Question

The wire in FIGURE P25.73 has linear charge density $\lambda$. What is the electric potential at the center of the semicircle?

Short answer

The electric potential at the center of the semicircle is$V=k\lambda \left(2In\right(3)+\mathrm{\pi })$.

Step-by-step solution

Given Information

We need to find the electric potential at the center of the semicircle.

Simplify

The potential at the point of interest will be the sum of potentials coming from the straight lines and the curved section. By symmetry, it is clear that the potential coming from the two straight segments are equal. So, we can calculate one of them, and the potential coming from both will be twice in our result.

Therefore, we can write

$V=k\left(2{\int }_0^{2R}\frac{\lambda dz}{x+R}+{\int }_0^{\mathrm{\pi R}}\frac{\lambda dx}{R}\right)$

Factorizing the linear charge density, we can simplify

$V=k\lambda \left({\int }_0^{2R}\frac{dx}{x+R}+\frac{1}{R}{\int }_0^{\mathrm{\pi R}}dx\right)$

The result will be

$k\lambda \left(2\left[In\left|3R\right|-In\left|R\right|\right]+\mathrm{\pi }\right)=k\lambda \left(2In\left(3\right)+\mathrm{\pi }\right)$