Question

The electric field at the centre of a uniformly charged ring is zero. What is the electric field at the centre of a half ring if the charge on it be Q and its radius be R?

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{Q}{\pi R^2}$
• B. $\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R^2}$
• C. $\frac{1}{4\pi {\epsilon }_0}\frac{2Q}{\pi R^2}$
• D. $\frac{1}{4\pi {\epsilon }_0}\frac{2Q}{R^2}$

Answer 1

The correct option is C $\frac{1}{4\pi {\epsilon }_0}\frac{2Q}{\pi R^2}$

Due to symmetry the x-component of field at the centre of half ring cancel out . So the field due to element of charge $dq=\lambda Rd\theta$ at O is

$d{E}_O=d{E}_y=dE\sin \theta =\frac{dq}{4\pi {ϵ}_0{R}^2}\sin \theta =\frac{\lambda Rd\theta }{4\pi {ϵ}_0{R}^2}\sin \theta$ where $Q=\lambda \frac{\pi }{2}R,\lambda =\frac{2Q}{\pi R}$, line charge density.

$E_O=\frac{1}{4\pi {ϵ}_0}\frac{2Q}{\pi R^2}{\int }_0^{\pi /2}\sin \theta d\theta =\frac{1}{4\pi {ϵ}_0}\frac{2Q}{\pi R^2}$