P19B2014 – Semicircular Charged Ring

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 the center 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 the center, we conclude that their horizontal components cancels and that, therefore, the net electric field at the center has no horizontal component and is vertical (downward). We therefore only seek to compute the $y$-component of the electric field at the center.

To derive the net electric field at the center, we consider one of the slices with charge $dQ$ 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 the center is equal to

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