Question

A insulating rod is bent into a semicircular arc of radius a, and a total charge Q is distributed uniformly along the rod. Calculate the electric potential at the center of curvature of the arc if the potential is assumed to be zero at infinity.

Answer 1

Radius=a

Charge=Q

$V=\smallint \frac{dq}{4 \pi\epsilon_0 r}$

$dq=\lambda dl$

$dl=ad\theta$

$\lambda=\frac{\theta}{\pi a}$

$dq=\frac{\theta}{\pi a} (ad\theta)$

$dq=\frac{\theta d\theta}{\pi}$

Limit

$\theta=0°$ to $\theta=\pi$

$V=\frac{1}{4\pi\epsilon_0}\int\frac{dq}{a}$

$V=\frac{1}{4\pi\epsilon_0}\smallint _{0}^ \pi \frac{Q d\theta}{\pi a}$

$V=\frac{Q}{4 \pi\epsilon_0 a}$