Question

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)?

Short answer

(c)The change in potential energy of the system is $-1.824\times {10}^{-16}\text{J}$.

Step-by-step solution

Given information

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

Change in Potential

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.

Step 3(c): The change in potential energy of the system

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 $-1.824\times {10}^{-16}\text{J}$.