91Ó°ÊÓ

Given that, R is the radius of the hemispherical bowl, is the charge density of the hemispherical bowl.

Calculate the potential at the center of hemispherical bowl.

${\mathit{V}}_{\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{e}\mathbf{r}}\mathbf{=}\frac{\mathbf{σ}\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}{\mathbf{∫}}_{\mathbf{r}}\mathbf{-}\mathit{d}\mathit{a}$ …… (1)

Then, ${\mathit{V}}_{\mathbf{center}}\mathbf{=}\frac{\mathbf{σ}\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}{\mathbf{∫}}_{}\mathbf{-}\mathit{d}\mathit{a}$

Here, $\mathbf{∫}\mathit{d}\mathit{a}$is the surface area of hemisphere. $\mathbf{∫}\mathit{d}\mathit{a}\mathbf{=}\mathbf{2}\mathit{π}{\mathit{R}}^{\mathbf{2}}$.

Thus, the potential at the center of hemispherical bowl is,

$$\begin{gathered} {\mathit{V}}_{\mathbf{center}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\frac{\mathbf{σ}}{\mathbf{R}}\mathbf{2}\mathit{Ï€}{\mathit{R}}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ¸é}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}} \end{gathered}$$

${\mathit{V}}_{\mathbf{center}}\mathbf{=}\frac{\mathbf{σ¸é}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}$ ………. (2)

Write the expression for potential at the North Pole using equation (1)

${\mathit{V}}_{\mathbf{p}\mathbf{o}\mathbf{l}\mathbf{e}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{∫}\frac{\mathbf{σ}}{\mathbf{r}}\mathit{d}\mathit{a}$

Here, it is not necessary to integrate the term with respect to $\mathit{θ}$. Now consider the pole. According to the diagram the pole is overhead of the point of consideration. It makes the angle $\mathit{θ}$to 0.

Considering pole,

$\mathit{d}\mathit{a}\mathbf{=}\mathbf{2}\mathit{π}{\mathit{R}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathit{d}\mathit{θ}$

$$\begin{gathered} {\mathit{r}}^{\mathbf{2}}\mathbf{=}{\mathit{R}}^{\mathbf{2}}\mathbf{+}{\mathit{R}}^{\mathbf{2}}\mathbf{-}\mathbf{2}{\mathit{R}}^{\mathbf{2}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{θ} \\ {\mathit{r}}^{\mathbf{2}}\mathbf{=}\mathbf{2}{\mathit{R}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{}\mathbf{θ}\mathbf{)} \\ \mathbf{}\mathit{r}\mathbf{=}\mathit{R}\sqrt{\mathbf{2}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{³¦´Ç²õθ}\mathbf{)}} \end{gathered}$$

Therefore, the pole is calculated as,

$$\begin{gathered} {\mathit{V}}_{\mathbf{pole}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{0}}}\frac{\mathbf{σ}\mathbf{\left(}\mathbf{2}{\mathbf{Ï€¸é}}^{\mathbf{2}}\mathbf{\right)}}{\mathbf{R}\sqrt{\mathbf{2}}}\underset{\mathbf{0}}{\overset{\mathbf{Ï€}\mathbf{/}\mathbf{2}}{\mathbf{∫}}}\frac{\mathbf{²õ¾\pm ²Ôθ»åθ}}{\sqrt{\mathbf{1}\mathbf{-}\mathbf{³¦´Ç²õθ}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ¸é}}{\sqrt{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}}{\mathbf{\left(}\mathbf{2}\sqrt{\mathbf{1}\mathbf{-}\mathbf{³¦´Ç²õθ}}\mathbf{\right)}}_{\mathbf{0}}^{\mathbf{Ï€}\mathbf{/}\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ¸é}}{\sqrt{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{0}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ¸é}}{\sqrt{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}} \end{gathered}$$

Therefore, the north pole is $\frac{\mathbf{σ¸é}}{\sqrt{\mathbf{2}}{\mathbf{Î\mu }}_{\mathbf{0}}}$.

$$\begin{gathered} {\mathit{V}}_{\mathbf{p}\mathbf{o}\mathbf{l}\mathbf{e}}\mathbf{-}{\mathit{V}}_{\mathbf{p}\mathbf{o}\mathbf{l}\mathbf{e}}\mathbf{=}\frac{\mathbf{σ}\mathbf{R}}{\sqrt{\mathbf{2}}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{-}\frac{\mathbf{σ}\mathbf{R}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ}\mathbf{R}}{\sqrt{\mathbf{2}}{\mathbf{Î\mu }}_{\mathbf{0}}}\left[1-\frac{1}{\sqrt{2}}\right] \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{σ¸é}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)} \end{gathered}$$

Hence, the potential difference between the "north pole" and the center is $\frac{\mathbf{σ¸é}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}$.