Figure 24-64 shows a ring of outer radius $\mathbf{R}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{0}\mathbf{\text{cm}}$, inner radius $\mathbf{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{R}$ , and uniform surface charge density $\mathbf{\sigma }\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{pC}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$ . With $\mathbf{V}\mathbf{=}\mathbf{0}$at infinity, find the electric potential at point $\mathbf{P}$ on the central axis of the ring, at distance $\mathbf{z}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{R}$ from the center of the ring.

• Outer radius of the disk,$R=13.0\text{cm or 0.13m}$
• Inner radius of the disk, $r=0.200R$
• Uniform surface charge density,$\sigma =6.20pC/m^2$role="math" localid="1662627662240" $\text{or}6.20\times {10}^{-12}$$C/m^2$
• The potential at infinity,$V=0$
• The distance from the center of the ring, $z=2.00R$

According to the superposition law, in a particular situation when one or more bodies have potential value on a distant point, the consequent potential value is equal to the total of the potentials resulting from the bodies that are now present in the system.

Here, you are given a point that is at a distance from the center of the disk with a smaller hole in its radius. Thus, if you consider that this case is a mixture of two disks with oppositely charge densities, you can calculate the required potential value for the disk problem using the superposition theorem.

Formula:

The potential due to a charged disk on a point at a given distance$(z>0)$,

$V=\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-z\right)$ ….. (i)

Where, $\sigma$ is the surface charge density of the disk, $R$ is the radius of the disk, ${\epsilon }_0$ is the permittivity in vacuum, $Z$ is the distance of the point from the center of the disk.

The given system is treated as a superposition of a disk of surface charge density $\sigma$ and radius $R$with a smaller, oppositely charged, the disk of surface charge density $-\sigma$ and radius $r$ as the smaller disk creates a hole.

Thus, the net electric potential at the point P can be given using equation (i) as follows:

$$\begin{gathered} V_{net}=\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-z\right)+\frac{\left(-\sigma \right)}{2{\epsilon }_0}\left(\sqrt{z^2+r^2}-z\right) \\ =\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{z^2+R^2}-\sqrt{z^2+r^2}\right) \end{gathered}$$

As required this equation vanishes as $r\to \infty$.

Substitute $2R$for $z$ and $0.2R$ for $r$ in the above equation.

role="math" localid="1662629432710" $$\begin{gathered} V_{net}=\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{{\left(2R\right)}^2+R^2}-\sqrt{{\left(2R\right)}^2+{\left(0.2R\right)}^2}\right) \\ =\frac{\sigma }{2{\epsilon }_0}\left(\sqrt{5R^2}-\sqrt{4.04R^2}\right) \\ =\frac{\sigma R}{{\epsilon }_0}\left(\frac{\sqrt{5}-\sqrt{4.04}}{2}\right) \end{gathered}$$

Substitute known numerical values in the above equation, and you get

$$\begin{gathered} V_{net}=\frac{\left(6.20\times {10}^{-12}C/m^2\right){\displaystyle \left(0.13m\right)}}{\left(8.85\times {10}^{-12}{C}^2/N.m^2\right)}\left(\frac{2.236-2.009}{2}\right) \\ =\left(0.0911\times 0.1135\right)\text{V} \\ =1.03\times {10}^{-2}\text{V} \end{gathered}$$

Hence, the required value of the electric potential at point $P$ is $1.03\times {10}^{-2}\text{V}$.