A nonconducting ring of radius R has uniformly distributed positive charge Q. A small part of the ring, of length d, is removed (d << R). The electric field at the centre of the ring will now be

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directed away from the gap, inversely proportional to R²

directed towards the gap, inversely proportional to R³

directed towards the gap, inversely proportional to R²

directed away from the gap, inversely proportional to R³

answer is A.

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Detailed Solution

Here shaded part is removed and electric field on it due to part diametrically opposite to it

$\begin{array}{l} \frac{∆ Q}{d} = \frac{Q}{2 πR} \Rightarrow ∆ Q = \frac{Qd}{2 πR} \\ dE = \frac{K ∆ Q}{R^{2}} = \frac{KQd}{(2 πR) R^{2}} \\ \Rightarrow E = \frac{KQd}{2 {πR}^{3}} \Rightarrow E \propto \frac{1}{R^{3}} \end{array}$

$∴$ directed towards the gap,

E inversely proportional to R³

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