A positive test charge of 5.5 10-4 C is near a point charge that exerts a force of 2.1 10-4 N on it.?

The test charge is moved to a distance four times as far from the charge. What is the magnitude of the force that the field exerts on the test charge now?

A positive test charge of 5.5 10-4 C is near a point charge that exerts a force of 2.1 10-4 N on it.?
F= q/(4pi(epsilon)r^2)


where r is the distance between the charges


so


2.1*10^-4 = q/(4pi(epsilon)r^2).............1


So now let the new distance be R


so R = 4r


so the new force is


Fn = q/(4pi(epsilon)R^2)


Fn= q/(4pi(epsilon)(4r)^2)


=q/(4pi(epsilon)16r^2)


from eqn 1


Fn = F/16


=(2.1*10^-4)/16


=0.13 * 10^-4N