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

Option 2: 2d

Let's analyze the problem using Coulomb's Law.

Coulomb's Law states that the force (F) between two point charges (q1 and q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (d) between them:

F = k * (q1 * q2) / d²

Where k is Coulomb's constant.

Initially, the force is:

F = k * (q1 * q2) / d²

Now, each charge is doubled, so the new charges are 2q1 and 2q2. We want to find the new distance (d') such that the force remains the same (F).

F = k * (2q1 * 2q2) / (d')²

Since the force must remain the same, we can equate the two expressions:

k * (q1 * q2) / d² = k * (4q1 * q2) / (d')²

We can cancel k, q1, and q2 from both sides:

1 / d² = 4 / (d')²

Cross-multiplying:

(d')² = 4d²

Taking the square root of both sides:

d' = √(4d²)

d' = 2d

Therefore, the distance between the charges should be changed to 2d to keep the force unchanged.

The correct answer is:

Option 2: 2d