In a particular region there is a uniform electric field of$<-760,380,0>\mathbf{}\mathit{V}\mathbf{/}\mathit{m}$. Location A is $<0.2,0.1,0>\mathbf{}\mathit{m}$, location B is $<0.7,0.1,0>$, and location C is $<0.7,-0.4,0>\mathbf{}\mathit{m}$. (a) What is the change in the potential along a path from B to A? (b) What is the change in the potential along a path from A to C? (c) An alpha particle (two protons and two neutrons) moves from A to C. What is the change in potential energy of the system (alpha + source charges)?

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

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

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

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

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,

$$\begin{gathered} \Delta U_{\text{System}}=\Delta U_{\text{Alpha}}+\Delta U_{\text{Proton}}+\Delta U_{\text{Electron}} \\ \Delta U_{\text{System}}=\left(+2e·\Delta V_{AC}\right)+\left(+e·\Delta V_{AC}\right)+\left(-e·\Delta V_{AC}\right) \\ \Delta U_{\text{System}}=2e·\Delta V_{AC}+e·\Delta V_{AC}-e·\Delta V_{AC} \\ \Delta U_{\text{System}}=2e·\Delta V_{AC} \end{gathered}$$

Putting the values,

$$\begin{gathered} \Delta U_{\text{System}}=2\left(1.6\times {10}^{-19}\text{C}\right)\left(-570\text{V}\right)\times \frac{1\text{J}}{1\text{C}·\text{V}} \\ \Delta U_{\text{System}}=-1.824\times {10}^{-16}\text{J} \end{gathered}$$

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