Question

A circular wire loop of radius ' $r$ ' carries a total charge '$Q$' distributed uniformly over its length. A small length dl of the wire is cut off. The electric field at the centre due to the remaining wire :-

• A. $\frac{\text { Qd } \ell}{8 \pi^{2} \in_{0} r^{3}}$
• B. $\frac{\text { Qd } \ell}{2 \pi^{2} \in_{0} r^{3}}$
• C. $\frac{\text { Qd } \ell}{8 \pi \in_{0} r^{3}}$
• D. $\frac{ Qd \ell}{4 \pi^{2} \in_{0} r ^{3}}$

Answer 1

Answer: (A)

We know that electric field at the centre of the charged circular wire is zero. Here we may conclude that electric field due to the element of length $d \ell$ and the rest of the wire are equal and opposite. So we have merely to determine the electric field due to the charged element of length $d \ell$
$E =\frac{1}{4 \pi \in_{0}} \frac{\lambda d \ell}{ r ^{2}}$
(where $\left.\lambda=\frac{Q}{2 \pi r}\right)$
$=\frac{1}{4 \pi \in_{0}} \frac{Q}{2 \pi r} \frac{ d \ell}{r^{2}}$
$=\frac{Q d \ell}{8 \pi^{2} \in_{0} r^{3}}$