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

`v = (kQ)/(a)`

Step-by-step explanation:

We define the linear density of charge as:

`\lambda = (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 = (kdq)/(r)`

where k is the coulomb's constant

r is the distance from dq to center of the circle

Thus.

`v = \int_{}^{}(kdq)/(a)`

`v = (k)/(a)\int_{}^{}dq`

`v = (kQ)/(a)` Potential at the center of the semicircle