Question

A circular wire of radius $R$ carries a total charge $Q$ distributed uniformly over its circumference. A small length of the wire subtending angle $\theta$ at the centre is cut off. Find the electric field at the centre due to the remaining portion

• A. $\frac{Q}{4{\pi }^2{\epsilon }_0{R}^2}\sin (\theta )$
• B. $\frac{Q}{4{\pi }^2{\epsilon }_0{R}^2}\sin \left(\frac{\theta }{2}\right)$
• C. $\frac{Q}{4{\pi }^2{\epsilon }_0{R}^2}\sin \left(\frac{\theta }{4}\right)$
• D. $\frac{Q}{4{\pi }^2{\epsilon }_0{R}^2}\sin \left(\frac{\theta }{8}\right)$

Answer 1

Answer: (B)

Electric field due to an arc at its centre is
$\frac{k\lambda }{R}2\sin \left(\frac{\theta }{2}\right),$
Where $k=\frac{1}{4\pi {\epsilon }_0}$
$\theta =$ angle subtended by the wire at the centre,
$\lambda =$ Linear density of charge.
Let $E$ be the electric field due to remaining portion.
Since intensity at the centre due to the circular wire is zero.
Applying principle of superposition.

$\frac{k\lambda }{R}2\sin \left(\frac{\theta }{2}\right)\hat{n}+\vec{E}=0$
$|\vec{E}|=\frac{1}{4\pi {\epsilon }_{0}R}\cdot \frac{Q}{2\pi R}\cdot 2\sin \left(\frac{\theta }{2}\right)$
$=\frac{Q}{4{\pi }^2{\epsilon }_0{R}^2}\sin \left(\frac{\theta }{2}\right)$