In a flat horizontal capacitor, a charged drop of mercury is in equilibrium.

In a flat horizontal capacitor, a charged drop of mercury is in equilibrium. The electric field between the plates is 30,000 N / C. Determine the mass of the drop if its charge is 8 * 10 ^ -19 C

A drop of mercury is acted upon by the force of gravity Ft = mg and the force of the electric field Fe = Eq. In the formulas, m is the droplet mass, g is the acceleration of gravity, E is the field strength, q is the magnitude of the charge.

If the drop is in equilibrium, it means that the sum of these forces is zero (the forces are directed in opposite directions). The absolute values of these forces are equal:

Ft = Fe;

mg = Eq;

m = Eq / g = 30,000 N / C * 8 * 10 ^ -19 C / (10 m / s ^ 2) = 24 * 10 ^ 4 * 10 ^ (- 20) ((kg * m) / s ^ 2) * (s ^ 2 / m) = 24 * 10 ^ (- 16) kg = 2.4 * 10 ^ (- 15) kg.

Answer: m = 2.4 * 10 ^ (- 15) kg.