Question

A uniformly charged rod of length l is bent into the shape of a semicircle..ıt has uniform charge density labmda and radius R. Find the elektric potential at the center of the semicircle

Answer 1

A uniformly charged rod of length $l$ is bent into the shape of a semicircle. It has uniform charge density lambda and radius $R$. Find the electric potential at the center of the semicircle

Electric potential from small part of semicircle:

$$d\varphi = \frac{1}{4\pi\varepsilon_0} \frac{dq}{R} = \frac{1}{4\pi\varepsilon_0} \frac{\lambda dl}{R}$$

Sum potential from all small parts:

$$\varphi = \int_0^l \frac{1}{4\pi\varepsilon_0} \frac{\lambda dl}{R} = \frac{1}{4\pi\varepsilon_0} \frac{\lambda l}{R}$$

Answer: $\varphi = \frac{1}{4\pi\varepsilon_0} \frac{\lambda l}{R}$