91Ó°ÊÓ

Given that, R is the radius of the hemispherical bowl, $\mathrm{σ}$is the charge density of the hemispherical bowl.

Calculate the potential at the center of hemispherical bowl.

$$\begin{gathered} {\mathit{V}}_{\mathbf{C}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{e}\mathbf{r}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï„}\mathbf{Ï„}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{∫}\frac{\mathbf{σ}}{\mathbf{r}}\mathbf{}\mathit{d}\mathit{a} \\ \mathrm{Then}{\mathit{V}}_{\mathbf{C}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{e}\mathbf{r}}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï„}\mathbf{Ï„}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{\mathbf{σ}}{\mathbf{R}}\mathbf{∫}\mathit{d}\mathit{a} \end{gathered}$$ .......(1)

Here, $∫da$is the surface area of hemisphere. $∫da=2{\mathrm{Ï€¸é}}^2$.

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

role="math" localid="1657536588790" $$\begin{gathered} V_{center}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{σ}}{R}2{\mathrm{Ï€¸é}}^2 \\ =\frac{\mathrm{σ¸é}}{2{\mathrm{Î\mu }}_0} \\ {\mathbf{V}}_{\mathbf{center}}\mathbf{=}\frac{\mathbf{σ¸é}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{0}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}.............\left(2\right) \end{gathered}$$

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

$V_{pole}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}∫{\frac{\mathrm{σ}}{r}}_{}^{}da$

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

Considering pole,

$$\begin{gathered} da=2{\mathrm{Ï€¸é}}^2\sin \mathbf{θ}\mathrm{d}\mathbf{θ} \\ {\mathrm{r}}^2={\mathrm{R}}^2+{\mathrm{R}}^2-2{\mathrm{R}}^2\cos \mathbf{θ} \\ {\mathrm{r}}^2=2{\mathrm{R}}^2(1-\cos \mathbf{θ}) \\ \mathrm{r}=\mathrm{R}\sqrt{2(1-\cos \mathbf{θ})} \end{gathered}$$

Therefore, the pole is calculated as,

role="math" localid="1657538218889" $$\begin{gathered} V_{pole}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{σ}\left(2{\mathrm{Ï€¸é}}^2\right)}{R\sqrt{2}}{∫}_0^{\mathrm{Ï€}/2}\frac{\sin \mathrm{θ}d\mathrm{θ}}{\sqrt{1-\cos \mathrm{θ}}} \\ =\frac{\mathrm{σ}R}{\sqrt{2}{\mathrm{Î\mu }}_0}{\left(2\sqrt{1-\cos \mathrm{θ}}\right)}_0^{\mathrm{Ï€}/2} \\ =\frac{\mathrm{σ}R}{\sqrt{2}{\mathrm{Î\mu }}_0}(1-0) \\ =\frac{\mathrm{σ}R}{\sqrt{2}{\mathrm{Î\mu }}_0} \end{gathered}$$

Therefore, the north pole is $\frac{\mathrm{σ}R}{\sqrt{2}{\mathrm{Î\mu }}_0}$.

$$\begin{gathered} V_{pole}-V_{center}=\frac{\mathrm{σ}R}{\sqrt{2}{\mathrm{Î\mu }}_0}-\frac{\mathrm{σ}R}{2{\mathrm{Î\mu }}_0} \\ =\frac{\mathrm{σ}R}{\sqrt{2}{\mathrm{Î\mu }}_0}\left[1-\frac{1}{\sqrt{2}}\right] \\ =\frac{\mathrm{σ}R}{2{\mathrm{Î\mu }}_0}(\sqrt{2}-1) \end{gathered}$$

Hence, the potential difference between the "north pole" and the center is $\frac{\mathrm{σ}R}{2{\mathrm{Î\mu }}_0}(\sqrt{2}-1)$.