Two charges attract each other with a force of 1.9 n. what will be the force if the distance between them is reduced to one-sixth of its original value?
From Couloumbs's law we have:
F = q1q2/r^2
So 1.9 = q1q2/r^2 ----------(i)
When the distance is one-sixth we have.
F = q1q2/(r/6)^2
This can be written as
F = 36 x q1q2/r^2 ----------(2)
F = 36 x 1.9. Recall 1.9 = q1q2/r^2 from (1)
So F = 68.4N
