91Ó°ÊÓ

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

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

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

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

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

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

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

$\mathit{Δ}\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{A}}\mathbf{-}{\mathit{V}}_{\mathbf{B}}$ …… (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.

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

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

Calculate the potential at point $A$.

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

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

Calculate the potential at point $B$.

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

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

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).

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

Calculate the change in the electrical potential.

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

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

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