Question

A circular wire loop of radius R carries a total charge q distributed uniformly over its length. A small length x(<<R)of the wire is cut off. The electric field at the centre due to the remaining wire is $\frac{qx}{n{\pi }^2{\epsilon }_o{R}^3}$. The value of $n$ is:

Answer 1

Initially the field at the center is zero, i.e., field due to all the elements of the circular wire cancel each other.
field at the center after removing "x" length of wire = field at center due to "x" length of wire
= $\frac{1}{4\pi {ϵ}_o}{q}_x{R}^2$

= $\frac{1}{4\pi {ϵ}_o}\frac{qx}{2\pi R{R}^2}$ $(\because q_x=\frac{qx}{2\pi R})$

= $\frac{qx}{8\pi {ϵ}_o{R}^3}$