Question

A circular wire-loop of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre due to the remaining wire.

Answer 1

We know that the electric field is zero at the centre of a uniformly charged circular wire.
That is, the electric field due to a small element dl of wire + electric field due to the remaining wire = 0
Let the charge on the small element dl be dq. So,
$dq=\frac{Q}{2\pi a}dL$
Electric field due to a small element at the centre,
$$\begin{gathered} E=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{dq}{a^2} \\ \Rightarrow E=\frac{1}{4\mathrm{\pi }{\epsilon }_0{\mathrm{a}}^2}\frac{Q}{2\pi a}dL=\frac{QdL}{8{\pi }^2{\epsilon }_0{a}^3} \end{gathered}$$
So, the electric field at the centre due to the remaining wire = $\frac{QdL}{8{\mathrm{\pi }}^2{\epsilon }_0{a}^3}$ (Opposite the direction of the electric field due to the small element)