Question

what could you do to increase the electric potential energy between two positively charged particles by a factor of 16?
a. reduce the distance by a factor of 4.
b. increase the distance by a factor of 4.
c. increase the distance by a factor of 16.
d. reduce the distance by a factor of 16.

Explanation:

Step1: Recall the formula for electric potential energy

The electric potential energy \(U\) between two charged particles is given by \(U=\frac{kq_1q_2}{r}\), where \(k\) is a constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them.

Step2: Analyze the effect of changing distance on potential energy

Let the initial potential energy be \(U_1=\frac{kq_1q_2}{r_1}\). If we want \(U_2 = 16U_1\), then \(16U_1=\frac{kq_1q_2}{r_2}\). Substituting \(U_1=\frac{kq_1q_2}{r_1}\) into \(16U_1=\frac{kq_1q_2}{r_2}\), we get \(16\times\frac{kq_1q_2}{r_1}=\frac{kq_1q_2}{r_2}\). Canceling out \(kq_1q_2\) from both sides gives \(16\times\frac{1}{r_1}=\frac{1}{r_2}\), or \(r_2=\frac{r_1}{16}\). This means the distance is reduced by a factor of 16.

Answer:

D. Reduce the distance by a factor of 16.