Question

One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value.

How will the electrical force between the charges compare with the original force?

It will increase to six times the original force.
It will increase to thirty-six times the original force.
It will decrease to one-sixth the original force.
It will decrease to one-thirty-sixth the original force.

Answer 1

New force = k*(q1/3)*(q2/2)/r^2 = (1/6) * k*q1*q2/r^2, so it decreases to one-sixth of the original force.