91Ó°ÊÓ

The electric field vectoris, $\stackrel{→}{E}=\left(-760\hat{i}+380\hat{j}+0\hat{k}\right)\text{V/m}$.

The location vector of A is, $\stackrel{→}{A}=\left(0.2\hat{i}+0.1\hat{j}+0\hat{k}\right)\text{m}$.

The location vector of B is, $\stackrel{→}{B}=\left(0.7\hat{i}+0.1\hat{j}+0\hat{k}\right)\text{m}$.

The location vector of C is, $\stackrel{→}{C}=\left(0.7\hat{i}-0.4\hat{j}+0\hat{k}\right)\text{m}$.

A charged particle moving through a uniform electric field causes the change in the electric potential of the particle.

The change in the electric potential of a charged particle relies upon the magnitude as well as the direction of the electric field and the travel distance of the particle.

The electric charge of an alpha particle having two protons and two neutrons is $+2e$.

The formula for the change in potential energy of the system having one alpha charge and source charges (proton and electron) is given by,

$$\begin{gathered} \mathrm{Δ}{U}_{\text{System}}=\mathrm{Δ}{U}_{\text{Alpha}}+\mathrm{Δ}{U}_{\text{Proton}}+\mathrm{Δ}{U}_{\text{Electron}} \\ \mathrm{Δ}{U}_{\text{System}}=\left(+2e·\mathrm{Δ}{V}_{AC}\right)+\left(+e·\mathrm{Δ}{V}_{AC}\right)+\left(-e·\mathrm{Δ}{V}_{AC}\right) \\ \mathrm{Δ}{U}_{\text{System}}=2e·\mathrm{Δ}{V}_{AC}+e·\mathrm{Δ}{V}_{AC}-e·\mathrm{Δ}{V}_{AC} \\ \mathrm{Δ}{U}_{\text{System}}=2e·\mathrm{Δ}{V}_{AC} \end{gathered}$$

Putting the values,

$$\begin{gathered} \mathrm{Δ}{U}_{\text{System}}=2\left(1.6×{10}^{-19}\text{C}\right)\left(-570\text{V}\right)×\frac{1\text{J}}{1\text{C}·\text{V}} \\ \mathrm{Δ}{U}_{\text{System}}=-1.824×{10}^{-16}\text{J} \end{gathered}$$

Hence, the change in potential energy of the system (alpha + source charges) is $-1.824×{10}^{-16}\text{J}$.