Question

3 equal charges A B C are placed at a distance between AC is 2cm and CD is 1cm . the force on C due to A is 3x10-6 . calculate force on charge C due to charge B​

Answer 1

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Given:

Repulsive force of magnitude 6 × 10−3 N

Charge on the first sphere, q1 = 2 × 10−7 C

Charge on the second sphere, q2 = 3 × 10−7 C

Distance between the spheres, r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation:

F = (1/4πε0). (q1q2)/ (r2)

Where, ε0 = Permittivity of free space and

(1/4πε0) =9 × 109 Nm2C−2

Therefore, force F = (9 × 109 × 2 × 10−7)/ ((0.3)2)

= 6 × 10−3N

Hence, force between the

two small charged spheres is 6 × 10−3 N.

The charges are of same nature. Hence,

force between them will be repulsive.