To find the change in electric potential energy (ΔU) when moving the charge $q = 1.31 \times 10^{-18} \, C$ a distance $d = 0.400 \, m$ closer to either of the two 1.90-µC charges, we can use the concept of electric potential energy in a system of point charges.

Step 1: Define the electric potential energy formula for point charges.

The electric potential energy $U$ between two charges $q_1$ and $q_2$, separated by a distance $r$, is given by:

$U = k_e \frac{q_1 q_2}{r}$

Where:

• $k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2$ is Coulomb's constant.
• $q_1$ and $q_2$ are the magnitudes of the charges.
• $r$ is the distance between the charges.

Step 2: Calculate the initial potential energy.

Let the initial distance between the charge $q$ and one of the 1.90-µC charges be $r_i$, and the final distance after moving closer by 0.400 m be $r_f = r_i - 0.400 \, m$.

The initial potential energy between $q$ and one of the 1.90-µC charges is:

$U_i = k_e \frac{q \times 1.90 \times 10^{-6}}{r_i}$

Step 3: Calculate the final potential energy.

The final potential energy after moving the charge closer by 0.400 m will be:

$U_f = k_e \frac{q \times 1.90 \times 10^{-6}}{r_f}$

Where $r_f = r_i - 0.400 \, m$.

Step 4: Find the change in potential energy.

The change in electric potential energy $\Delta U$ is given by the difference between the final and initial energies:

$\Delta U = U_f - U_i = k_e q \times 1.90 \times 10^{-6} \left( \frac{1}{r_f} - \frac{1}{r_i} \right)$

Step 5: Substitute values.

• $q = 1.31 \times 10^{-18} \, C$
• $1.90 \times 10^{-6} \, C$ is the charge of each particle.
• $d = 0.400 \, m$

Now, we need the initial distance $r_i$ between the charge $q$ and one of the 1.90-µC charges. If this is given in the figure, we can substitute it in and calculate the result.

Do you know the initial distance $r_i$ from the figure?