Question

Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are doubled and their separation is also doubled, what happens to the electrical force that each charge exerts on the other one?

A. It increases by a factor of 2
B. It increases by a factor of the square root of 2
C. It increases by a factor of 4
D. It remains the same
E. It is reduced by a factor of the square root of 2

Answer 1

The electrical force between two point charges that have both been doubled in magnitude, with their separation also doubled, remains the same, according to Coulomb's Law.

Step-by-step explanation:

The question involves Coulomb's Law, which describes the electrostatic force between two point charges. According to Coulomb's Law, the electrostatic force (F) between two point charges is directly proportional to the product of the magnitudes of the charges (Q1 and Q2) and inversely proportional to the square of the separation between them (R2). Mathematically, F is proportional to (Q1*Q2)/R2.

When both charges are doubled, the product of their magnitudes becomes 4 times greater since (2*Q1) * (2*Q2) = 4 * Q1 * Q2. If the separation is also doubled, the square of the separation becomes 4 times greater as well since (2*R)2 = 4 * R2. As a result, the factor increase in the product of charges (which increases the force by a factor of 4) is exactly cancelled out by the factor increase in the square of the separation (which decreases the force by a factor of 4). Thus, the force remains the same.

The correct answer is D. It remains the same.