91Ó°ÊÓ

Potential energy is given by:

$\mathit{U}\mathbf{=}\mathit{k}\underset{\mathbf{i}\mathbf{<}\mathbf{j}}{\mathbf{∑}}\frac{{\mathbf{q}}_{\mathbf{i}}{\mathbf{q}}_{\mathbf{j}}}{{\mathbf{r}}_{\mathbf{i}\mathbf{j}}}$

Where k is a constant and the system is made of n charges as ${\mathbf{q}}_{\mathbf{1}}\mathbf{,}{\mathbf{q}}_{\mathbf{2}}\mathbf{,}{\mathbf{q}}_{\mathbf{3}}\mathbf{,}{\mathbf{q}}_{\mathbf{4}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathbf{q}}_{\mathbf{n}}$,

and the distance between these charges are $$\begin{gathered} {\mathbf{r}}_{\mathbf{12}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{ }{\mathbf{r}}_{\mathbf{1}}\mathbf{ }\mathbf{and}\mathbf{ }{\mathbf{r}}_{\mathbf{2}}\mathbf{)}\mathbf{,}{\mathbf{r}}_{\mathbf{23}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{ }{\mathbf{r}}_{\mathbf{2}}\mathbf{ }\mathbf{and}\mathbf{ }{\mathbf{r}}_{\mathbf{3}}\mathbf{)}\mathbf{,} \\ {\mathbf{r}}_{\mathbf{34}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{ }{\mathbf{r}}_{\mathbf{3}}\mathbf{ }\mathbf{and}\mathbf{ }{\mathbf{r}}_{\mathbf{4}}\mathbf{)}\mathbf{,}{\mathbf{r}}_{\mathbf{45}}\mathbf{(}\mathbf{distance}\mathbf{}\mathbf{between}\mathbf{ }{\mathbf{r}}_{\mathbf{4}}\mathbf{ }\mathbf{and}\mathbf{ }{\mathbf{r}}_{\mathbf{5}}\mathbf{)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.} \end{gathered}$$.

The image is drawn from the given data:

The electric potential is due to multiple point charge is:

$U=k\underset{i<j}{∑}\frac{q_i{q}_j}{r_{ij}}$

For this system, the electric potential at the center can be calculated:

$$\begin{gathered} {\mathrm{U}}_{\mathrm{center}}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_{\mathrm{Î\pm }}}{\mathrm{d}}+\frac{{\mathrm{q}}_2{\mathrm{q}}_{\mathrm{Î\pm }}}{\mathrm{d}}+\frac{{\mathrm{q}}_3{\mathrm{q}}_{\mathrm{Î\pm }}}{\mathrm{d}}+\frac{{\mathrm{q}}_4{\mathrm{q}}_{\mathrm{Î\pm }}}{\mathrm{d}} \\ \mathrm{d}=\frac{1}{2}\sqrt{{10}^2+{10}^2}=7.071\mathrm{nm} \\ {\mathrm{q}}_1={\mathrm{q}}_2={\mathrm{q}}_3={\mathrm{q}}_4=\mathrm{q} \end{gathered}$$

Now the above expression can be calculated as:

$U_{ceneter}=k\frac{4q{q}_{\mathrm{Î\pm }}}{7.071nm}$

For this system, the electric potential at the side can be calculated:

$$\begin{gathered} {\mathrm{U}}_{\mathrm{side}}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_{\mathrm{Î\pm }}}{{\mathrm{r}}_1}+\frac{{\mathrm{q}}_2{\mathrm{q}}_{\mathrm{Î\pm }}}{{\mathrm{r}}_2}+\frac{{\mathrm{q}}_3{\mathrm{q}}_{\mathrm{Î\pm }}}{{\mathrm{r}}_2}+\frac{{\mathrm{q}}_4{\mathrm{q}}_{\mathrm{Î\pm }}}{{\mathrm{r}}_1} \\ {\mathrm{r}}_1=5\mathrm{nm}\mathrm{and}{\mathrm{r}}_2=\sqrt{{\left(10\mathrm{nm}\right)}^2+{\left(5\mathrm{nm}\right)}^2}=11.18\mathrm{nm} \\ {\mathrm{q}}_1={\mathrm{q}}_2={\mathrm{q}}_3={\mathrm{q}}_4=\mathrm{q} \end{gathered}$$

Now the above expression can be calculated as:

$$\begin{gathered} {\mathrm{U}}_{\mathrm{side}}={\mathrm{kqq}}_{\mathrm{Î\pm }}\left(\frac{2}{{\mathrm{r}}_1}+\frac{2}{{\mathrm{r}}_2}\right) \\ =2{\mathrm{kqq}}_{\mathrm{Î\pm }}\left(\frac{1}{5\mathrm{nm}}+\frac{1}{11.18\mathrm{nm}}\right) \\ =2{\mathrm{kqq}}_{\mathrm{Î\pm }}\left(\frac{11.18\mathrm{nm}+5\mathrm{nm}}{5×11.18{\left(\mathrm{nm}\right)}^2}\right) \\ =2{\mathrm{kqq}}_{\mathrm{Î\pm }}\left(\frac{16.18}{55.9\mathrm{nm}}\right) \end{gathered}$$

Work is needed to move the alpha particle to the midpoint of one of the sides of the square :

$$\begin{gathered} \mathrm{W}={\mathrm{U}}_{\mathrm{side}}-{\mathrm{U}}_{\mathrm{center}} \\ =2{\mathrm{kqq}}_{\mathrm{Î\pm }}\left(\frac{16.18}{55.9\mathrm{nm}}\right)-\mathrm{k}\frac{4{\mathrm{qq}}_{\mathrm{Î\pm }}}{7.071\mathrm{nm}} \\ =\left(2×9×{10}^9×1.6×{10}^{-19}×2×1.6×{10}^{-19}\left(\frac{16.18}{55.9×{10}^{-9}}\right)\right) \\ =\left(9×{10}^9×-1.6×{10}^{-19}×2×1.6×{10}^{-19}×4×\frac{1}{7.071×{10}^{-9}}\right) \\ =-26.67×{10}^{-20}+26.06×{10}^{-20} \\ =-0.60×{10}^{-20}\mathrm{J} \\ =-6.00×{10}^{-21}\mathrm{J} \end{gathered}$$

Therefore,$-6×{10}^{-21} \mathrm{J}$ is needed to move the alpha particle to the midpoint of one of the sides of the square.