A thin spherical shell with radius ${\mathit{R}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}$cm is concentric with a larger thin spherical shell with radius ${\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}$cm. Both shells are made of insulating material. The smaller shell has charge ${\mathit{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{00}$nC distributed uniformly over its surface, and the larger shell has charge ${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{00}$nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) $\mathit{r}\mathbf{=}\mathbf{0}$; (ii) $\mathit{r}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}$cm; (iii) $\mathit{r}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}$cm? (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

Potential inside the sphere:

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{R}}$ (1)

Where $\mathit{R}$is the radius of the sphere.

Potential outside the sphere:

$\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{r}}$ (2)

Where $\mathit{r}$is the distance where the potential is measured.

${\mathit{V}}_{\mathbf{r}\mathbf{=}\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{q_1}{R_1}+\frac{q_2}{R_2}\right]$

Plug the values,

${\mathit{V}}_{\mathbf{r}\mathbf{=}\mathbf{0}}\mathbf{=}\left(9.0*{10}^9\right)\left[\frac{6*{10}^{-9}}{0.03}+\frac{-9*{10}^{-9}}{0.05}\right]\mathbf{=}\mathbf{180}\mathit{V}$

${\mathit{V}}_{\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{q_1}{r}+\frac{q_2}{R_2}\right]$

Plug the values,

${\mathit{V}}_{\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}}\mathbf{=}\left(9.0*{10}^9\right)\mathbf{(}\frac{\mathbf{6}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}{\mathbf{0}\mathbf{.}\mathbf{04}}\mathbf{+}\frac{\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{00}\mathbf{*}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}{\mathbf{0}\mathbf{.}\mathbf{05}}\mathbf{]}\mathbf{=}\mathbf{-}\mathbf{270}\mathit{V}$

${\mathit{V}}_{\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{06}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{q_1}{r}+\frac{q_2}{r}\right]$

Plug the values,

${\mathit{V}}_{\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{06}}\mathbf{=}\left(9.0*{10}^9\right)\left[\frac{6*{10}^{-9}}{0.06}+\frac{-9*{10}^{-9}}{0.06}\right]\mathbf{=}\mathbf{-}\mathbf{450}\mathit{V}$

$\mathbf{∆}\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{1}}\mathbf{-}{\mathit{V}}_{\mathbf{2}}$

For ${\mathit{V}}_{\mathbf{1}}$, get the potential by using equation (1) and plug the values into equation (Calculated before) so ${\mathit{V}}_{\mathbf{1}}\mathbf{=}\mathbf{180}\mathit{V}$.

For ${\mathit{V}}_{\mathbf{2}}$, get the potential and plugging our values into equation, where $\mathit{r}\mathbf{=}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathit{m}$

${\mathit{V}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\left[\frac{q_1}{r}+\frac{q_2}{R_2}\right]\mathbf{=}\left(9*{10}^9\right)\left[\frac{6*{10}^{-9}}{0.05}+\frac{-9*{10}^{-9}}{0.05}\right]\mathbf{=}\mathbf{-}\mathbf{540}\mathit{V}$

Therefore, the potential difference is

$\mathbf{∆}\mathit{V}\mathbf{=}\mathbf{180}\mathbf{-}\left(-540\right)\mathbf{=}\mathbf{720}\mathit{V}$

As shown by the result, the potential at the inner sphere is higher than the potential at the outer sphere.

Thus, the electric potential due to the two shells at $\mathit{r}\mathbf{=}\mathbf{0}\mathrm{is}\mathbf{}\mathbf{180}\mathit{V}$. The electric potential due to the two shells at $\mathit{r}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathit{c}\mathit{m}\mathrm{is}\mathbf{}\mathbf{-}\mathbf{270}\mathit{V}$. The electric potential due to the two shells at $\mathit{r}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathit{c}\mathit{m}\mathrm{is}\mathbf{}\mathbf{-}\mathbf{450}\mathit{V}$. The magnitude of potential difference between the surfaces of the two shells is$\mathbf{720}\mathit{V}$and the potential at the inner sphere is higher than the potential at the outer surface.