question 14 of 25\ntwo point charges are separated by a distance d. the first has a charge of +2e, and the…

question 14 of 25\ntwo 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?\n(pe = \\(\\frac{kq_1q_2}{r}\\), k = 9.00×10^9 n·m²/c²)\n\na. the new electric potential energy is 4 times as strong as the original electric potential energy.\nb. the new electric potential energy is \\(\\frac{1}{4}\\) as strong as the original electric potential energy.\nc. the new electric potential energy is 16 times as strong as the original electric potential energy.\nd. the new electric potential energy is \\(\\frac{1}{16}\\) as strong as the original electric potential energy.

question 14 of 25\ntwo 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?\n(pe = \\(\\frac{kq_1q_2}{r}\\), k = 9.00×10^9 n·m²/c²)\n\na. the new electric potential energy is 4 times as strong as the original electric potential energy.\nb. the new electric potential energy is \\(\\frac{1}{4}\\) as strong as the original electric potential energy.\nc. the new electric potential energy is 16 times as strong as the original electric potential energy.\nd. the new electric potential energy is \\(\\frac{1}{16}\\) as strong as the original electric potential energy.

Answer

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.