Question

A charged wire is bent in the form of a semicircular arc of radius a. If charge per unit length is $\lambda$ coulomb/ metre, the electric field at the centre O is :

• A. $\frac{\lambda }{2\pi a^2{\epsilon }_0}$
• B. $\frac{\lambda }{4\pi a^2{\epsilon }_0}$
• C. $\frac{\lambda }{2\pi a{\epsilon }_0}$
• D. zero

Answer 1

Answer: (C)

$\frac{\lambda }{2\pi a{\epsilon }_0}$

Considering symmetric elements each of length dl at A and B, we note that electric fields
perpendicular to PO are cancelled and those along PO are added. The electric field due to an
element of length $dl\left(ad\theta \right)$ along PO
$dE=\frac{1}{4\pi E_0}\frac{dq}{a^2}cos\theta$
$=\frac{1}{4\pi E_0}\frac{\lambda dl}{a^2}cos\theta$
$=\frac{1}{4\pi E_0}\frac{\lambda \left(ad\theta \right)}{a^2}cos\theta$

Net electric field at O
$E={\int }_{-\pi /2}^{\pi /2}dE=2{\int }_0^{\pi /2}\frac{1}{4\pi E_0}\frac{\lambda acos\theta d\theta }{a^2}$
$2.\frac{1}{4\pi E_0}\frac{\lambda }{a}{\left[sin\theta \right]}_0^{\pi /2}=2.\frac{1}{4\pi E_0}.\frac{\lambda }{a}.1=\frac{\lambda }{2\pi E_{0}a}$