Questions: Two charged particles exert an electrostatic force of 20 N on each other. What will the magnitude of the electrostatic force be if the distance between the two charges is reduced to one-third of the original distance? (A) 170 N (B) 180 N (C) 190 N

Solution Steps

Coulomb's Law describes the electrostatic force between two point charges. It is given by the formula:

$$F = k \frac{|q_1 q_2|}{r^2}$$

where:

• $F$ is the electrostatic force,
• $k$ is Coulomb's constant,
• $q_1$ and $q_2$ are the magnitudes of the charges,
• $r$ is the distance between the charges.

The problem states that the distance between the charges is reduced to one-third of the original distance. If the original distance is $r$, the new distance becomes $\frac{r}{3}$.

Using Coulomb's Law, the new force $F'$ when the distance is reduced to $\frac{r}{3}$ is:

$$F' = k \frac{|q_1 q_2|}{\left(\frac{r}{3}\right)^2} = k \frac{|q_1 q_2|}{\frac{r^2}{9}} = 9 \cdot k \frac{|q_1 q_2|}{r^2} = 9F$$

Given that the original force $F$ is 20 N, the new force $F'$ is:

$$F' = 9 \times 20 = 180 \, \text{N}$$

Final Answer