Question

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

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

Answer 1

Answer: (A)

From figure $dl=Rd\theta$
charge on $dl=\lambda Rd\theta$
$\left\{\lambda =\frac{q}{\pi R}\right\}$
Electric field at centre due to $dl$ is
$dE=\frac{k\cdot \lambda Rd\theta }{R^2}$

We need to consider only the component $dE\cos \theta$, as the component $dE\sin \theta$ will cancel out because of the field at $C$ due to the symmetrical element $d{l}^{\prime }$
Total field at centre $=2\underset{0}{\overset{\pi /2}{\int }}dE\cos \theta$
$=\frac{2k\lambda }{R}\underset{0}{\overset{\pi /2}{\int }}\cos \theta d\theta$
$=\frac{2k\lambda }{R}\frac{q}{2{\pi }^2{\epsilon }_0{R}^2}$