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From a conducting ring of radius R which carries a charge Q (uniformly distributed) along it periphery a small length dl is cut-off. The electric field at the centre due to the remaining wire is

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zero

$\frac{Q d l}{8 π^{2} ε_{0} R^{3}}$

$\frac{Q d l}{4 π^{2} ε_{0} R^{3}}$

$\frac{Q}{4 π ε_{0} R^{2}}$

answer is A.

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Detailed Solution

if it was a complete ring electric field at the center would be zero. but now there will be an electric field due to the removed part and that is

$\frac{Q}{2 π R} \frac{d l}{4 π ε_{0} R^{2}}$

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