Question

Charge Q is distributed uniformly on length $l$ of a wire. It is bent in the form of a semicircular ring. Find the electric field at the centre of the ring.

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

Answer 1

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

Consider two small elements of length $dl$ each charge $dq$ on each element. If $\lambda$ is the line charge density, $Q=\lambda l=\lambda .\pi r$. as wire is bent in the form of semicircular ring.
Hence , $dq=\lambda dl=\frac{Q}{\pi r}.\left(rd\theta \right)=\frac{Q}{\pi }d\theta$.
Here sine component of the field at center cancel out and only cosine will contribute the filed.
thus, $d{E}_c=2dE\cos \theta =2\frac{dq}{4\pi {ϵ}_0{r}^2}\cos \theta =\frac{Q}{2{\pi }^2{ϵ}_0{r}^2}\cos \theta d\theta$
or $E_c=\frac{Q}{2{ϵ}_0(\pi r{)}^2}{\int }_0^{\pi /2}\cos \theta d\theta =\frac{Q}{2{ϵ}_0(\pi r{)}^2}(\sin \frac{\pi }{2}-\sin 0)=\frac{Q}{2{ϵ}_0{l}^2}$ as $(l=\pi r)$