remains the same
Step 1: The electrostatic potential energy ($U$) between two point charges $q_1$ and $q_2$ separated by distance $r$ is given by: \[ U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r} \] Step 2: Since both charges are protons, we have: \[ q_1 = q_2 = e = 1.6 \times 10^{-19} { C} \]
Step 3: Since both charges are positive, they repel each other. As the distance between them decreases, $r$ decreases.
Step 4: Since $U \propto \frac{1}{r}$, decreasing $r$ increases $U$, meaning the electrostatic potential energy of the system increases.
Step 5: Therefore, the correct answer is (B). \bigskip
Two charges, \( q_1 = +3 \, \mu C \) and \( q_2 = -4 \, \mu C \), are placed 20 cm apart. Calculate the force between the charges.