Question

The electrostatic force on a small sphere of charge 0.4 $\mu$C due to another small sphere of charge –0.8 $\mu$C in air is 0.2 N.
What is the distance between the two spheres?
What is the force on the second sphere due to the first?

Answer 1

Let us consider two charges $q_1$ and $q_2$. According to the question, $q_1=0.4\mu C,q_2=-0.8\mu C$



Let the distance between two charges be r. Force on the charge $q_1\left(0.4\mu C\right)$ due to another charge $q_2(-0.8\mu C)is$
F=0.2N
We have to find the value of r. Using Coulomb's law, the force between two charges is
$F=\frac{1}{4\pi {\epsilon }_0}.\frac{q_1{q}_2}{r^2}$
Putting the values of F, $\frac{1}{4\pi {\epsilon }_0},q_{1}and{q}_2,weget$
$0.2=9\times {10}^9\times \frac{0.4\times {10}^{-6}\times 0.8\times {10}^{-6}}{r^2}\Rightarrow r^2=16\times 9\times {10}^{-4}\Rightarrow r=4\times 3\times {10}^{-2}\Rightarrow r=12\times {10}^{-2}m\Rightarrow r=12cm$
Here, the charge $q_2$ is negative in nature and $q_1$ is positive in nature. So, the force between $q_1$ and $q_2$ will be attractive in nature as unlike charges attract each other.

The force attraction on the second sphere due to the first sphere is same as the force between the two charges i.e., 0.2N. Electrostatic force between two charges obeys the Newton's third law of action-reaction law.