Refer to figure 6.86. Calculate the change in electric energy along the two different paths in moving charge away from charge $\mathit{Q}$from A to B along a radial path, then to C along a circle centered on , then to D along a radial path. Also calculate the change in energy in going directly from A to D along a circle centered at $\mathit{Q}$. Specifically. What are ${\mathbf{U}}_{\mathbf{B}}\mathbf{-}{\mathbf{U}}_{\mathbf{A}}\mathbf{,}\mathbf{}{\mathbf{U}}_{\mathbf{C}}\mathbf{-}{\mathbf{U}}_{\mathbf{B}}\mathbf{,}\mathbf{}{\mathbf{U}}_{\mathbf{D}}\mathbf{-}{\mathbf{U}}_{\mathbf{C}}$and their sum? What is ${\mathbf{U}}_{\mathbf{D}}\mathbf{-}{\mathbf{U}}_{\mathbf{A}}$? Also, calculate the round-trip difference in the electrical energy when moving charge along the path from A to B to C to D to A

Two charges q and Q are given.

The radii of two circular paths AD and BC are $r_1\text{and}{r}_2$respectively.

The total work done in carrying a positive charge q from infinity to a point inside an electric field due to another charge positive charge $\mathit{Q}$is known as the potential energy of the system. The difference in potential energy between two points gives the work done in carrying the charge q between these points.

The potential energy at a point is mathematically given as:

$\mathbf{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{qQ}}{\mathbf{r}}$

The potential energy at points A, B, C, and D, for the system of charges$Qandq$ are:

role="math" localid="1668506217310" $$\begin{gathered} U_A=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_1} \\ {U}_B=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2} \\ {U}_C=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2} \\ {U}_D=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_1} \end{gathered}$$

Now, change in electrical energy along AB is given as:

$$\begin{gathered} U_B-U_A=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2}-\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_1} \\ =\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \end{gathered}$$

The change in electrical energy along BC is given as:

$$\begin{gathered} U_C-U_B=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2}-\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2} \\ =\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_2}-\frac{1}{r_2}\right) \\ =0 \end{gathered}$$

The change in electrical energy along CD is given as:

$$\begin{gathered} U_D-U_C=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_1}-\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2} \\ =\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \end{gathered}$$

The change in electrical energy along AD is given as:

$$\begin{gathered} U_D-U_A=\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2}-\frac{1}{4\pi {\epsilon }_o}\frac{Qq}{r_2} \\ =\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_2}-\frac{1}{r_2}\right) \\ =0 \end{gathered}$$

The total change in electrical energy along the path ABCD is given as:

$$\begin{gathered} \left(U_B-U_A\right)+\left(U_C-U_B\right)+\left(U_D-U_C\right)=\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)+0+\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \\ =0 \end{gathered}$$

The total change in electrical energy along the complete round trip ABCDA will be:

$$\begin{gathered} \left(\left(U_B-U_A\right)+\left(U_C-U_B\right) \\ +\left(U_D-U_C\right)+\left(U_A-U_D\right)\right)=\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)+0+\frac{Qq}{4\pi {\epsilon }_o}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)-0 \\ =0 \end{gathered}$$

The change in electrical energy along the path ABCD is 0, and the change in electrical energy for the path AD is also $0$. The change in electrical energy for the one round trip ABCDA is$0$