Question

A wire of length $l$ is bent to form a semicircle. If it has charge $Q$, then the electric field intensity at the centre of the ring is :

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

Answer 1

Answer: (D)

$\frac{Q}{2{ϵ}_0{l}^2}$

length of the wire $l=\pi r$

$\therefore RadiusofwireR=\frac{l}{\pi }$

In a semicircle the horizontal components of E get cancelled

$\therefore$ E due to a small element dl. at angle $d\theta$ is

${\int }_0^{E}dE={\int }_0^{\pi }\frac{KQ\cdot Rd\theta }{l{R}^2}sin\theta$

$E=\frac{KQ}{lR}[-cos\theta {]}_0^{\pi }$

$E=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{lR}2(-\hat{y})$

$E=\frac{Q}{2{\epsilon }_0{l}^2}(-\hat{y})$