By thinking about the physical situation, predict the magnitude of the electric field at the center of a uniformly charged ring of radius $\mathit{R}$ carrying a charge role="math" localid="1668494008173" $\mathbf{}\mathbf{+}\mathbf{}\mathit{Q}$ . Then use the equation derived in the text to confirm this result.

A uniformly charged ring of radius $R$ carrying a charge $+Q$.

The electric field at a distance $\mathit{z}$ on the axis of a ring of radius $\mathit{R}$and carrying a uniformly distributed charge role="math" localid="1668494241402" $\mathit{Q}$ is

role="math" localid="1668494255657" $\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{Q}\mathbf{z}}{{\left(R^2+z^2\right)}^{\mathbf{3}\mathbf{/}\mathbf{2}}}$ …(i)

Here, role="math" localid="1668494329799" ${\mathit{\epsilon }}_{\mathbf{0}}$ is the permittivity of free space.

If the ring is divided into infinitesimal parts of equal lengths, the field from one such part will be equal in magnitude but oppositely directed to the field due to a part diametrically opposite. Since the net field at the centre is the vector sum of fields due to the individual parts, all fields due to diametrically opposite parts will cancel each other. Thus the net field will be zero at the centre.

For the field at the centre, $z$ will be zero in equation (i). Thus

$$\begin{gathered} E\left(z=0\right)=\frac{1}{4\pi {\epsilon }_0}\frac{Q\times 0}{{\left(R^2+0^2\right)}^{3/2}} \\ =0 \end{gathered}$$

Hence, it is proved that the field at the centre is zero both with physical arguments and from the formula for electric field.