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

Answer 1

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

From figure $dl=Rd\theta$;
Charge on $dl=\lambda Rd\theta \{\lambda =\frac{q}{\pi R}\}$
Electric field at centre due to dl is $dE=k.\frac{\lambda Rd\theta }{R^2}$.

We need to consider only the component $dEcos\theta$, as the component $dEsin\theta$ will cancel out because of the field at C due to the symmetrical element dl'.
Total field at centre $=2{\int }_0^{\pi /2}dEcos\theta$
$=\frac{2k\lambda }{R}{\int }_0^{\pi /2}cos\theta d\theta =\frac{2k\lambda }{R}=\frac{q}{2{\pi }^2{ϵ}_0{R}^2}$