Question

how does the electric force between two charged particles change if the distance between them is increased by a factor of 3? a. it is increased by a factor of 9. b. it is reduced by a factor of 3. c. it is increased by a factor of 3. d. it is reduced by a factor of 9.

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law is \( F = k\frac{q_1q_2}{r^2} \), where \( F \) is the electric force, \( k \) is a constant, \( q_1\) and \( q_2\) are the charges, and \( r \) is the distance between the charges.

Step2: Analyze the effect of changing the distance

Let the original distance be \( r_1=r \), and the new distance be \( r_2 = 3r \). The original force \( F_1=k\frac{q_1q_2}{r_1^2}=k\frac{q_1q_2}{r^{2}} \). The new force \( F_2=k\frac{q_1q_2}{r_2^2}=k\frac{q_1q_2}{(3r)^{2}}=k\frac{q_1q_2}{9r^{2}}=\frac{F_1}{9} \).

Answer:

D. It is reduced by a factor of 9.