Q51P - Introduction to Electrodynamics

Question

Find the potential on the rim of a uniformly charged disk (radius R 002C

charge density u ).

Step-by-Step Solution

Answer

The potential due to uniformly charge disk on is rim is  V=$\frac{\sigma R}{\pi {\epsilon }_0}$.

Step 1: Define functions

Consider the below figure,

                                     

Here, dr is the small element of wedge at a distance r from point A .

Now, write the expression for charge contained by this element.

$\mathbf{dq}\mathbf{=}\mathbf{\sigma rd\theta dr}$                                                              …… (1)

Here, $\sigma$ is the charge density of the uniformly charged disk.

Therefore, write the expression for potential at the point due to small element on the wedges.

${\mathbf{dV}}_{\mathbf{w}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{dq}}{\mathbf{r}}$                                                         …… (2)

Step 2: Determine potential at point due to entire wedge

Substitute equation (1) in equation (2)

$d{V}_{\mathit{w}}\mathit{=}\frac{\mathit{1}}{\mathit{4}\mathit{\pi }{\mathit{\epsilon }}_{\mathit{0}}}\frac{\mathit{\sigma }\mathit{r}\mathit{d}\mathit{\theta }\mathit{d}\mathit{r}}{\mathit{r}}$                             …… (3)

         $=\frac{\sigma d\theta dr}{4\pi {\epsilon }_0}$

Now, integrate the equation (3) to find out the potential at point A due to entire wedge.

  $V_w={\int }_0^a\frac{\sigma d\theta dr}{4\pi {\epsilon }_0}$

        $=\frac{\sigma d\theta }{4\pi {\epsilon }_0}{\int }_0^{a}dr$

        $=\frac{\sigma d\theta }{4\pi {\epsilon }_0}\left(a-0\right)$

        $=\frac{\sigma a}{4\pi {\epsilon }_0}d\theta$

     $V=\frac{\sigma a}{4\pi {\epsilon }_0}d\theta$                                  ........(4)

Step 3: Determine potential

Find the value of from the above figure.

$a=2R\cos \theta \dots \dots \left(5\right)$

 Substitute the equation (5) in equation (4)

$V_w=\frac{\sigma }{4\pi {\epsilon }_0}2R\cos \theta d\theta .........\left(6\right)$

         $\frac{\sigma R}{2\pi {\epsilon }_0}\cos \theta d\theta$

Now integrate the equation (6) from to $\mathit{-}\frac{\mathit{\pi }}{\mathit{2}}\mathrm{to}\frac{\mathit{\pi }}{\mathit{2}}$

$V={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\sigma R}{2\pi {\epsilon }_0}\cos \theta d\theta$

  $=\frac{\sigma R}{2\pi {\epsilon }_0}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\cos \theta d\theta$

  $$\begin{gathered} =\frac{\sigma R}{2\pi {\epsilon }_0}{\left[\sin \theta \right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \\ =\frac{\sigma R}{2\pi {\epsilon }_0}(1-(-1\left)\right) \end{gathered}$$

Solve further as,

$$\begin{gathered} V=\frac{\sigma R}{2\pi {\epsilon }_0}\left(2\right) \\ =\frac{\sigma R}{\pi {\epsilon }_0} \end{gathered}$$

Hence, the potential due to uniformly charge disk on is rim is  $V=\frac{\sigma R}{\pi {\epsilon }_0}$.