Question

Two charges of magnitude +q are separated by a distance d. Assuming there are no changes to anything else, but the charges are separated by 3 times their current distance, what will happen to the Coloumb force between them? (1 point)

Answer 1

Answer:

The Coulomb force between them will decrease by a factor of 1/9

Explanation:

The Coulomb's force acting between two particles of equal charge, q, separated by a distance, d, is given as:

$F = \frac{kq^2}{d^2}$

where k = Coulomb's constant

If they are separated by 3 times their current distance, the new distance becomes 3d. Hence, the Coulomb force acting between them becomes:

$F_n = \frac{kq^2}{(3d)^2}\\ \\\\F_n = \frac{kq^2}{9d^2} \\\\\\F_n = \frac{1}{9} F$

Hence, the Coulomb force decreases by a factor of 1/9.

Answer 2

Answer: 1/9

Explanation:

Coulomb's law states that "the force acting between two charged bodies is directly proportional to the product of the charge on the two bodies and inversely proportional to the square of the separation(distance) between two bodies."

Now, it is known that the only factor changing, is the distance between the two bodies, thus, we can say from Coulomb's law that the ratio of both forces(initial and final) is equal to the square of the distance of the distance between the 2 i.e

 F(final) / F(initial) = d(initial)² / d(final)²;

where d is the distance between the bodies.

From the question, we know that ratio of the distances is 3, i.e d(final) / d(initial) = 3, so that,

F(final) / F(initial) = 1/9.

The final coulomb force between the charges will be 1/9th of the initial