Question

How does the magnitude of the electrical force compare between a pair of charged particles when they are brought to half their original distance of separation? To one-quarter their original distance? To four times their original distance? (What law guides your answers?)

Answer 1

Answer:

a) 4 times larger. b) 16 times larger. c) 16 times smaller. d) Coulomb´s Law

Explanation:

Between any pair of charged particle, there exists a force, acting on the line that join the charges (assuming they can assimilated to point charges) directed from one to the other, which is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them.

F= k q1q2 / (r12)2

a) If the distance is reduced to the half of the original distance of separation, and we introduce this value in the force equation, we get:

F(r/2) = k q1q2 / (r12/2)2 = k q1q2 /((r12)2/4) = 4 F(r)

b) By the same token, if r= r/4, we will have F(r/4) = 16 F(r)

c) If the distance increases 4 times, as the force is inversely proportional to the square of the distance, the force will be the original divided by 16, i.e., 16 times smaller.

The empirical law that allows to find out easily these values, is the Coulomb´s Law.

Explanation: