Physics - Electrostatics Question with Solution | TestHub

The electrostatic potential energy of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r}$. Given $q_1=+3\text{ nC}=3\times 10^{-9}\text{ C}$ and $q_2=-5\text{ nC}=-5\times 10^{-9}\text{ C}$. The distance $r$ between the charges is $4\text{ m}$. We use the value of $k=\frac{1}{4\pi {ϵ}_0}=9\times 10^9{\text{ Nm}}^2/{\text{C}}^2$.$U=(9\times 10^9)\frac{(3\times 10^{-9})\times (-5\times 10^{-9})}{4}$ $U=9\times \frac{-15\times 10^{-9}}{4}=-\frac{135}{4}\times 10^{-9}\text{ J}=-33.75\times 10^{-9}\text{ J}$. Converting to nanojoules (nJ): $U=-33.75\text{ nJ}$. Rounding to the nearest integer, the electrostatic potential energy is $-34\text{ nJ}$.