91影视

There is a molecule of 60 carbon atoms approximated as a conducting spherical

shell of radius $\mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathit{A}\mathbf{掳}$. A nearby electron on average-smears itself out uniformly over the surface.

The force of attraction on a point charge q at a distance afrom a neutral conducting sphere of radius R is

$\mathit{F}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}{\mathit{R}}^{\mathbf{3}}\frac{\mathbf{2}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{{\left(r^2-R^2\right)}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

Here, ${\mathit{蔚}}_{\mathbf{0}}$is the permittivity of free space.

The force of repulsion between two similar point charges q at a distance a from each other is

$\mathit{F}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

From equations (1) and (2), the net force on the incoming electron is

$$\begin{gathered} \mathit{F}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}{\mathit{R}}^{\mathbf{3}}\frac{\mathbf{2}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{{\left(r^2-R^2\right)}^{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(r-R^3\frac{2r^2-R^2}{{\left(r^2-R^2\right)}^2}\right) \end{gathered}$$

For the net force to be zero,

$$\begin{gathered} {\mathit{r}}_{\mathbf{0}}\mathbf{-}{\mathit{R}}^{\mathbf{3}}\frac{\mathbf{2}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}}{{\mathbf{(}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{r}}_{\mathbf{0}}{\mathbf{(}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathit{R}}^{\mathbf{3}}\mathbf{(}\mathbf{2}{\mathbf{r}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{-}{\mathbf{R}}^{\mathbf{2}}\mathbf{)} \end{gathered}$$

Solving this in Mathematica,

${\mathit{r}}_{\mathbf{0}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{663}\mathbf{}\mathit{A}\mathbf{掳}$

Thus, the distance at which force is zero is $\mathbf{5}\mathbf{.}\mathbf{663}\mathbf{}\mathit{A}\mathbf{掳}$.

The work done to bring the electron from infinity to ${\mathit{r}}_{\mathbf{0}}$is

$$\begin{gathered} \mathit{W}\mathbf{=}\mathbf{-}\underset{\mathbf{鈭�}}{\overset{{\mathbf{r}}_{\mathbf{0}}}{\mathbf{鈭�}}}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\left(r-R^3\frac{2r^2-R^2}{{\left(r^2-R^2\right)}^2}\right)\mathit{d}\mathit{r} \\ \mathbf{=}\mathbf{-}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}\mathbf{蟺}{\mathbf{蔚}}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{3}}}\underset{\mathbf{鈭�}}{\overset{{\mathbf{r}}_{\mathbf{0}}}{\mathbf{鈭�}}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{3}}}\left(r-R^3\frac{2r^2-R^2}{{\left(r^2-R^2\right)}^2}\right)\mathit{d}\mathit{r} \end{gathered}$$

Substitute $\mathit{x}\mathbf{=}\frac{\mathbf{r}}{\mathbf{R}}$

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}\mathbf{R}}\underset{{\mathbf{x}}_{\mathbf{0}}}{\overset{\mathbf{鈭�}}{\mathbf{鈭�}}}\mathbf{(}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}{\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{2}}}\mathbf{)}\mathit{d}\mathit{x} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}\mathbf{R}}\mathbf{\left[}\frac{\mathbf{1}\mathbf{+}\mathbf{2}{\mathbf{x}}_{\mathbf{0}}\mathbf{-}\mathbf{2}{\mathbf{x}}_{\mathbf{0}}^{\mathbf{3}}}{\mathbf{2}{\mathbf{x}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{x}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{)}}\mathbf{\right]} \end{gathered}$$

Substitute value of ${\mathit{r}}_{\mathbf{0}}$to get

$$\begin{gathered} \mathit{W}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}\mathbf{R}}\mathbf{脳}\frac{\mathbf{1}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{蟺蔚}}_{\mathbf{0}}\mathbf{R}} \end{gathered}$$

Substitute

$$\begin{gathered} \mathit{q}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathit{C} \\ \frac{\mathbf{1}}{\mathbf{4}{\mathbf{蟺蔚}}_{\mathbf{0}}}\mathbf{9}\mathbf{脳}{\mathbf{10}}^{\mathbf{9}}\frac{\mathbf{N}\mathbf{路}{\mathbf{m}}^{\mathbf{2}}}{{\mathbf{C}}^{\mathbf{2}}} \\ \mathit{R}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathit{A}\mathbf{掳} \end{gathered}$$

and get,

$\mathit{W}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{}\mathit{e}\mathit{V}$

Thus, the work done is $\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{}\mathit{e}\mathit{V}$.