Question

Refer to figure 6.86. Calculate the change in electric energy along the two different paths in moving charge $\mathit{q}$away from charge $\mathit{Q}$ from A to B along a radial path, then to C along a circle centeredon $\mathit{Q}$ , 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 ${\mathit{U}}_{\mathbf{B}}\mathbf{-}{\mathit{U}}_{\mathbf{A}\mathbf{,}\mathbf{}}{\mathit{U}}_{\mathbf{C}}\mathbf{-}{\mathit{U}}_{\mathbf{B}\mathbf{,}\mathbf{}}{\mathit{U}}_{\mathbf{D}}\mathbf{-}{\mathit{U}}_{\mathbf{C}}$and their sum? What is ${\mathit{U}}_{\mathbf{D}}\mathbf{-}{\mathit{U}}_{\mathbf{A}}$? 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.

Short answer

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

Step-by-step solution

Given Data

Two charges q and Q are given.

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

Concept

The total work done in carrying a positive charge from infinity to a point inside an electric field due to another charge positive charge 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-

$\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}\mathbf{Q}}{\mathbf{r}}$

Calculation

The potential energy at points A, B, C, and D, for the system of charges $q\mathrm{and}Q$ are-

$$\begin{gathered} U_A=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_1} \\ {U}_B=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_2} \\ {U}_C=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_2} \\ {U}_D=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_1} \end{gathered}$$

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

$$\begin{gathered} U_B-U_A=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_2}-\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_1} \\ =\frac{qQ}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \end{gathered}$$

The change in electrical energy along BC is given as-

$$\begin{gathered} U_C-U_B=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_2}-\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_2} \\ =0 \end{gathered}$$

The change in electrical energy along CD is given as-

$$\begin{gathered} U_D-U_C=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_1}-\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_2} \\ =\frac{qQ}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \end{gathered}$$

The change in electrical energy along AD is given as-

$$\begin{gathered} U_D-U_A=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_1}-\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{qQ}{r_1} \\ =0 \end{gathered}$$

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

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

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

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

Conclusion

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