Question

calculate the electrostatic force between two charges of 300 nano coulomb and 200 nano coulomb placed 0.05m appart in air ?​

Answer 1

Explanation:

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.