Question

A thin glass rod is bent into a semicircular shape of radius R. A charge $+Q$ is uniformly distributed along the upper half and a charge $-Q$ is distributed uniformly along the lower half as shown. The electric field at the centre P is:

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

Answer 1

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

Consider two element of length $Rd\theta$ for negative and positive charge. Due to symmetry sine component of electric field cancel each other and only cosine component will contribute the electric field at P.

The charge on element , $dq=\lambda Rd\theta$ where $\lambda =$ line charge density.

The electric field due to element at P is $d{E}_P=2dE\cos \theta =2\left(\frac{dq}{4\pi {ϵ}_0}\right)\cos \theta =2\left(\frac{\lambda Rd\theta }{4\pi {ϵ}_0}\right)\cos \theta$

Thus, $E_P=\frac{2\lambda }{4\pi {ϵ}_{0}R}{\int }_0^{\pi /2}\cos \theta d\theta =\frac{2\left(2Q/\pi R\right)}{4\pi {ϵ}_{0}R}=\frac{Q}{{\pi }^2{ϵ}_0{R}^2}$ where $Q=\lambda \frac{\pi }{2}R$