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 di= Rd $\theta$
charge on dl $=\lambda Rd\theta \{\lambda =\frac{q}{\pi R}\}$
E electric field at centre due to di is dE = $\frac{k.\lambda Rd\theta }{R^2}$
We need to consider only the component dE cos $\theta$ , as the component dE sin $\theta$ will cancel out.
Total field at centre = $2\underset{0}{\overset{\pi /2}{\int }}dEcos\theta$
= $\frac{2k\lambda }{R}\underset{0}{\overset{\pi /2}{\int }}cosd\theta$
= $\frac{2k\lambda }{R}=\frac{q}{2{\pi }^2{\epsilon }_0{R}^2}$