Question

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed to 1. 2d. 2. d. 3. d/2. 4. d/ d/4.

Answer 1

Answer:

2.$d'=\frac{d}{\sqrt2}$

Explanation:

The electric force between two point charges is given by Coulomb's law:

$F=\frac{kq_1q_2}{d^2}$

Here k is the Coulomb constant, $q_1$ and $q_2$  are the signed magnitudes of the charges and d the distance of separation between them. We need $F'=2F$. So:

$F'=\frac{kq_1q_2}{d'^2}\\2F=\frac{kq_1q_2}{d'^2}\\2\frac{kq_1q_2}{d^2}=\frac{kq_1q_2}{d'^2}\\2d'^2=d^2\\d'=\frac{d}{\sqrt2}$