Question

Two charged particles, placed at a distance d apart in vacuum, exert a force F on each other. Now, each of the charges is doubled. To keep the force unchanged, the distance between the charges should be changed to_____
Fill in the blank with the correct answer from the options given below.

• A. 4d
• B. 2d
• C. d
• D. \(\frac{d}{2}\)

Answer 1

The Correct Option is B

To address this problem, we start with Coulomb's Law, which defines the electrostatic force \( F \) between two point charges \( q_1 \) and \( q_2 \) at a distance \( d \) as:

\( F = \frac{k \cdot q_1 \cdot q_2}{d^2} \)

Here, \( k \) is Coulomb's constant. Initially, the charges exert a force \( F \) at distance \( d \).

When each charge is doubled, the new charges are \( 2q_1 \) and \( 2q_2 \). The resulting new force, \( F' \), is calculated as:

\( F' = \frac{k \cdot (2q_1) \cdot (2q_2)}{d'^2} = \frac{4k \cdot q_1 \cdot q_2}{d'^2} \)

To maintain the force at its original value, we equate \( F' \) and \( F \):

\( \frac{4k \cdot q_1 \cdot q_2}{d'^2} = \frac{k \cdot q_1 \cdot q_2}{d^2} \)

By canceling \( k \cdot q_1 \cdot q_2 \) from both sides, we get:

\( \frac{4}{d'^2} = \frac{1}{d^2} \)

Solving for \( d' \), we determine:

\( d'^2 = 4d^2 \)

\( d' = 2d \)

Consequently, for the force to remain constant when both charges are doubled, the distance must be increased to 2d.