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