Question

question 4 of 10
how does the electric potential energy between two negatively charged
particles change if one particles charge is reduced by a factor of 3?

a. it is reduced by a factor of 3.
b. it is increased by a factor of 3.
c. it is decreased by a factor of 9.
d. it is increased by a factor of 9.

Explanation:

Step1: Recall the formula for electric potential energy

The formula for electric potential energy \(U\) between two charged particles is \(U = k\frac{q_1q_2}{r}\), where \(k\) is a constant, \(q_1\) and \(q_2\) are the charges of the two particles, and \(r\) is the distance between them.

Step2: Analyze the change when one charge is reduced

Let the original charges be \(q_1\) and \(q_2\), so \(U_1=k\frac{q_1q_2}{r}\). If one charge (say \(q_1\)) is reduced by a factor of 3, the new charge is \(q_1'=\frac{q_1}{3}\). Then the new potential energy \(U_2 = k\frac{\frac{q_1}{3}q_2}{r}=\frac{1}{3}(k\frac{q_1q_2}{r})\).

Answer:

A. It is reduced by a factor of 3.