Question

A uniformly charged ring has linear charge density $4\frac{C}{m}$. What is the electric potential at its centre?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let’s look at the situation first.

This diagram represents the uniformly charged ring with linear charge density . The radius however is not provided in the question.We can calculate the potential due to a point charge so let’s divide the ring into infinitesimally small parts and try to find potential at O due to one such part.

So here, we take a small part dl located at an angle $\theta$from some reference line subtending $d\theta$ at the centre. The charge contained in the small element dl, dQ=$\lambda$ dl and its potential at the centre will be $\frac{kdQ}{r}$
So $dV=\frac{kdQ}{r}=\frac{k\lambda dl}{r}$. Since potential is a scalar quantity we don’t have to worry about directions like we did in case of electric field. All we need to do now is integrate this to get the answer
\(\cdot \textstyle \int dV = V = \textstyle \int_0^{2\pi}k\lambda \frac{dl}{r}\)
Now in this integral k, \( \lambda \) and r are constants. So the integral becomes
$\cdot V=\frac{k\lambda }{r}{\int }_0^{2\pi }dl=\frac{k\lambda }{r}.2\pi r=\frac{1}{4\pi {ϵ}_0}\lambda 2\pi =\frac{\lambda }{2{ϵ}_0}$
Good thing that we got rid of the radius as it was not provided in the first place.