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.

$\mathit{d}\mathit{q}\mathit{d}\mathit{σ}\mathbf{=}\mathbf{0}\mathit{d}\mathit{r}$ …… (1)

Here, $\mathit{σ}$is the charge density of the uniformly charged disk.

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

$\mathit{d}{\mathit{V}}_{\mathbf{w}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{d}\mathbf{q}}{\mathbf{r}}$ …… (2)

Substitute equation (1) in equation (2)

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

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

$$\begin{gathered} {\mathit{V}}_{\mathbf{w}}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{a}}{\mathbf{∫}}}\frac{\mathbf{σ}\mathbf{d}\mathbf{θ}\mathbf{d}\mathbf{r}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{d}\mathbf{θ}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\underset{\mathbf{0}}{\overset{\mathbf{a}}{\mathbf{∫}}}\mathit{d}\mathit{r} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{d}\mathbf{θ}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\left(a-0\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{a}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathit{d}\mathit{θ} \end{gathered}$$

${\mathit{V}}_{\mathbf{w}}\mathbf{=}\frac{\mathbf{σ²¹}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\mathit{d}\mathit{θ}$ …… (4)

Find the value of a from the above figure.

$\mathit{a}\mathbf{=}\mathbf{2}\mathit{R}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}$ …… (5)

Substitute the equation (5) in equation (4)

$$\begin{gathered} {\mathit{V}}_{\mathbf{w}}\mathbf{=}\frac{\mathbf{σ}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{2}\mathit{R}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}\mathit{θ} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{R}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}\mathit{d}\mathit{θ}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$ …… (6)

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

$$\begin{gathered} \mathit{V}\mathbf{=}\underset{\mathbf{-}\frac{\mathbf{Ï€}}{\mathbf{2}}}{\overset{\frac{\mathbf{Ï€}}{\mathbf{2}}}{\mathbf{∫}}}\frac{\mathbf{σ}\mathbf{R}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}\mathit{d}\mathit{θ} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{R}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\underset{\mathbf{-}\frac{\mathbf{Ï€}}{\mathbf{2}}}{\overset{\frac{\mathbf{Ï€}}{\mathbf{2}}}{\mathbf{∫}}}\mathit{c}\mathit{o}\mathit{s}\mathit{θ}\mathit{d}\mathit{θ} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{R}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}{\left[sin\mathrm{θ}\right]}_{\frac{\mathbf{Ï€}}{\mathbf{2}}}^{\frac{\mathbf{Ï€}}{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{R}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\left(1-\left(-1\right)\right) \end{gathered}$$

Solve further as,

$$\begin{gathered} \mathit{V}\mathbf{=}\frac{\mathbf{σ¸é}}{\mathbf{2}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\left(2\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ¸é}}{{\mathbf{Ï€Î\mu }}_{\mathbf{0}}} \end{gathered}$$

Hence, the potential due to uniformly charge disk on is rim is $\mathit{V}\mathbf{=}\frac{\mathbf{σ¸é}}{{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}$.