Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of their electrical charges q1 and q2 and inversely as the square of the distance d between the particles. Find the effect on F of doubling q1 and q2 and halving the distance between them.

Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d² ... (1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k (2q1) (2q2) / (d/2) ²

F2 = 4kq1q2 / (d²/4)

F2 = 16kq1q2/d² ... (2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² : 16kq1q2/d²

F1/F2 = kq1q2/d² * d²/16kq1q2

F1/F2 = 1/16

F1 = 1/16F2

This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved