P19B2015 – Charged Circular Ring

We consider a ring of radius $R$ and linear charge density $\lambda $ and the contribution of an infinitesimal arc length $d\ell $ with infinitesimal charge $dQ=\lambda d\ell $ to the electric field at point $P$ (located a height $h$ above the center of the ring).

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}}

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

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 $dQ=\sigma dA$ to the electric field at point $P$.

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{co}\mathrm{s} \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}}

To create an infinite plane with surface charge density $\sigma $, we can start with a disk of radius $R$ and surface charge density $\sigma $ (uniform) and then let $R\to +\infty $.

Given our answer to question 2., we conclude that the electric field created by an infinite plane with surface charge density $\sigma $ should be equal to

E_{plane}={\mathop{\mathrm{lim}}_{R\to +\infty } E_{disk}\ }

Recall that

{\mathop{\mathrm{lim}}_{R\to +\infty } \frac{1}{\sqrt{R^2+h^2}}\ }=0

Thus, we conclude that the electric field created by an infinite plane with surface charge density $\sigma $ is

\boxed{E_{plane}={\mathop{\mathrm{lim}}_{R\to +\infty } 2\pi kh\sigma \cdot \left[\frac{1}{h}-\frac{1}{\sqrt{R^2+h^2}}\right]\ }=2\pi kh\sigma \cdot \frac{1}{h}=2\pi k\sigma =\frac{\sigma }{2{\varepsilon }_0}}

This consistent with the result derived in lecture for the electric field created by an infinite plane (see Gauss’s Law).