An inverted hemispherical bowl of radius Rcarries a uniform surface charge density .Find the potential difference between the "north pole" and the center.

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

Calculate the potential at the center of hemispherical bowl.

${\mathit{V}}_{\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{r}\mathbf{e}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{\sigma }}{\mathbf{r}}\mathit{d}\mathit{a}$ ......(1)

Then, ${\mathit{V}}_{\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{r}\mathbf{e}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{\sigma }}{\mathbf{R}}\mathbf{\int }\mathit{d}\mathit{a}$

Here,role="math" localid="1657685113411" $\int da$ is the surface area of hemisphere. .role="math" localid="1657685129991" $\int da=2\pi R^2$

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

role="math" localid="1657685639169" $$\begin{gathered} V_{centre}=\frac{1}{4\pi {\epsilon }_0}\frac{\sigma }{R}2\pi R^2 \\ =\frac{\sigma R}{2{\epsilon }_0} \\ {\mathit{V}}_{\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{r}\mathbf{e}}\mathbf{=}\frac{\mathbf{\sigma }\mathbf{R}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}} \end{gathered}$$

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

$V_{pole}=\frac{1}{4\pi {\epsilon }_0}\int \frac{\sigma }{r}da$

Here, it is not necessary to integrate the term with respect to role="math" localid="1657685924325" $\mathit{\theta }$. Now consider the pole. According to the diagram the pole is overhead of the point of consideration. It makes the angle $\mathit{\theta }$to 0.

Considering pole,

$$\begin{gathered} da=2\pi R^2\sin \mathit{\theta }d\mathit{\theta } \\ {r}^2=R^2+R^2-2R^2\cos \mathit{\theta } \\ {r}^2=2R^2\left(1-\cos \mathit{\theta }\right) \\ r=R\sqrt{2\left(1-\cos \mathit{\theta }\right)} \end{gathered}$$

Therefore, the pole is calculated as,

$$\begin{gathered} V_{pole}=\frac{1}{4\pi {\epsilon }_0}\frac{\sigma \left(2\pi R^2\right)}{R\sqrt{2}}{\int }_0^{\pi /2}\frac{\sin \mathit{\theta }d\mathit{\theta }}{\sqrt{1-\cos \mathit{\theta }}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}=\frac{\sigma R}{\sqrt{2{\mathit{\epsilon }}_0}}{\left(2\sqrt{1-\cos \theta }\right)}_0^{\pi /2} \\ =\frac{\sigma R}{\sqrt{2{\epsilon }_0}}\left(1-0\right) \\ =\frac{\sigma R}{\sqrt{2{\mathit{\epsilon }}_0}} \end{gathered}$$

Therefore, the north pole is$\frac{\sigma R}{\sqrt{2{\epsilon }_0}}$.

$$\begin{gathered} V_{pole}-V_{centre}=\frac{\mathit{\sigma }R}{\sqrt{2}{\mathit{\epsilon }}_0}-\frac{\mathit{\sigma }R}{2{\mathit{\epsilon }}_0} \\ =\frac{\mathit{\sigma }R}{\sqrt{2{\mathit{\epsilon }}_0}}\left[1-\frac{1}{\sqrt{2}}\right] \\ =\frac{\mathit{\sigma }R}{2{\mathit{\epsilon }}_{\mathbf{0}}}\left(\sqrt{2}-1\right) \end{gathered}$$

Hence, the potential difference between the "north pole" and the center is$\frac{\sigma R}{2{\epsilon }_0}\left(\sqrt{2}-1\right)$