P19-110 – Computation: semicircle of charge

Electric Field Created by a Uniform Semicircular Loop of Charge

We consider a semicircle of radius $R$ carrying a charge $Q$ distributed uniformly along its length.

Because the semicircle is uniformly charged, we define the linear charge density $\lambda $ which represents the charge per unit length of the rod and is equal to

\lambda =\frac{Q}{\pi R}

The charge $Q$ is continuously distributed along the semicircular rod and we therefore consider the contribution of an infinitesimal arc length $d\ell $ with infinitesimal charge $dQ=\lambda d\ell $.

By symmetry, we conclude that the net electric field at point $P$ is directed along the $y$-axis (downward) and has no horizontal component. Indeed, for any (blue) infinitesimal charge $dQ$ located on the left side of the semicircular rod, an equivalent (red) infinitesimal charge $dQ$ can be found on the right side of the rod. When considering the infinitesimal electric fields they create at point $P$, we conclude that their horizontal components cancels and that, therefore, the net electric field at point $P$ has no horizontal component and is vertical (downward). We therefore only seek to compute the $y$-component of the electric field at point $P$.

To derive the net electric field at point $P$, we consider one of the charges $dQ$ and argue that it creates an infinitesimal field $d\overrightarrow{E}$ with a $y$-component $dE_y$ given by

dE_y=dE{\mathrm{sin} \left(\theta \right)\ }=\frac{kdQ}{R^2}{\mathrm{sin} \left(\theta \right)\ }

Recalling that $dQ=\lambda d\ell $ and $d\ell =Rd\theta $, we conclude that the $y$-component of the infinitesimal electric field $d\overrightarrow{E}$ is equal to

dE_y=\frac{k\lambda d\ell }{R^2}{\mathrm{sin} \left(\theta \right)\ }=\frac{k\lambda }{R}{\mathrm{sin} \left(\theta \right)\ }d\theta

To find the magnitude $E_y$ created by the semicircular rod, we integrate over the entire semicircular wire from $\theta =0$ to $\theta =\pi $ which yields

E_y=\int^{\theta =\pi }_{\theta =0}{\frac{k\lambda }{R}{\mathrm{sin} \left(\theta \right)\ }d\theta }=-\frac{k\lambda }{R}\cdot {\left[{\mathrm{cos} \left(\theta \right)\ }\right]}^{\pi }_0=-\frac{k\lambda }{R}\left[\left(-1\right)-1\right]=\frac{2k\lambda }{R}

The electric field created by the semicircular rod at point $P$ is equal to

\boxed{\overrightarrow{E}=\frac{2k\lambda }{R}\hat{y}=\frac{2kQ}{\pi R^2}\hat{y}}