A microscopic oil drop of mass 3.2 10⁻¹² g is held stationary in a downward vertical uniform electric field of magnitude 10^4 N C ⁻¹ . The number of excess electrons on the drop is : (Given g = 10 m s ⁻² , elementary charge e = 1.6 10⁻¹⁹ C )

Step-by-step solution

For the oil drop to remain stationary, the net force acting on it must be zero. The upward electric force must perfectly balance the downward gravitational force. Since the electric field is directed downwards and the electric force must be upwards, the charge on the drop must be negative (due to excess electrons). Equating the magnitudes of the forces: qE = mg Using the quantization of charge, q = ne , where n is the number of excess electrons: neE = mg n = mg eE Converting the mass into SI units (kg): m = 3.2 10⁻