91Ó°ÊÓ

According to the problem, we have three charges: \(q_1 = 4.0 \times 10^{-6}\,\text{C}\) (the "+" sign indicates a positive charge) at position \((0,0)\) m; \(q_2 = 3.0 \times 10^{-6}\,\text{C}\) at position \((4.0,3.0)\) m; \(q_3 = -1.0 \times 10^{-6}\,\text{C}\) (the "-" sign indicates a negative charge) at position \((0,3)\) m.

Next, we need to calculate the distance between each pair of charges using the distance formula: \(r_{12} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 0)^2 + (3 - 0)^2} = 5\) m; \(r_{13} = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2} = \sqrt{(0 - 0)^2 + (3 - 0)^2} = 3\) m; \(r_{23} = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} = \sqrt{(0 - 4)^2 + (3 - 3)^2} = 4\) m.

Next, we will calculate the electric potential energy for each pair of charges using the formula mentioned in the Analysis section: \(U_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}} = \frac{1}{4\pi(8.85 \times 10^{-12} \,\text{C}^{-2}\text{N m}^2)} \frac{(4\times 10^{-6}\,\text{C})(3\times 10^{-6}\,\text{C})}{5\,\text{m}} \approx 8.543 \times 10^{-4}\) J; \(U_{13} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_3}{r_{13}} = \frac{1}{4\pi(8.85 \times 10^{-12} \,\text{C}^{-2}\text{N m}^2)} \frac{(4\times 10^{-6}\,\text{C})(-1\times 10^{-6}\,\text{C})}{3\,\text{m}} \approx -1.885 \times 10^{-4}\) J; \(U_{23} = \frac{1}{4\pi\epsilon_0} \frac{q_2 q_3}{r_{23}} = \frac{1}{4\pi(8.85 \times 10^{-12} \,\text{C}^{-2}\text{N m}^2)} \frac{(3\times 10^{-6}\,\text{C})(-1\times 10^{-6}\,\text{C})}{4\,\text{m}} \approx -2.989 \times 10^{-4}\) J.

Finally, we sum up the potential energies for each pair of charges to get the total electric potential energy: \(U = U_{12} + U_{13} + U_{23} = (8.543 - 1.885 - 2.989) \times 10^{-4}\,\text{J} \approx 3.669 \times 10^{-4}\,\text{J}\). The total electric potential energy of the array of charges is approximately \(3.669 \times 10^{-4}\) J.