Question

Calculate the intensity of electric field due to the sphere of the following points at the surface of the sphere , at an external point 0.8m from the surface of the sphere, at a point 0.4m from the centre of the sphere

Answer 1

Explanation:

Given:

Repulsive force of magnitude 6 × 10−3N

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

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.

 Question2.

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Answer:

(a) Electrostatic force on the first sphere, F = 0.2 N

Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C

Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C

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, r2 = ((1/4πε0)). ((q1q2)/ (F))

= (0.4 × 10−6 × 8 × 10−6 × 9 × 109)/ (0.2)

= 144 × 10−4

⇒ r = √ (144 × 10−4)

 = 12 × 10−2

= 0.12 m

The distance between the two spheres is 0.12 m.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N.