P19B2016 – Pair of Charged Rings

The charge density $\lambda $ is uniform and we therefore consider the contribution of an infinitesimal arc length $d\ell $ with infinitesimal charge $dQ=\lambda d\ell $ as shown below.

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

dE_x=dE{\mathrm{cos} \left(\phi \right)\ }=\frac{kdQ}{R^2+x^2}\cdot {\mathrm{cos} \left(\phi \right)\ }

Recalling that ${\mathrm{cos} \left(\phi \right)\ }=x/\sqrt{R^2+x^2}$ and $d\ell =Rd\theta $, we conclude that the $x$-component of the infinitesimal electric field $d\overrightarrow{E}$ is equal to

dE_x=\frac{kdQ}{R^2+x^2}\cdot {\mathrm{cos} \left(\phi \right)\ }=\frac{\lambda kxR}{{\left(R^2+x^2\right)}^{3/2}}d\theta

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

E_x=\int^{2\pi }_0{\frac{\lambda kxR}{{\left(R^2+x^2\right)}^{3/2}}d\theta }=\frac{2\pi \lambda kxR}{{\left(R^2+x^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 kRx}{{\left(R^2+x^2\right)}^{3/2}}\ \hat{x}}

The ring on the left creates an electric field $\overrightarrow{E_+}\left(x\right)$ at an arbitrary location on the $x$-axis given by

\boxed{\overrightarrow{E_+}\left(x\right)=\frac{2\pi \lambda kR\left(x+\ell \right)}{{\left(R^2+{\left(x+\ell \right)}^2\right)}^{3/2}}\ \hat{x}}

At that same location, the ring on the right creates an electric field $\overrightarrow{E_-}\left(x\right)$ given by

\boxed{\overrightarrow{E_-}\left(x\right)=-\frac{2\pi \lambda kR\left(x-\ell \right)}{{\left(R^2+{\left(x-\ell \right)}^2\right)}^{3/2}}\ \hat{x}}

By superposition, the net electric field at an arbitrary location $x$ on the symmetry axis is given by

\boxed{\overrightarrow{E}\left(x\right)=\overrightarrow{E_+}\left(x\right)+\overrightarrow{E_-}\left(x\right)=\left[\frac{2\pi \lambda kR\left(x+\ell \right)}{{\left(R^2+{\left(x+\ell \right)}^2\right)}^{3/2}}-\frac{2\pi \lambda kR\left(x-\ell \right)}{{\left(R^2+{\left(x-\ell \right)}^2\right)}^{3/2}}\right]\ \hat{x}}

As $x\to +\infty $, the system begins to look like two point charges $Q_+=2\pi R\lambda $ and $Q_-=-2\pi R\lambda $ and therefore the electric field it creates is equivalent to the electric field created by an electric dipole.