An oil drop with a mass of 8.3 grams and a charge of + 2.2 mC is floating 7.68 meters above a positively charged floor. What is the magnitude of the charge on the floor?

Correct answer: Q = 0.247 μC

Explanation:

Since the oil drop floats, this means that the weight of the drop and the Coulomb force are equal.

F = m · g drop's weight

Fc = k q Q / r² Coulomb force

where it was given:

k = 9 · 10⁹ N m²/C² dielectric constant,

q = 2.2 mC = 2.2 · 10⁻³ C drop's charge,

Q - floor's charge,

r = 7.68 m distance between drop and floor

m = 8.3 grams = 8.3 · 10⁻³ kg drop's weight

and g = 10 m/s²

F = Fc ⇒ k q Q / r² = m · g ⇒ Q = m · g · r² / k · q

Q = 8.3 · 10⁻³ · 10 · 7.68² / 9 · 10⁹ · 2.2 · 10⁻³

Q = 24.72 · 10⁻⁸ C = 0.2472 · 10⁻⁶ C = 0.2472 μC

Q = 0.247 μC

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