(a) Characteristics of equipotential surface :

(i) Potential remains the same at all the points on equipotential surface.

(ii) No work is required to move a charge within an equipotential surface.

(b) Point $\mathit{P}$ is on theaxis of the ring at a distance $\mathit{r}$ from the centre.
The field $\mathrm{dE}$ due to a small element $\mathrm{dl}$ is considered.

Linear charge density $=\mathit{\lambda }$
So, charge of $\mathrm{dl}$ element is $\mathrm{dq}=\mathit{\lambda }\mathrm{dl}$

The perpendicular field component for the whole ring will be zero.

So, $\mathrm{dE}=\frac{\mathit{k}\mathit{d}\mathit{q}}{{\mathit{d}}^2}\cos \mathit{\theta }=\frac{\mathit{k}\mathit{d}\mathit{q}}{{\mathit{d}}^2}$ $\times \frac{\mathit{x}}{\mathit{d}}$
Or, $\mathrm{dE}=\frac{\mathit{k}\mathit{x}\mathit{d}\mathit{q}}{{\left({\mathit{x}}^2+{\mathit{r}}^2\right)}^{3∕2}}$
Or, $\mathrm{E}=\int \frac{\mathit{k}\mathit{x}\mathit{d}\mathit{q}}{{\left({\mathit{x}}^2+{\mathit{r}}^2\right)}^{3∕2}}$ $=\int \frac{\mathit{k}\mathit{x}\mathit{\lambda }\mathit{d}\mathit{l}}{{\left({\mathit{x}}^2+{\mathit{r}}^2\right)}^{3∕2}}$
$\therefore \mathrm{E}=\frac{\mathit{k}\mathit{x}\mathit{\lambda }]}{{\left({\mathit{x}}^2+{\mathit{r}}^2\right)}^{3∕2}}$ $=\frac{\mathit{k}\mathit{x}\mathrm{Q}}{{\left({\mathit{x}}^2+{\mathit{r}}^2\right)}^{3∕2}}$
When $\mathit{x}>>\mathit{r}$
$\mathrm{E}=\frac{\mathit{k}\mathit{x}\mathit{Q}}{{\left({\mathit{x}}^2\right)}^{3∕2}}$ $=\frac{\mathit{k}\mathit{Q}}{{\mathit{x}}^2}$
This reduces to a simple Coulomb field. In this case, the charged ring looks like a point charge.
OR
(a) Gauss's Law: The total of the electric flux out of a closed surface is equal to the chargeenclosed divided by the permittivity.
$\mathit{\Phi }=\frac{\mathit{Q}}{{\mathit{\epsilon }}_0}$
Electric field due to an infinitely long straight uniformly charged wire : charge density $=\mathit{\lambda }$
$\mathit{P}$ is a point where the electric field is to be calculated.
$\mathit{r}=$ distance of point $\mathit{P}$from the wire
$\mathrm{E}=$ electric field at the point $\mathrm{P}$

A cylinder of length 1 , radius $\mathit{r}$ , closed at each end is imagined as Gaussian surface.
$\mathit{d}\mathit{s}=\mathit{a}$ very small area on the Gaussian surface By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. $\mathrm{E}$ and ds are along the same direction.Theelectric flux $\left(\mathit{\varphi }\right)$ through curved surface
$=\oint \mathrm{Eds}\cos \mathit{\theta }$
$\mathit{\varphi }=\oint \mathrm{Eds}\left[\because \mathit{\theta }=0^{\circ }\right]$
$\mathit{\varphi }=\mathrm{E}\left(2\mathrm{r}\mathit{\pi }\mathrm{l}\right)$
$\therefore \mathrm{Q}=\mathit{\lambda }\mathrm{l}=$ the net charge enclosed by Gaussian surface is

$\mathit{\varphi }=\frac{\mathit{Q}}{{\mathit{\epsilon }}_0}$

$\therefore \mathrm{E}=\frac{\mathit{\lambda }}{2\mathit{\pi }\mathrm{r}{\mathit{\epsilon }}_0}$
(b) $\mathrm{E}=-\frac{\mathit{d}\mathit{V}}{\mathit{d}\mathit{r}}$
$\therefore \left|\mathrm{V}\right|=\underset{1}{\overset{10}{\int }}\mathit{E}\mathit{d}\mathit{r}$




Thus, electric potential changes by $540\mathrm{V}$.