An electron is initially at rest. It is moved from a location $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$from a proton to a location $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathit{m}$ from the proton. What is the change in electric potential energy of the system of proton and electron?

Initial distance of electron from proton,$r_1=4\times {10}^{-10}\text{m}$

After distance, the distance of electron from proton,$r_2=6\times {10}^{-10}\text{m}$

Charge on proton,$q_1=+1.6\times {10}^{-19}\text{C}$

Charge of electron, $q_2=-1.6\times {10}^{-19}\text{C}$

The expression to calculate the potential at distance $r$ due to charge $q$ is given as follows.

$\mathit{V}\mathbf{=}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{\times }\frac{\mathbf{q}}{\mathbf{r}}$ …… (i)

The expression to calculate the change in the potential is given as follows.

$\mathit{\Delta }\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{A}}\mathbf{-}{\mathit{V}}_{\mathbf{B}}$ …… (ii)

Here,$V_B$is the potential at the point B and $V_A$ is the potential at point A.

The expression to calculate the change in the electrical potential is given as follows.

$\mathit{\Delta }\mathit{U}\mathbf{=}\left|q_2\right|\mathit{\Delta }\mathit{V}$ …… (iii)

Consider the system of the electron and proton as shown below.

Calculate the potential at point $A$.

Substitute $4\times {10}^{-10}\text{m}$for $r_1$and $1.6\times {10}^{-19}\text{C}$for into equation (i).

$$\begin{gathered} V_A=9\times {10}^9\times \frac{1.6\times {10}^{-19}}{4\times {10}^{-10}} \\ {V}_A=\frac{14.4\times {10}^{-10}}{4\times {10}^{-10}} \\ {V}_A=3.6\text{V} \end{gathered}$$

Calculate the potential at point $B$.

Substitute $6\times {10}^{-10}\text{m}$ for $r_2$and $1.6\times {10}^{-19}\text{C}$ for $q_1$into equation (i).

$$\begin{gathered} V_B=9\times {10}^9\times \frac{1.6\times {10}^{-19}}{6\times {10}^{-10}} \\ {V}_B=\frac{14.4\times {10}^{-10}}{6\times {10}^{-10}} \\ {V}_B=2.4\text{V} \end{gathered}$$

Calculate the change in the potential.

Substitute $3.6\text{V}$ for $V_A$ and $2.4\text{V}$for $V_B$ into equation (ii).

$$\begin{gathered} \Delta V=3.6-2.4 \\ \Delta V=1.2\text{V} \end{gathered}$$

Calculate the change in the electrical potential.

Substitute $1.2\text{V}$ for $\Delta V$ and $-1.6\times {10}^{-19}\text{C}$ for $q_2$ into equation (iii).

$$\begin{gathered} \Delta U=\left|-1.6\times {10}^{-19}\right|\times 1.2 \\ \Delta U=1.92\times {10}^{-19}\text{J} \end{gathered}$$

Hence the change in the electrical potential energy is $1.92\times {10}^{-19}\text{J}$.