Question

A ring shaped conductor or radius $R$ carries a total charge $q$ uniformly distributed around it. Find the electric field at a point $P$ that lies on the axis of the ring at a distance $x$ from its centre.

Answer 1

Consider a differential element of the ring of length $ds$. Charge on this element is $dq=\left(\frac{q}{2\pi R}ds\right)$
This element sets up a differential electric field $d\overrightarrow{E}$ at point $P$.
The resultant field $\overrightarrow{E}$ at $P$ is found by integrating the effects of all the elements that make up up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of $d\overrightarrow{E}$ parallel to this axis contributes to the final result.
To find the total x-component $E_x$ of the field at $P$, we integrate this expression over all segments of the ring.
$E_x=\int dE\cos \theta =\frac{1}{4\pi {\in }_0}\frac{qx}{2\pi R(R^2+x^2{)}^{\frac{3}{2}}}\int ds$
The integral is simply the circumference of the ring which is equal to $2\pi R$
$\therefore E=\frac{1}{4\pi {\in }_0}\frac{qx}{2\pi R(R^2+x^2{)}^{\frac{3}{2}}}$
As $q$ is a positive charge, hence field is directed away from the centre of the ring along its axis.