Question

A uniformly charged semicircular arc of radius $R$ has a linear charge density $\lambda$ . The electric field at its centre is :

• A. $\frac{\lambda }{4{ϵ}_0}$
• B. $\frac{2{ϵ}_0}{\lambda }$
• C. $\frac{\lambda }{4{ϵ}_{0}R}$
• D. $\frac{2\pi {ϵ}_0}{\lambda }$

Answer 1

The correct option is C $\frac{\lambda }{4{ϵ}_{0}R}$

The electric field due to the small length of the semicircle is given as

$E=\frac{dq}{4\pi {\epsilon }_0{R}^2}$

$E=\frac{\lambda .\pi R}{4\pi {\epsilon }_0{R}^2}$

By solving the above equation we get

$E=\frac{\lambda }{4{\epsilon }_{0}R}$