Question

Two point charges of 25μC and -36μC of charges are separated by a distance of 4m in air. Determine the position of the point at which the resultant electric field is zero and find the magnitude and sign of charge to be placed at that point so that the system will be in equilibrium? `(20m 500µC)

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.

Explanation:

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