P19-130 – Computation: disk of charge

Electric Field Created by a Uniform Disk of Charge

We consider a disk of radius $R$ carrying a charge $Q$ distributed uniformly over its surface.

Because the disk is uniformly charged, we define the surface charge density $\sigma $ which represents the charge per unit length of the disk and is equal to

\sigma =\frac{Q}{\pi R^2}

The charge $Q$ is continuously distributed over the disk and we therefore consider the contribution of an infinitesimal area $dA=rd\theta dr$ with infinitesimal charge $dQ=\sigma dA$.

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 $dA=rdrd\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{kh\sigma }{{\left(r^2+h^2\right)}^{3/2}}rdrd\theta

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

\begin{aligned}
E_y&=\int^{\theta =2\pi }_{\theta =0}{\int^{r=R}_{r=0}{\frac{kh\sigma }{{\left(r^2+h^2\right)}^{3/2}}rdrd\theta }} = kh\sigma \cdot {\left[-\frac{1}{\sqrt{r^2+h^2}}\right]}^R_0\cdot {\left[\theta \right]}^{2\pi }_0 = 2\pi kh\sigma \cdot \left[\frac{1}{h}-\frac{1}{\sqrt{R^2+h^2}}\right]
\end{aligned}

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

\boxed{\overrightarrow{E}=2\pi kh\sigma \cdot \left[\frac{1}{h}-\frac{1}{\sqrt{R^2+h^2}}\right]\ \hat{y}=\frac{2khQ}{R^2}\cdot \left[\frac{1}{h}-\frac{1}{\sqrt{R^2+h^2}}\right]\ \hat{y}}

Note: in the calculation above, we used the following antiderivative

\int{\frac{x}{{\left(x^2+a^2\right)}^{3/2}}}dx=-\frac{1}{\sqrt{x^2+a^2}}