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,

\[E = \frac{1}{4\pi \epsilon_0}\frac{dq}{a^2}\] 

\[ \Rightarrow E = \frac{1}{4\pi \epsilon_0 a^2}\frac{Q}{2\pi a}dL = \frac{QdL}{8 \pi^2 \epsilon_0 a^3}\]

So, the electric field at the centre due to the remaining wire = \[\frac{QdL}{8 \pi^2 \epsilon_0 a^3}\]  (Opposite the direction of the electric field due to the small element)