A microscopic oil drop of mass 1.1×10-12 kg and charge 3e is suspended in the gap between two parallel metal plates.

Question 27

A microscopic oil drop of mass 1.1×10^-12 kg and charge 3e is suspended in the gap between two parallel metal plates. The separation between the plates is 10.68 mm. Calculate the potential difference between the plates, in kV (i.e. V x 10³), to the nearest kV.

Reference: ???

Solution:

The forces acting on the oil drop should balance for the drop to be suspended. The two forces acting on the drop are the weight and the electric force.

Weight = mg

Electric force = E × q              where E is the electric field strength and q is the charge

Electric field strength = V / d             where V is the p.d. and d is the separation of the plates

Electric force = (V/d) × 3e

Electric force = Weight

V × 3e / d = mg

V = mgd / 3e

V = (1.1×10^-12 × 9.81 × 10.68×10^-3) / (3 × 1.6×10^-19)

p.d. V = 2.4×10⁵ V = 240 kV