Q 1.12) (i) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×107 C6.5×107 C? The radii of A and B are negligible compared to the distance of separation.

(ii) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?


(a)

Charge on the sphere A, qA=6.5×107 CA, qA=6.5×107 C
Charge on sphere B, qB=6.5×107 CB, qB=6.5×107 C
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between the two spheres, F=14πε0qAqBr2F=14πε0qAqBr2
F=9×109×(6.5×107)2(0.5)2=1.52×102NF=9×109×(6.5×107)2(0.5)2=1.52×102N

(b)

When the charges are doubled and the distance is halved,
The force of repulsion between the two spheres,
F=14πε02qA×2qB(r2)2=16×qA×qBr2=16×1.52×102=0.243NF=14πε02qA×2qB(r/2)2=16×qA×qBr2=16×1.52×102=0.243N