Question

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

Answer 1

Answer:

$v = \frac{kQ}{a}$  

Explanation:

We define the linear density of charge as:

$\lambda = \frac{Q}{L}$

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

$dv = \frac{kdq}{r}$

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

$v = \int_{}^{}\frac{kdq}{a}$  

$v = \frac{k}{a}\int_{}^{}dq$

$v = \frac{kQ}{a}$     Potential at the center of the semicircle