(a) Two insulated charged copper spheres A and B have their centers speparated by a distance of 50 cm. What is the mutal force of electrostatic repulsion if the charge on each is `6.5xx10^(-7) C` ? The radill of A and B are negalible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved ?

Explanation

(a) The radii of the charged spheres are very small, hence we can treat them like point charges. Using Coulomb.s aw for the force between spheres, we get the following:
`F=1/(4piepsilon_0)(q_1q_2)/r^2`
`rArr F=9xx10^9xx(6.5xx10^(-7))^2/(0.5)^2`
`=1.521xx10^(-2)` N
(b) We know that the magnitude of the electrostatic force between two charges is directly proportional to the product of the charges, therefore by making the charge double in both the spheres, the force should become four times its original value. Further, the same force is inversely proportional to the square of the distance between the charges, which means that by making the distance between the charges half of the original value, the force again becomes four times its original value. Thus the electrostatic force acting between the charges will be 16 times the original value when the changes are made. The magnitude of the electrostatic force can be written as follows:
`F_"new"=16xx1.521xx10^(-2)` =0.243 N