The Total solution for NCERT class 6-12
(a) Twoinsulated charged copper spheres A and B have their centers separated by adistance of 50 cm. What is the mutual force of electrostatic repulsion if thecharge on each is 6.5 × 10−7 C? The radii of A and B arenegligible compared to the distance of separation.
(a) Charge onsphere A, qA = Charge on sphere B, qB =6.5 × 10−7 C
Distance between thespheres, r = 50 cm = 0.5 m
Where,
∈0 = Free space permittivity
= 9 × 109 N m2 C−2
= 1.52 × 10−2 N
Therefore, the force between thetwo spheres is 1.52 × 10−2 N.
(b) Afterdoubling the charge, charge on sphere A, qA =Charge on sphere B, qB = 2 × 6.5 × 10−7 C= 1.3 × 10−6 C
The distance between the spheresis halved.
Force of repulsion between thetwo spheres,
= 16 × 1.52 × 10−2
= 0.243 N
Therefore, the force between thetwo spheres is 0.243 N.