Q.(a) Give the relation between electric field and potential.

The electric field is minus the spatial rate of change of potential; a dipole's potential at a general point falls off as $1/r^2$ (not $1/r$ as for a point charge); and the given point charge produces a potential of $4\times10^4$ V at 9 cm.

1. Relation between E and V: The electric field is the negative gradient of the electrostatic potential: $\vec{E} = -\nabla V, \qquad \text{or along one direction: } E = -\dfrac{dV}{dr}$ The field points in the direction of the steepest decrease of potential; equivalently, $V(r) = -\displaystyle\int \vec{E}\cdot d\vec{r}$.
2. Potential due to an electric dipole: Let a dipole have $-q$ at A and $+q$ at B, separated by $2a$ (moment $p=q\cdot2a$), and let P be a point at distance $r$ from the centre O, at angle $\theta$ to the dipole axis. For $r \gg a$, the distances from P to the two charges can be approximated as: $r_1 \text{(to } +q\text{)} \approx r - a\cos\theta, \qquad r_2 \text{(to } -q\text{)} \approx r + a\cos\theta$ The net potential at P (potentials are scalars, so they simply add): $V = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q}{r_1}-\dfrac{q}{r_2}\right) = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{r_2-r_1}{r_1r_2}$ Using $r_2 - r_1 \approx 2a\cos\theta$ and $r_1r_2\approx r^2$ (since $r\gg a$): $V \approx \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{q(2a\cos\theta)}{r^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{p\cos\theta}{r^2}$ …