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

The correct option is A $\frac{q}{2{\pi }^2{\epsilon }_0{R}^2}$

$dl=Rd\theta$

Charge on $dl$ is $\lambda Rd\theta$ where $\lambda$ is linear charge density

Electric field at center due to $dl$

$dE=k\times \frac{\lambda Rd\theta }{R^2}$

We need to consider only the component $dEcos\theta$ as the component $dEsin\theta$ will cancel out

Therefore total field at center = $2{\int }_0^{\frac{\pi }{2}}dEcos\theta$

$2{\int }_0^{\pi /2}\frac{k\lambda Rcos\theta }{R^2}d\theta =\frac{2k\lambda }{R}{\int }_0^{\pi /2}cos\theta d\theta =\frac{q}{2{\pi }^2{ϵ}_o{R}^2}$