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