P19-120 – Computation: ring of charge

Electric Field Created by a Uniform Circular Ring of Charge

We consider a ring of radius $R$ carrying a charge $Q$ distributed uniformly.

Because the ring 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}{2\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 (upward) and has no horizontal component. Indeed, for any (blue) infinitesimal charge $dQ$ located on the left side of the ring, an equivalent (red) infinitesimal charge $dQ$ can be found on the right side of the ring. 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 (upward). 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{cos} \left(\phi \right)\ }=\frac{kdQ}{R^2+h^2}\cdot {\mathrm{cos} \left(\phi \right)\ }

Recalling that ${\mathrm{cos} \left(\phi \right)\ }=h/\sqrt{R^2+h^2}$ 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{kdQ}{R^2+h^2}\cdot {\mathrm{cos} \left(\phi \right)\ }=\frac{\lambda khR}{{\left(R^2+h^2\right)}^{3/2}}d\theta

To find the magnitude $E_y$ of the electric field, we integrate $dE_y$ over $\theta $ from $\theta =0$ to $\theta =2\pi $ which yields

E_y=\int^{2\pi }_0{\frac{\lambda khR}{{\left(R^2+h^2\right)}^{3/2}}d\theta }=\frac{2\pi \lambda khR}{{\left(R^2+h^2\right)}^{3/2}}

The electric field created by the circular ring at point $P$ is equal to

\boxed{\overrightarrow{E}=\frac{2\pi \lambda khR}{{\left(R^2+h^2\right)}^{3/2}}\ \hat{y}=\frac{kQh}{{\left(R^2+h^2\right)}^{3/2}}\ \hat{y}}

Note: if $h=0$ we find that $E=0$ at the center of the ring which we expect from symmetry.