Q48P - Introduction to Electrodynamics

Question

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

Step-by-Step Solution

Answer

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

Step 1: Define functions

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.

 $$\begin{gathered} {\mathit{V}}_{\mathbf{C}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{e}\mathbf{r}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{\int }\frac{\mathbf{\sigma }}{\mathbf{r}}\mathbf{ }\mathit{d}\mathit{a} \\ \mathrm{Then} {\mathit{V}}_{\mathbf{C}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{e}\mathbf{r}}\mathbf{ }\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{\sigma }}{\mathbf{R}}\mathbf{\int }\mathit{d}\mathit{a} \end{gathered}$$                   .......(1)

                                                 

Here, $\int da$is the surface area of hemisphere. $\int da=2{\mathrm{\pi R}}^2$.

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

 

$$\begin{gathered} V_{center}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\sigma }{R}2{\mathrm{\pi R}}^2 \\ =\frac{\mathrm{\sigma R}}{2{\mathrm{\epsilon }}_0} \\ {\mathbf{V}}_{\mathbf{center}}\mathbf{=}\frac{\mathbf{\sigma R}}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ } .............\left(2\right) \end{gathered}$$

Step 2: Determine potential at the North Pole

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

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

 

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

 

Considering pole,

 

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

 

Therefore, the pole is calculated as,

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

 

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

Step 3: Determine potential difference between the North Pole and center

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

 

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