(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 $\vec{E}$ is perpendicular to the plane. A Gaussian surface is considered through $P$ like a cylinder of flat caps parallel to the plane and one cap passing through $P$. The plane being the plane of symmetry for the Gaussian surface. $\oint \vec{E} \cdot \vec{ds} = \underset{\text{through caps}}{\overset{\rightarrow}{\int E \cdot \vec{ds}}}$ $\vec{E} \perp \vec{ds}$ for all over curved surface and hence $\vec{E} \cdot \vec{ds} = 0$ $\int\limits_{\text{caps}} E \, ds = 2E \Delta s$ $\Delta s = \text{ area of each cap }$ By Gauss' law $\oint \vec{E} \cdot \vec{ds} = \frac{q}{\varepsilon_0} = \frac{\sigma \Delta s}{\varepsilon_0}$ $\therefore 2E \Delta s = \frac{\sigma \Delta s}{\varepsilon_0}$ $E = \frac{\sigma}{2\varepsilon_0}$ If $\sigma$ is positive $\vec{E}$ points normally outwards/ away from the sheet If $\sigma$ is $(-)$ ve $\vec{E}$ points normally inwards/towards the sheet $U_s = \frac{1}{2} C_s V_s^2$ $U_p = \frac{1}{2} C_p V_p^2$ $\Rightarrow \frac{V_{\text{series}}}{V_{\text{parallel}}} = \sqrt{\frac{C_{\text{equivalent parallel}}}{C_{\text{equivalent series}}}}$ $= \sqrt{\frac{\frac{C_1+C_2}{C_1 C_2}}{C_1+C_2}}$ $= \frac{C_1+C_2}{\sqrt{C_1 C_2}} = \frac{3}{\sqrt{2}}$ 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 $\vec{E} = \vec{E}_1 + \vec{E}_2$ $= \frac{\sigma+\sigma}{2E_0} = \frac{\sigma}{E_0}$ Outside $\vec{E} = \vec{E}_2 - \vec{E}_1$ $= \frac{\sigma-\sigma}{2\varepsilon_0} = 0$ (b) Potential difference between plates $V = Ed = \frac{1}{\varepsilon_0} \frac{Qd}{A}$ (c) Capacitance $C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}$ (ii) As potential on and inside a charged sphere is given $V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{4\pi r^2 \sigma}{r}$ $\therefore V \propto r$ Hence, the bigger sphere will be at higher potential, so charge will flow from bigger sphere to smaller sphere.