Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200
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Answer:
(a) The potential difference between the spheres is 750 kVA
(b) The charge on the smaller sphere is 6. μC
(c) The charge on the smaller sphere, Q₁ = 13. μC
Explanation:
(a) The given parameters are;
The charge on the inner sphere, q = 5.00 μC
The radius of the inner sphere, r = 3.00 cm = 0.03 m
The charge on the larger sphere, Q = 15.0 μμC
The radius of the larger sphere, R = 6.00 cm = 0.06 m
The potential difference between two concentric spheres is given according to the following equation;
Where;
R = The radius of the larger sphere = 0.06 m
r = The radius of the inner sphere = 0.03 m
q = The charge of the inner sphere = 5.00 × 10⁻⁶ C
Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C
k = 9 × 10⁹ N·m²/C²
Therefore, by plugging in the value of the variables, we have;
The potential difference between the spheres, = 750,000 N·m/C = 750 kVA
(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;
= Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C
= 20 × 10⁻⁶ C
From which we have;
Q₁/Q₂ = R/r
Where;
Q₁ = The new charge on the on the larger sphere
Q₂ = The new charge on the on the smaller sphere
= 20 × 10⁻⁶ C = Q₁ + Q₂
∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂
∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2
20 μC - Q₂ = 2·Q₂
20 μC = 3·Q₂
Q₂ = 20 μC/3
The charge on the smaller sphere, Q₂ = 20 μC/3 = 6. μC
(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3
The charge on the smaller sphere, Q₁ = 40 μC/3 = 13. μC.
