(i) Derivation for electric field due to infinite plane Sheet of charge Directions of field
(ii) Formula
Calculation and result Symmetry of situation suggests that $\overrightarrow{\mathit{E}}$ is perpendicular to the plane. A Gaussian surface is considered through $\mathrm{P}$ like a cylinder of flat caps parallel to the plane and one cap passing through $\mathrm{P}$. The plane being the plane of symmetry for the Gaussian surface.


$\overrightarrow{\mathit{E}}\perp \overrightarrow{\mathit{d}\mathit{s}}$ for all over curved surface and hence $\overrightarrow{\mathit{E}}\cdot \overrightarrow{\mathit{d}\mathit{s}}=0$
$\underset{\mathit{c}\mathit{a}\mathit{p}\mathit{s}}{\int }\mathit{E}\mathit{d}\mathit{s}=2\mathit{E}∆\mathit{s}$

By Gauss' law
$\oint \overrightarrow{\mathit{E}}\cdot \overrightarrow{\mathit{d}\mathit{s}}=\frac{\mathit{q}}{{\mathit{\epsilon }}_0}=\frac{\mathit{\sigma }∆\mathit{s}}{{\mathit{\epsilon }}_0}$
$\therefore \mathrm{}2\mathit{E}∆\mathit{s}=\frac{\mathit{\sigma }∆\mathit{s}}{{\mathit{\epsilon }}_0}$
$\mathit{E}=\frac{\mathit{\sigma }}{2{\mathit{\epsilon }}_0}$
If $\mathit{\sigma }$ is positive $\overrightarrow{\mathit{E}}$ points normally outwards/ away from the sheet
If $\mathit{\sigma }$ is $(-)$ ve $\overrightarrow{\mathit{E}}$ points normally inwards/towards the sheet
${\mathit{U}}_{\mathit{s}}=\frac{1}{2}{\mathit{C}}_{\mathit{s}}{\mathit{V}}_{\mathit{s}}^2$
${\mathit{U}}_{\mathit{p}}=\frac{1}{2}{\mathit{C}}_{\mathit{p}}{\mathit{V}}_{\mathit{p}}^2$

$=\sqrt{\frac{\frac{{\mathit{C}}_1+{\mathit{C}}_2}{{\mathit{C}}_1{\mathit{C}}_2}}{{\mathit{C}}_1+{\mathit{C}}_2}}$
$=\frac{{\mathit{C}}_1+{\mathit{C}}_2}{\sqrt{{\mathit{C}}_1{\mathit{C}}_2}}=\frac{3}{\sqrt{2}}$
OR(i) Deriving the expression for field between the plate & outside
Direction of electric field inside and outside
Potential difference between the plates
Capacitance
(ii) Direction of flow of charge

Inside
$\overrightarrow{\mathit{E}}={\overrightarrow{\mathit{E}}}_1+{\overrightarrow{\mathit{E}}}_2$
$=\frac{\mathit{\sigma }+\mathit{\sigma }}{2{\mathit{E}}_0}=\frac{\mathit{\sigma }}{{\mathit{E}}_0}$
Outside
$\overrightarrow{\mathit{E}}={\overrightarrow{\mathit{E}}}_2-{\overrightarrow{\mathit{E}}}_1$
$=\frac{\mathit{\sigma }-\mathit{\sigma }}{2{\mathit{\epsilon }}_0}=0$
(b) Potential difference between plates
$\mathit{V}=\mathit{E}\mathit{d}=\frac{1}{{\mathit{\epsilon }}_0}\frac{\mathrm{Q}\mathit{d}}{\mathit{A}}$
(c) Capacitance
$\mathit{C}=\frac{\mathit{Q}}{\mathit{V}}=\frac{{\mathit{\epsilon }}_0\mathit{A}}{\mathit{d}}$
(ii) As potential on and inside a charged sphere is given
$\mathit{V}=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{\mathit{q}}{\mathit{r}}=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\cdot \frac{4\mathit{\pi }{\mathit{r}}^2\mathit{\sigma }}{\mathit{r}}$
$\therefore \mathit{V}\propto \mathit{r}$
Hence, the bigger sphere will be at higher potential, so charge will flow from bigger sphere to smaller sphere.