Question

A pith ball of mass 9*10^(-5) kg carries charge 5 micro coulomb. What must be the charge on another pithball placed at 2cm directly above the first so that the 1st one will remain in equilibrium?

Answer 1

Here, q1=5 μC=5×10−6C, q2=?

       r=2 cm=2×10−2 mm=9×10−5 kg

As,        F=mg

So,        14πε0q1q2r2=mg

or,        9×109×5×10−6 q2(2×10−2)2=9×10−5×9.8

So,          q2=7.84×10−12C