Respuesta :
Answer:
Part A;
The charges are;
Q₁ = 1.32687 μC and Q₂ = 92.67313 μC
Part B
Q₁ = 95.29042 μC, Q₂ = -1.29042 μC
Explanation:
Part A
The total charge on the two nonconducting spheres = 94.0 μC
The force exerted by each on the other when placed 31 cm apart = 11.5 N
Let Q₁ = x represent the charge on one of the spheres and let Q₂ represent the charge on the other sphere
The force, 'F', exerted by a charge is given as follows;
[tex]F = k \times \dfrac{\left | Q_1 \right | \cdot \left | Q_2 \right | }{d^2}[/tex]
Where;
d = The distance between the spheres = 31 cm = 0.31 m
k = 8.9875 × 10⁹ N·m²/C²
Where F = 11.5 N
Q₁ + Q₂ = 94.0 μC
∴ Q₂ = 94.0 - Q₁ = 94.0 - x
We get;
[tex]F =11.5 = 8.9875 \times 10^9 \times \dfrac{x\times (94.0-x) \times 10^{-12}}{0.31^2}[/tex]
Therefore;
94·x - x² - 122.965 = 0
x² - 94·x + 122.965
x = (94 ± √((-94)² - 4×1×122.965))/(2 × 1)
Solving gives;
x ≈ 1.32687 × 10⁻⁶ C or x = 92.67313 × 10⁻⁶ C
Therefore, the charges are;
Q₁ = 1.32687 × 10⁻⁶ C and Q₂ = 92.67313 × 10⁻⁶ C
Q₁ = 1.32687 μC and Q₂ = 92.67313 μC
Part B
For attractive force, we have;
Q₁ + Q₂ = 94 × 10⁻⁶...(1)
[tex]11.5 = 8.9875 \times 10^9 \times \dfrac{-x\times (94.0-x)}{0.31^2} = 8.9875 \times 10^9 \times \dfrac{-Q_1\times Q_2}{0.31^2}[/tex]
-Q₁ × Q₂ = 11.5 × 0.31²/(8.9875 × 10⁹) = 1.2296523 × 10⁻¹⁰...(2)
∴ Q₂ = -1.2296523 × 10⁻¹⁰/(Q₁)
Q₁ + Q₂ = Q₁ - 1.2296523 × 10⁻¹⁰/(Q₁) = 94 × 10⁻⁶
Q₁² - 94 × 10⁻⁶·Q₁ - 1.2296523 × 10⁻¹⁰ = 0
∴ Q₁ = (94 × 10⁻⁶ ± √((-94 × 10⁻⁶)² - 4 × 1 × 1.2296523 × 10⁻¹⁰))/(2×1)
Q₁ = 9.529042 × 10⁵ C or -1.29042 × 10⁻⁶ C
Therefore, Q₁ = 9.529042 × 10⁵ C and Q₂ = -1.29042 × 10⁻⁶ C
Q₁ = 95.29042 μC and Q₂ = -1.29042 μC
