(a) Using Eq. 24-32, show that the electric potential at a point on the central axis of a thin ring (of charge q and radius R ) and at distance Z from the ring is,$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{o}}}\mathbf{\cdot }\frac{\mathbf{q}}{\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}}}$

(b) From this result, derive an expression for the electric field magnitude

E at points on the ring’s axis; compare your result with the calculation of E in Module 22-4.

The electric potential at a point on the central axis of a thin ring and at distance z from the ring is$V=\frac{1}{4\pi {\epsilon }_o}\underset{}{\overset{}{}}\frac{dq}{r}$ .

Figure drawn below.Consider a ring to be lying in an $\mathbf{x}\mathbf{-}\mathbf{y}$plane.

Now, to find the value of the electric potential at pon the z-axis. Let qbe the charge that is uniformly distributed in the ring.

Formula:

The electric potential is define by,

$V=\frac{1}{4\pi {\epsilon }_o}\cdot \frac{q}{R}$

Where, V is electric potential, R is distance between the point charges, q is the charge, and ${\epsilon }_0$ is the permittivity of free space.

The electric potential at p is given as,

$V=\frac{1}{4\pi {\epsilon }_o}\underset{}{\overset{}{}}\frac{dq}{r}$

Here,

The distance,$r=\sqrt{R^2+z^2}$

Therefore, the electric potential will be,

$\begin{array}{c}V=\frac{1}{4\pi {\epsilon }_o}\frac{1}{\sqrt{R^2+z^2}}\int dq\\ =\frac{1}{4\pi {\epsilon }_o}\frac{q}{\sqrt{R^2+z^2}}\end{array}$

Hence, proved.

According to the definition of the electric potential gradient,

$$\begin{gathered} \begin{array}{c}E=-\frac{dV}{dz}\\ =-\frac{d}{dz}\left(\frac{1}{4\pi {\epsilon }_o}\frac{q}{\sqrt{R^2+z^2}}\right)\\ =-\frac{1}{4\pi {\epsilon }_o}\{-\frac{1}{2}{(R^2+z^2)}^{-3/2}\left(2z\right)q\}\end{array} \\ E=\frac{1}{4\pi {\epsilon }_o}\frac{qz}{\sqrt{{(R^2+z^2)}^3}} \end{gathered}$$

Hence, expression for electric potential gradient is $E=\frac{1}{4\pi {\epsilon }_o}\frac{qz}{\sqrt{{(R^2+z^2)}^3}}$.