Respuesta :
Answer:
(a) The negative charge on one of the charges is -8.79630245 × 10⁻⁷C
(b) The positive charge on one of the other charges is 8.79630245 × 10⁻⁷C
Explanation:
The given parameters are;
The force of attraction between the two spheres = 0.0988 N
The distance between their centers = 44.5 cm = 0.445 m
Therefore, we have;
[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{d^2}[/tex]
Therefore, we have;
[tex]0.0988 \ N = -\dfrac{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}\cdot q_1 \cdot q_2}{(0.445 \ m)^2}[/tex]
Therefore, we have;
q₁·q₂ = -0.0988 N × (0.445 m)²/(8.99 × 10⁹ N·m²·C⁻²) = -2.17629255 × 10⁻¹² C²
q₁·q₂ = -2.17629255 × 10⁻¹² C²...(1)
When the two charges are connected, we get;
[tex]F_2 = \dfrac{k \cdot \left (\dfrac{q_1 + q_2}{2} \right) ^2}{d^2}[/tex]
Therefore, we have;
[tex]q_1 + q_2 = \sqrt{\dfrac{4 \cdot F_2 \cdot d^2}{k} }[/tex]
[tex]q_1 + q_2 = \sqrt{\dfrac{4 \times 0.0276 \ N \times(0.445 \ m)^2}{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}} } = 1.59446902743 \times 10^{-6} \ C[/tex]
q₁ + q₂ = 1.59446902743 × 10⁻⁶ C...(2)
From, equation (2), we have;
q₁ = 1.59446902743 × 10⁻⁶ C - q₂
Plugging in the value of q₁ in equation (1) givens;
q₁·q₂ = -2.17629255 × 10⁻¹²
Therefore, we have;
(1.59446902743 × 10⁻⁶ - q₂) × q₂ = -2.17629255 × 10⁻¹²
Which gives;
-q₂² + 1.59446902743 × 10⁻⁶·q₂+2.17629255 × 10⁻¹² = 0
Solving, with a graphing calculator, we get;
q₂ = 2.4741×10⁻⁶ C, or -8.79630245 × 10⁻⁷C
q₁ = 8.79630245 × 10⁻⁷C or -2.4741×10⁻⁶ C
Therefore, we have;
(a) The negative charge on one of the charges = -8.79630245 × 10⁻⁷C
(b) The positive charge on one of the other charges = 8.79630245 × 10⁻⁷C
