Question

Find the electric potential energy for the following array of charges: charge \(q_{1}=+4.0 \mu \mathrm{C}\) is located at \((x, y)=(0.0,0.0) \mathrm{m} ;\) charge \(q_{2}=+3.0 \mu \mathrm{C}\) is located at \((4.0,3.0) \mathrm{m} ;\) and charge \(q_{3}=-1.0 \mu \mathrm{C}\) is located at (0.0,3.0) \(\mathrm{m}\)

Short answer

Answer: The total electric potential energy of the array of charges is approximately \(3.669 \times 10^{-4}\) J.

Step-by-step solution

Determine charge magnitudes and positions

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.

Calculate the distances between charges

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.

Calculate the electric potential energy between each pair of charges

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.

Calculate the total electric potential energy

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.