Answer:
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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