Question

The linear charge density on a ring which varies with angle $\theta$ can be represented as $\lambda =Kcos\frac{\theta }{2},where$k = 2 cm^{-1}$and$\theta$ is the angle subtended by the radius of the ring with the horizontal. The potential at the centre of the ring is

Answer 1

Take a small element $dl$ subtending an angle $d\theta$ at the centre.


Charge on the element $dq=\lambda dl=\lambda Rd\theta =2cos\frac{\theta }{2}Rd\theta$

Potential due to this element at the centre
$dV=\frac{kdq}{R}=\frac{K.2cos\frac{\theta }{2}Rd\theta }{R}$
$=k.2cos\frac{\theta }{2}d\theta$

Potential at the centre due to the whole ring can be found by integrating this over the circumference of the ring.

$V={\int }_0^{2\pi }k.2cos\frac{\theta }{2}d\theta$

$=4k.(sin\pi -sin0)=0$