Question

A rod with linear charge density $\lambda$ is bent in the shape of circular ring. The electric potential at the center of the circular ring is :

• A. $\frac{\lambda }{4{\epsilon }_0}$
• B. $\frac{\lambda }{2{\epsilon }_0}$
• C. $\frac{\lambda }{{\epsilon }_0}$
• D. $\frac{3\lambda }{4{\epsilon }_0}$

Answer 1

The correct option is B $\frac{\lambda }{2{\epsilon }_0}$

let L be the length of nod
$\Rightarrow 2\pi R=L$
We know potential $=\frac{kQ}{R}$

$Q=\lambda \times totallength$

$V=\frac{k\lambda \times L}{\left(\frac{L}{2\pi }\right)}$

$=2\pi \lambda k$

And $k=\frac{1}{4\pi {ϵ}_0}$

$V=\frac{\lambda }{2{ϵ}_0}$