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Potential inside the sphere:

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

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

Potential outside the sphere:

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

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

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

Plug the values,

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

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

Plug the values,

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

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

Plug the values,

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

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

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 $\mathit{r}\mathbf{=}{\mathit{R}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathit{m}$

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

Therefore, the potential difference is

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

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.