Question

Calculate the formula of amount of work done in moving a charge of 5C from one point to
another on equatorial plane of a dipole of system of two charges of 10C
separated by a distance of 2mm.​

Answer 1

Answer:

What is the force between two small charged spheres having charges of 2 ×10-7C and 3 × 10–7C placed 30 cm apart in air?

Answer:

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.

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