Chapter 19, Problem 12TP

To determine

Potential energy stored in system consisting three-point charges when last two charges are interchanged

Answer to Problem 12TP

Potential energy of system consisting three-point charges decreases when last two charges are interchanged

Explanation of Solution

Given info:

Charges

  $\begin{array}{l}{q}_1=1\mu C\\ q_2=-2\mu C\\ q_3=3\mu C\end{array}$

Formula used:

The potential energy is given by

  $PE=\frac{-k{q}_1{q}_2}{d}$

Where k is constant $k=9\times {10}^9{\text{ Nm}}^2/{\text{C}}^2$

  $q_1,q_2$ are two charges, which are distance $d$ apart

Calculation:

For normal condition, substituting the given values, we get

  $\begin{array}{l}PE=\frac{-k{q}_1{q}_2}{d_{12}}+\frac{-k{q}_2{q}_3}{d_{23}}+\frac{-k{q}_1{q}_3}{d_{13}}\\ PE=-k\left(\frac{q_1{q}_2}{d_{12}}+\frac{q_2{q}_3}{d_{23}}+\frac{q_1{q}_3}{d_{13}}\right)\\ PE=-9\times {10}^9\left(\frac{1\times {10}^{-6}\times -2\times {10}^{-6}}{0.3}+\frac{-2\times {10}^{-6}\times 3\times {10}^{-6}}{0.4}+\frac{1\times {10}^{-6}\times 3\times {10}^{-6}}{0.7}\right)\\ PE=0.156\text{ J}\end{array}$

When the charges $q_{2,}{q}_3$ are interchanged, the potential energy will be

  $\begin{array}{l}{q}_1=1\mu C\\ q_2=3\mu C\\ q_3=-2\mu C\end{array}$

now

  $\begin{array}{l}P{E}_2=\frac{-k{q}_1{q}_2}{d_{12}}+\frac{-k{q}_2{q}_3}{d_{23}}+\frac{-k{q}_1{q}_3}{d_{13}}\\ PE=-k\left(\frac{q_1{q}_2}{d_{12}}+\frac{q_2{q}_3}{d_{23}}+\frac{q_1{q}_3}{d_{13}}\right)\\ PE=-9\times {10}^9\left(\frac{1\times {10}^{-6}\times 3\times {10}^{-6}}{0.3}+\frac{3\times {10}^{-6}\times -2\times {10}^{-6}}{0.4}+\frac{1\times {10}^{-6}\times -2\times {10}^{-6}}{0.7}\right)\\ PE=0.07\text{ J}\end{array}$

As compared to first, the potential energy decreases when the charges are traded.

Conclusion:

Potential energy of system consisting three-point charges decreases when last two charges are interchanged