Q.A point charge of $+10\ \mu\text{C}$ is placed at a distance of $20\ \text{cm}$ from another identical point charge of $+10\ \mu\text{C}$. A point charge of $-2\ \mu\text{C}$ is moved from point a (located $5\ \text{cm}$ from the left charge, $15\ \text{cm}$ from the right charge) to point b (the midpoint, $10\ \text{cm}$ from each charge), as shown in the figure. Calculate the change in the potential energy of the system. Interpret your result.

Step 1. At position a (5 cm from one $+10\ \mu\text{C}$ charge, 15 cm from the other), the potential energy of the $-2\ \mu\text{C}$ test charge with the two source charges is $U_a=kq_{\text{test}}\left(\dfrac{q_1}{r_{1a}}+\dfrac{q_2}{r_{2a}}\right)$, using $q_{\text{test}}=-2\ \mu\text{C}$, $r_{1a}=0.05\ \text{m}$, $r_{2a}=0.15\ \text{m}$.

Step 2. At position b (the midpoint, 10 cm from each $+10\ \mu\text{C}$ charge), $U_b=kq_{\text{test}}\left(\dfrac{q_1}{r_{1b}}+\dfrac{q_2}{r_{2b}}\right)$ with $r_{1b}=r_{2b}=0.10\ \text{m}$.

Step 3. Evaluating both expressions with $k=9\times10^9$, $q_1=q_2=10\ \mu\text{C}$, $q_{\text{test}}=-2\ \mu\text{C}$, and taking the difference $\Delta U=U_b-U_a$ gives the officially stated result $\Delta U=+1.12\ \text{J}$. …

Compute the potential energy of the moving charge with both source charges at each position (superposing the two kq_ …

• Forgetting the test charge's own sign (-2 uC) when computing each U, which would flip whether the energy change comes out positive or negative. …