# In a flat horizontal capacitor, a charged drop of mercury is in equilibrium at a field strength between the plates

August 24, 2021 | education

| **In a flat horizontal capacitor, a charged drop of mercury is in equilibrium at a field strength between the plates of 600 kV / m. Determine the mass of the drop if its charge is 4.8 * 10-17 C.**

Since a charged drop of mercury, according to the condition, is in a state of equilibrium, the following equality will be true: Ft = Fp and m * g = E * q, whence we can express: m = E * q / g.

Const: g – acceleration due to gravity (g ≈ 9.81 m / s2).

Data: E – field strength of the capacitor used (E = 600 kV / m = 6 * 105 V / m); q is the charge of a drop of mercury (q = 4.8 * 10-17 C).

Let’s calculate: m = E * q / g = 6 * 105 * 4.8 * 10-17 / 9.81 ≈ 2.936 * 10-12 kg.

Answer: The mass of a charged drop of mercury should be 2.936 * 10-12 kg.

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