Question

A total electric charge of $\mathbf{3}\mathbf{.}\mathbf{50}$nC is distributed uniformly over the surface of a metal sphere with a radius of $\mathbf{24}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) $\mathbf{48}\mathbf{.}\mathbf{0}$cm; (b) $\mathbf{24}\mathbf{.}\mathbf{0}$cm; (c) $\mathbf{12}\mathbf{.}\mathbf{0}$cm.

Short answer

1. The potential at $48$cm from the center of the sphere is$65.6V$.

2. The potential at$24$cm from the center of the sphere is$131.0V$.

3. The potential at $12.0$cm from the center of the sphere is $131.0V$.

Step-by-step solution

Formula used to solve the question

Potential inside the sphere:

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}}{\mathbf{R}}$ (1)

Where $R$ is the radius of the sphere.

Potential outside the sphere:

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}}{\mathbf{r}}$ (2)

Where $r$is the distance where the potential is measured.

Determine the potential at r=24.0 cm

a) Here, the distance $\mathit{r}\mathbf{<}\mathit{R}$, so the potential is given by

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}}{\mathbf{r}}$

Plug the values,

$\mathit{V}\mathbf{=}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathbf{)}\left(\frac{3.5*{10}^{-9}}{0.48}\right)\mathbf{=}\mathbf{65}\mathbf{.}\mathbf{6}\mathit{V}$

Determine the potential at r=24.0 cm

a) Here, the distance $\mathit{r}\mathbf{=}\mathit{R}$, so the potential is given by

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}}{\mathbf{r}}$

Plug the values,

$\mathit{V}\mathbf{=}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{*}{\mathbf{10}}^{\mathbf{9}}\mathbf{)}\mathbf{\left(}\frac{\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}{\mathbf{0}\mathbf{.}\mathbf{48}}\mathbf{\right)}\mathbf{=}\mathbf{65}\mathbf{.}\mathbf{6}\mathit{V}$

Determine the potential at r=12.0cm cm

Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus, the potential is the same at every point inside the sphere and is equal to the potential on the surface, and it will be the same as in part (b)

$\mathit{V}\mathbf{=}\mathbf{131}\mathbf{.}\mathbf{0}\mathit{V}$

Thus, the potential at$\mathbf{48}\mathbf{.}\mathbf{0}$cm from the center of the sphere is$\mathbf{65}\mathbf{.}\mathbf{6}\mathit{V}$. The potential at $\mathbf{24}\mathbf{.}\mathbf{0}$cm from the center of the sphere is$\mathbf{131}\mathbf{.}\mathbf{0}\mathit{V}$ . The potential at $\mathbf{12}\mathbf{.}\mathbf{0}$cm from the center of the sphere is$\mathbf{131}\mathbf{.}\mathbf{0}\mathit{V}$.