right)}^2$
$=\frac{1}{2}\times \mathit{C}\times {\left(\frac{\mathit{V}}{3}\right)}^2$
$=\frac{\mathit{C}{\mathit{V}}^2}{18}$
Finally energy stored in $\mathrm{B}=\frac{1}{2}2{\mathrm{CV}}^2$
$=\frac{1}{2}\times 2\mathit{C}\times {\left(\frac{\mathit{Q}}{3\mathit{C}}\right)}^2$
$=\frac{1}{2}\times 2\mathit{C}\times {\left(\frac{\mathit{V}}{3}\right)}^2$
$=\frac{\mathit{C}{\mathit{V}}^2}{9}$
Total energy stored in A and B
$=\frac{\mathit{C}{\mathit{V}}^2}{18}+\frac{\mathit{C}{\mathit{V}}^2}{9}$
$=\frac{\mathit{C}{\mathit{V}}^2}{6}$
So, the required ratio
$=\frac{\mathit{C}{\mathit{V}}^2∕6}{\mathit{C}{\mathit{V}}^2∕2}$
$=\frac{1}{3}$