A positive test charge of 5.5 10-4 C is near a point charge that exerts a force of 2.3 10-4 N on it. The tes

A positive test charge of 5.5 10-4 C is near a point charge that exerts a force of 2.3 10-4 N on it. The test charge is moved to a distance five 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 2.3 10-4 N on it. The tes
distance has an inverse square relationship with the force.





F = kQq / r^2





Original force:





Fo = 2.310^-4 = k (5.5 10-4) / r^2





New Force:


Fn = k (5.5 10-4) / (5r)^2


Fn = constant / 25r^2


25Fn = constant / r^2 = Fo





25Fn = Fo


Fn = 1/25Fo


Fn = 1/25( 2.310^-4)


Fn = 9.2x10^-6





Hence new Force is 25x smaller than the original force.
Reply:First, the force is positive so that means the test charge has the same sign as the point charge. Next use:





F = kq1q2/r^2 k =9x10^9 (Nt-m^2/coul^2), q1 - 5.5x10^-4 Coul, r= distance between charges





Now you know F at some unknow distance, r. Now increase the distance by a factor of 5. Let r2 = 5r





F2 = kq1q2/r2^2 = kq1q2/(25r^2 ) = F/25 = 2.3x10^-4/25





F2 = 9.2x10^-6 N