question 21 of 25\nhow does the electric potential energy between two positively charged particles change if…

Explanation:

Step1: Recall electric - potential - energy formula

The electric potential energy between two point charges is given by $U = \frac{kq_1q_2}{r}$, where $k$ is the Coulomb's constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between them.

Step2: Consider the initial and new situations

Let the initial distance be $r_1$ and the initial potential energy be $U_1=\frac{kq_1q_2}{r_1}$. When the distance is tripled, $r_2 = 3r_1$, and the new potential energy is $U_2=\frac{kq_1q_2}{r_2}=\frac{kq_1q_2}{3r_1}$.

Step3: Find the ratio of new to initial potential energy

$\frac{U_2}{U_1}=\frac{\frac{kq_1q_2}{3r_1}}{\frac{kq_1q_2}{r_1}}=\frac{1}{3}$. This means $U_2=\frac{1}{3}U_1$, so the electric - potential energy is reduced by a factor of 3.

Answer:

C. It is reduced by a factor of 3.