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The force between two small charged spheres having charges of 1 x 10-7 C and 2 x 10-7 C placed 20 cm apart in air is

  • a)
    4.5 x 10-2N

  • b)
    4.5 x 10-3 N

  • c)
    5.4 x 10-2 N

  • d)
    5.4 x 10-3

Correct answer is option 'B'. Can you explain this answer?
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Calculation of force between two charged spheres having charges of 1 x 10^-7 C and 2 x 10^-7 C placed 20 cm apart in air.

Given,
Charge on sphere 1, q1 = 1 x 10^-7 C
Charge on sphere 2, q2 = 2 x 10^-7 C
Distance between the spheres, r = 20 cm = 0.2 m
Permittivity of air, ε0 = 8.85 x 10^-12 C^2/Nm^2

Coulomb's Law:
According to Coulomb's law, the force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Mathematically, F = k(q1q2)/r^2, where k is the Coulomb's constant.

Coulomb's Constant:
The Coulomb's constant (k) is defined as the proportionality constant in Coulomb's law, which relates the electrostatic force between two charged particles to their charges and the distance between them.

k = (1/4πε0) = 9 x 10^9 Nm^2/C^2

Calculation:
Substituting the given values in the formula of Coulomb's law, we get

F = k(q1q2)/r^2
F = (9 x 10^9 Nm^2/C^2) x (1 x 10^-7 C) x (2 x 10^-7 C) / (0.2 m)^2
F = 4.5 x 10^-3 N

Therefore, the force between two charged spheres having charges of 1 x 10^-7 C and 2 x 10^-7 C placed 20 cm apart in air is 4.5 x 10^-3 N.

Hence, the correct option is (b) 4.5 x 10^-3 N.
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