Question

question 14 of 25
two point charges are separated by a distance d. the first has a charge of +2e, and the second has a charge of +3e. how does the electric potential energy of the +2e charge change if it is moved to a new position, so that it is separated from the second charge by a distance of 4d?
(pe = \\(\frac{kq_1q_2}{r}\\), k = 9.00×10^9 n·m²/c²)

a. the new electric potential energy is 4 times as strong as the original electric potential energy.
b. the new electric potential energy is \\(\frac{1}{4}\\) as strong as the original electric potential energy.
c. the new electric potential energy is 16 times as strong as the original electric potential energy.
d. the new electric potential energy is \\(\frac{1}{16}\\) as strong as the original electric potential energy.

Explanation:

Step1: Write original potential energy formula

The electric - potential energy formula is $PE=\frac{kq_1q_2}{r}$. Initially, $q_1 = 2e$, $q_2 = 3e$, and $r = d$, so the original potential energy $PE_1=\frac{k(2e)(3e)}{d}=\frac{6ke^2}{d}$.

Step2: Write new potential energy formula

When the distance $r$ is changed to $4d$, the new potential energy $PE_2=\frac{k(2e)(3e)}{4d}=\frac{6ke^2}{4d}$.

Step3: Find the ratio of new to original potential energy

Calculate $\frac{PE_2}{PE_1}=\frac{\frac{6ke^2}{4d}}{\frac{6ke^2}{d}}$. The $6ke^2$ terms cancel out, and we get $\frac{PE_2}{PE_1}=\frac{1}{4}$.

Answer:

B. The new electric potential energy is $\frac{1}{4}$ as strong as the original electric potential energy.