MODEL Because the ring is thin, we’ll assume the charge lies along a circle of radius R.

VISUALIZE FIGURE 25.30 illustrates the problem-solving strategy. We’ve chosen a coordinate system in which the ring lies in the xy-plane and point P is on the z-axis. We’ve then divided the ring into N small segments of charge $\text{Δ}Q$, each of which can be modeled as a point charge. The distance $r_i$ between segment i and point P is

Note that $r_i$ is a constant distance, the same for every charge segment.

FIGURE 25.30 Finding the potential of a ring of charge.

SOLVE The potential V at P is the sum of the potentials due to each segment of charge:

We were able to bring all terms involving z to the front because z is a constant as far as the summation is concerned. Surprisingly, we don’t need to convert the sum to an integral to complete this calculation. The sum of all the $\text{Δ}Q$ charge segments around the ring is simply the ring’s total charge, $\sum \left(\text{Δ}Q\right)=Q;$ hence the electric potential on the axis of a charged ring is

REVIEW From far away, the ring appears as a point charge Q in the distance. Thus we expect the potential of the ring to be that of a point charge when $z\gg R$. You can see that $V_{\text{ring}}\approx Q\text{/}4\pi {ϵ}_{0}z$ when $z\gg R$, which is, indeed, the potential of a point charge Q.

MODEL Model the disk as a uniformly charged disk of zero thickness, radius R, and charge Q. The disk has uniform surface charge density $\eta =Q\text{/}A=Q\text{/}\pi R^2$. We can take advantage of now knowing the on-axis potential of a ring of charge.

VISUALIZE Orient the disk in the xy-plane, as shown in FIGURE 25.31, with point P at distance z. Then divide the disk into rings of equal width $\text{Δ}r$. Ring i has radius $r_i$ and charge $\text{Δ}{Q}_i$.

FIGURE 25.31 Finding the potential of a disk of charge.

SOLVE

1. We can use the result of Example 25.10 to write the potential at distance z of ring i as

   $$V_i=\frac{1}{4\pi {ϵ}_0}\frac{\text{Δ}{Q}_i}{\sqrt{r_i{}^2+z^2}}$$

   The potential at P due to all the rings is the sum

   $$V={\displaystyle \sum _i{V}_i={\displaystyle \frac{1}{4\pi {ϵ}_0}}\underset{i=1}{\overset{N}{{{\displaystyle \sum }}^{\text{}}}}{\displaystyle \frac{\text{Δ}{Q}_i}{\sqrt{r_i^2+z^2}}}}$$

   The critical step is to relate $\text{Δ}{Q}_i$ to a coordinate. Because we now have a surface, rather than a line, the charge in ring i is $\text{Δ}{Q}_i=\eta \text{Δ}{A}_i$, where $\text{Δ}{A}_i$ is the area of ring i. We can find $\text{Δ}{A}_i$, as you’ve learned to do in calculus, by “unrolling” the ring to form a narrow rectangle of length $2\pi r_i$ and width $\text{Δ}r$. Thus the area of ring i is $\text{Δ}{A}_i=2\pi r_i\text{Δ}r$ and the charge is

   $$\text{Δ}{Q}_i=\eta \text{Δ}{A}_i={\displaystyle \frac{Q}{\pi R^2}}2\pi r_i\text{Δ}r={\displaystyle \frac{2Q}{R^2}}{r}_i\text{Δ}r$$

   With this substitution, the potential at P is

   $$V=\frac{1}{4\pi {ϵ}_0}\sum _{i=1}^N\frac{2Q}{R^2}\frac{r_i\text{Δ}{r}_i}{\sqrt{r_i{}^2+z^2}}\to \frac{Q}{2\pi {ϵ}_0{R}^2}{\displaystyle {\int }_0^R\frac{rdr}{\sqrt{r^2+z^2}}}$$

   where, in the last step, we let $N\to \infty$ and the sum become an integral. This integral can be found in Appendix A, but it’s not hard to evaluate with a change of variables. Let $u=r^2+z^2$, in which case $rdr=\frac{1}{2}du$. Changing variables requires that we also change the integration limits. You can see that $u=z^2$ when $r=0$, and $u=R^2+z^2$ when $r=R$. With these changes, the on-axis potential of a charged disk is

   $$\begin{array}{ll}{V}_{\text{disk}\text{on}\text{axis}}\hfill & =\frac{Q}{2\pi {ϵ}_0{R}^2}{{\displaystyle \int }}_{z^2}^{R^2+z^2}\frac{\frac{1}{2}du}{u^{1\text{/}2}}=\frac{Q}{2\pi {ϵ}_0{R}^2}{u}^{1\text{/}2}{|}_{z^2}^{R^2+z^2}\hfill \\ \hfill & =\frac{Q}{2\pi {ϵ}_0{R}^2}\left(\sqrt{R^2+z^2}-z\right)\hfill \end{array}$$

2. To calculate the potential energy, we first need to determine the potential of the disk at $z=0.0100\text{m}$. Using $R=0.0175\text{m}$ and $Q=5.00$ nC, you can calculate $V=2980\text{\hspace{0.17em}V}$. The electron’s charge is $q=-e=-1.60\times {10}^{-19}\text{C,}$ so the potential energy with an electron at $z=1.00\text{cm}$ is $U_{\text{elec}}=qV=-4.77\times {10}^{-16}\text{J}$.

REVIEW We earlier found that the potential 1 cm from a 1 nC point charge is 900 V. Thus a potential of 2980 V at the same distance from a 5 nC disk seems reasonable.