Question

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.

1. What is the distance between the two spheres?
2. What is the force on the second sphere due to the first?

Answer 1

Given: 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

(a) The electrostatic force between the spheres is given by the relation,

`F = (q_1q_2)/(4piin_0r^2)` and, `1/(4piin_0)`

= `9xx10^9 Nm^2C^-2`

Where, ∈₀ = Permittivity of free space

`r^2 = (q_1q_2)/(4piin_0F)`

`= (9 xx 10^9 xx 0.4 xx 10^-6 xx 0.8 xx 10^-6)/0.2`

`= (2.88 xx 10^-3)/0.2`

`= 14.4 xx 10^-3`

`= 144 xx 10^-4`

`r = sqrt(144xx10^-4)`

= 0.12 m

The distance between the two spheres is 0.12 m.

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