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?

#### Solution

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

Charge on this sphere, q_{1} = 0.4 μC = 0.4 × 10^{−6 }C

Charge on the second sphere, q_{2}_{ }= − 0.8 μC = − 0.8 × 10^{−6}^{ }C

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)`

`= (9xx10^9xx0.4xx10^-6xx0.8xx10^-6)/0.2`

`=144xx10^-4`

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