Question

A disk with a hole has inner radius Rin and outer radius Rout.The disk is uniformly charged with total charge Q. Find an expression for the on-axis electric potential at distance z from the center of the disk. Verify that your expression has the correct behavior when RinS0.

Short answer

The expression for the on-axis electric potential is$V=\frac{2Qk}{R_o^2-R_i^2}\sqrt{R_o^2+z^2}-\sqrt{R_i^2+z^2}$

Step-by-step solution

Given Information

We have given that

The inner radius of the desk is$R_{in}$,

the outer radius of the desk is$R_{out}$,

and the total charge is$Q$

Electric potential

Let's consider the charge distribution. Taking the ring infinitesimally small, that is, a line, the surface charge density $eta$andthe charge contained in this ring is given by

$dQ=2\mathrm{\pi rndr}$.

The surface charge density is

$n=\frac{Q}{\mathrm{\pi }({\mathrm{R}}_{\mathrm{o}}^2-{\mathrm{R}}_{\mathrm{i}}^2)}$

The expression for the charge contained in the ring as

$\frac{dQ}{dr}=\frac{2Q}{R_o^2-R_i^2}r$

Considering the geometry, from the pythagorean theorum, the distance of the ring from the point we're calculating the potential is given as

$\sqrt{r^2+z^2}$,

Whereas one might expect, the radius runs from $R_i$to $R_o$. Combining, we get the expression for the potential as

$V=k{\int }_{R_i}^{R_0}\frac{2Qr/\left(R_0^2-R_i^2\right)}{\sqrt{r^2+z^2}}dr$

simplify

Let us consider that the solution for such an integral will be

$\int \frac{xdx}{\sqrt{x^2\pm z^2}}=\sqrt{x^2+a^2}$

By so, our expression can be evaluated as

$V=\frac{2Qk}{R_0^2-R_i^2}{\int }_{R_i}^{R_0}\frac{rdr}{\sqrt{r^2+z^2}}=\frac{2Qk}{R_0^2-R_i^2}{\left[\sqrt{r^2+z^2}\right]}_{R_i}^{R_0}$

Putting in the values, we get

$V-\frac{2Qk}{R_o^2-R_i^2}\sqrt{R_o^2+z^2}-\sqrt{R_i^2+z^2}$

If we set $R_i=0$, one can see that the expression becomes

$\frac{V={\displaystyle \frac{2Qk}{R_o^2}}\sqrt{R_o^2+z^2}-z}{}$,

which is the expression that was derived in the book. So we can be sure about our solution.