91Ó°ÊÓ

Consider the below figure,

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

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

$\mathbf{dq}\mathbf{=}\mathbf{σ°ù»åθ»å°ù}$ …… (1)

Here,$\mathrm{σ}$ 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.

${\mathbf{dV}}_{\mathbf{w}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\frac{\mathbf{dq}}{\mathbf{r}}$ …… (2)

Substitute equation (1) in equation (2)

$d{V}_{\mathit{w}}\mathit{=}\frac{\mathit{1}}{\mathit{4}\mathit{Ï€}{\mathit{Î\mu }}_{\mathit{0}}}\frac{\mathit{σ}\mathit{r}\mathit{d}\mathit{θ}\mathit{d}\mathit{r}}{\mathit{r}}$ …… (3)

localid="1657512915252" $=\frac{\mathrm{σ}d\mathrm{θ}dr}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}$

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

$V_w={∫}_0^a\frac{\mathrm{σ}d\mathrm{θ}dr}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}$

$=\frac{\mathrm{σ}d\mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_0^{a}dr$

$=\frac{\mathrm{σ}d\mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(a-0\right)$

localid="1657513173454" $=\frac{\mathrm{σ}a}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}d\mathrm{θ}$

$V=\frac{\mathrm{σ}a}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}d\mathrm{θ}$ ........(4)

Find the value of from the above figure.

$a=2R\cos \mathrm{θ}……\left(5\right)$

Substitute the equation (5) in equation (4)

$V_w=\frac{\mathrm{σ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}2R\cos \mathrm{θ}d\mathrm{θ}.........\left(6\right)$

$\frac{\mathrm{σ}R}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\cos \mathrm{θ}d\mathrm{θ}$

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

$V={∫}_{-\frac{\mathrm{Ï€}}{2}}^{\frac{\mathrm{Ï€}}{2}}\frac{\mathrm{σ}R}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\cos \mathrm{θ}d\mathrm{θ}$

localid="1657513962264" $=\frac{\mathrm{σ}R}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{∫}_{-\frac{\mathrm{Ï€}}{2}}^{\frac{\mathrm{Ï€}}{2}}\cos \mathrm{θ}d\mathrm{θ}$

localid="1657514143650" $$\begin{gathered} =\frac{\mathrm{σ}R}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}{\left[\sin \mathrm{θ}\right]}_{-\frac{\mathrm{Ï€}}{2}}^{\frac{\mathrm{Ï€}}{2}} \\ =\frac{\mathrm{σ}R}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}(1-(-1\left)\right) \end{gathered}$$

Solve further as,

$$\begin{gathered} V=\frac{\mathrm{σ}R}{2\mathrm{Ï€}{\mathrm{Î\mu }}_0}\left(2\right) \\ =\frac{\mathrm{σ}R}{\mathrm{Ï€}{\mathrm{Î\mu }}_0} \end{gathered}$$

Hence, the potential due to uniformly charge disk on is rim is $V=\frac{\mathrm{σ}R}{\mathrm{Ï€}{\mathrm{Î\mu }}_0}$.