P19B2016 – Two Half Rings of Charge

If the goal is to derive the electric field created by a charge distribution, then Gauss’s Law can be applied if the charge distribution has sufficient symmetry. In the case of a ring of charge, Gauss’s Law fails such a distribution does not satisfy the key condition to apply Gauss’s Law: the electric field must be constant in magnitude and direction on a surface that is simple enough that its area can easily be computed.

In general, Gauss’s Law applies to the infinite line or cylinder of charge, the infinite sheet or slab of charge, and any distribution with a spherical symmetry (point charge, solid sphere, shell, etc.).

We consider the contribution of an infinitesimal arc length $d\ell $ with infinitesimal charge $dQ=\lambda d\ell $ on either side of the upper ring of charge.

By symmetry, we conclude that the net electric field at 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 wire, an equivalent (red) infinitesimal charge $dQ$ can be found on the right side. 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 upper ring of charge, 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 upper ring of charge at the center is equal to

\boxed{{\overrightarrow{E}}_+=-\frac{2k\lambda }{R}\ \hat{y}}

We repeat the above derivation with the lower half ring which creates an electric field that points downward at the center because it carries a negative charge density $-\lambda $.

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 lower ring of charge, 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 lower ring of charge at the center is equal to

\boxed{{\overrightarrow{E}}_-=-\frac{2k\lambda }{R}\ \hat{y}}

By superposition, the net electric field at the center of the charge distribution is equal to

\boxed{\overrightarrow{E}={\overrightarrow{E}}++{\overrightarrow{E}}-=-\frac{2k\lambda }{R}\ \hat{y}-\frac{2k\lambda }{R}\ \hat{y}=-\frac{4k\lambda }{R}\ \ \hat{y}}