A pith ball of mass $9\times {10}^{-5}$ kg carries a charge of $5\mu \mathrm{C}$. What must be charge in another pith ball placed directly $2$ cm above the given pith ball such that they are held in equilibrium ?

Given data: A $9\times {10}^{-5}$  pith ball with a $5\mu C$   charge is in motion.

Concept used:

Detailed solution:

We know the general formula for force is  $F=mg$

Using Coulomb’s law force can be expressed as:

 $F=\frac{K{q}_B{q}_A}{r^2}\left(r=0.02m_{1}k=\frac{1}{4\pi {\epsilon }_0}\right)$ 

 $\frac{K{q}_B{q}_A}{(0.02{)}^2}=9\times {10}^{-5}\times 9.8$

 $\frac{9\times {10}^9\times q_B\times 5\times {10}^{-6}}{4\times {10}^{-4}}=9\times {10}^{-5}\times 9.8$ 

 $q=7.84\times {10}^{-12}\mathrm{C}$