A pith ball A of mass 9×10-5 kg carries a charge of 5 μC. What must be the magnitude and sign of the charge on a pith ball B held 2 cm directly above the pith ball A, such that the pith ball A, remains stationary?

Here, charge on pitch ball A, q1=5μC=5×10−6CMass of pitch ball A, m1=9×10−5kgThe weight m1  of the pitch ball A acts vertically downwards.Let q2 be charge on the pitch ball B held 2 cm above the pitch ball A, so that the pitch ball A remains stationary. It can be possible only, if the charges on two pitch balls are of opposite signs,i.e if charge on pitch ball A is positive, charge on B must be negative. Then, the force on pitch ball A due to B, i.e. FAB will act vertically upwards (figure). For charge q1 to remain stationary,FAB=m1g or 14πε0⋅q1q2AB2=m1gHere, AB = 2 cm = 0.02 m∴ 9×109×5×10−6×q2(0.02)2=9×10−5×9.8 or  q2=7.84×10−12C