Question

Two charges +7 mC and -4 mC at coordinates (-7, 0, 0) and (7, 0, 0), respectively, find the electrostatic potential energy. $$\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \, \text{N}^{-1} \, \text{m}^{-2}.$$

Answer 1

The electrostatic potential energy $U$ between two point charges is given by the formula: $$U = \frac{k q_1 q_2}{r},$$ where: - $k = \frac{1}{4 \pi \epsilon_0}$ is Coulomb’s constant, - $q_1 = 7 \, \text{mC} = 7 \times 10^{-3} \, \text{C}$, - $q_2 = -4 \, \text{mC} = -4 \times 10^{-3} \, \text{C}$, - $r$ is the distance between the charges. We can calculate the distance $r$ between the charges using the distance formula: $$r = \sqrt{(7 - (-7))^2 + (0 - 0)^2 + (0 - 0)^2} = \sqrt{(14)^2} = 14 \, \text{m}.$$ Now, using the value of Coulomb's constant $k = \frac{1}{4 \pi \epsilon_0}$: $$k = \frac{1}{4 \pi \times 8.85 \times 10^{-12}} = 9 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2.$$ Substitute the values into the formula for potential energy: $$U = \frac{9 \times 10^9 \times (7 \times 10^{-3}) \times (-4 \times 10^{-3})}{14}.$$ $$U = \frac{9 \times 10^9 \times (-28 \times 10^{-6})}{14} = -18 \, \text{J}.$$ Thus, the electrostatic potential energy is $-18 \, \text{J}$.