(a) Derivation of $E$ along the axial line of dipole (b) Graph between $E$ vs $r$ (c) (i) Diagrams for stable and unstable equilibrium of dipole (ii) Torque on the dipole in the two cases Electric field at $P$ due to charge $+q\;\text{ is }\;E_1=\frac{1}{4\pi\varepsilon_o}\;\frac{1}{(r-a)^2}$ Electric field at $P$ due to charge $-q$ is $E_2=\frac{1}{4\pi\varepsilon_0}\;\frac{q}{(r+a)^2}$ Net electric Field at $P$ $E_1-E_2\;=\;\frac{1}{4\pi\varepsilon_{o}}\;\frac{q}{(r-a)^2}$ $-\frac{1}{4\pi\varepsilon_{o}}\;\frac{q}{(r-a)^2}$ $=\frac{1}{4\pi\varepsilon_0}\;\frac{2pr}{(r^2-a^2)^2}\;\;(p=q\cdot 2a)$ Its direction is parallel to $\vec{p}$. (For short Dipole $=\frac{1}{4\pi\varepsilon_0}\;\frac{2p}{r^3}$ without drawing the graph) (i) For stable Equilibrium: $\vec{p}$ is parallel to $\vec{E}$. (ii) For unstable equilibrium: $\vec{p}$ is antiparallel to $\vec{E}$ Torque $=0$ for (i) as well as case (ii). (Also accept, $\vec{\tau}=\vec{p}\times\vec{E}/\tau=pE\sin\;\theta$ OR (a) Using Gauss's theorem to find $E$ due to an infinite plane sheet of charge (b) Expression for the work done to bring charge $q$ from infinity to $r$ (a) $\oint E\cdot ds=\frac{q}{\varepsilon_0}$ The electric field $E$ points outwards normal to the sheet. The field lines are parallel to the Gaussian surface except for surfaces 1 and 2 . Hence the net flux $=\oint E\cdot ds=\frac{q}{\varepsilon_0}=\frac{\sigma A}{\varepsilon_0}=2EA$ where $A$ is the area of each of the surface 1 and 2. $\; \therefore \oint E\cdot ds=\frac{q}{\varepsilon_0}=\frac{\sigma A}{\varepsilon_0}=2EA$ $E=\frac{\sigma}{2\varepsilon_0}$ (b) $W=q\int\limits_{\infty}^{r} \vec{E}\cdot \vec{dr}$ $=q\int\limits_{\infty}^{r}(-E dr)$ $=-q\int\limits_{\infty}^{r} \left(\frac{\sigma}{2\varepsilon_0}\right) dr$ $=\frac{q\sigma}{2\varepsilon_0}\left|\infty-r\right|$ $=(\infty)$