blackberri9675
16.08.2020 •
Physics
Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on each? A. 4.0x10-6 C 4.0x10-6 C B. 7.4x10-6 C 0.6x10-6 C C. 6.6x10-6 C 1.4x10-6 C D. 5.0x10-6 C 3.0x10-6 C
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Ответ:
B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C
Explanation:
From Coulomb's Law the electrostatic repulsive force is given by the following formula:
F = kq₁q₂/r²
where,
F = Repulsive Force = 0.15 N
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = Magnitude of 1st Charge = ?
q₂ = Magnitude of 2nd Charge = ?
r = Distance between Charges = 0.5 m
Therefore,
0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²
q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)
q₁q₂ = 4.17 x 10⁻¹²
q₁ = (4.17 x 10⁻¹²)/q₂ equation (1)
The sum of charges is given as:
q₁ + q₂ = 8 μC
q₁ + q₂ = 8 x 10⁻⁶
using equation (1):
(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶
(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂
q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0
Solving this quadratic equation:
q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C
q₂ = 7.4 μC (OR) q₂ = 0.6 μC
Therefore,
q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)
q₁ = 0.6μC
Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.
Therefore, the correct option will be:
B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C
Ответ: