two 10.0 cm diameter metal plates are 1.00cm apart. they are charged to +/- 12.5nC. they are suddenly connected together by a 0.224mm diameter copper wire stretched taut from the center of one plate to the center of the other.
Q1: what is the maximum current in the wire?
Q2: what is the largest electric field in the wire?
Q3: does the current increase with time, decrease with time or remain steady? Explain.
Q4: what is the total amount of energy dissipated in the wire?
capacitor and current
2009-03-08 4:01 pm
回答 (1)
2009-03-08 5:40 pm
✔ 最佳答案
Resistivity of copper, ρ = 1.7 X 10^-8 ΩmArea of the copper wire, A = π(d/2)^2 = π[(0.224 X 10^-3)/2]^2 = 3.94 X 10^-8 m
Length of the copper wire, l = 0.01 m
So, resistance of the copper wire, R = ρl / A = (1.7 X 10^-8)(0.01) / (3.94 X 10^-8) = 4.31 X 10^-3 Ω
Now, capacitance of the metal plates, C
= εoA / d (where εo is the permittivity of vacuum)
= (8.85 X 10^-12)π[(0.1)/2]^2 / 0.01
= 6.95 X 10^-12 F
By Q = CV
12.5 X 10^-9 = 6.95 X 10^-12 V
Voltage across the two metal plates, V = 1800 V
1. Maximum current occurs when t = 0,
Max. current = V / R
= 1800 / 4.31 X 10^-3
= 4.18 X 10^5 A
2. Largest electric field, E
= V / d
= 1800 / 0.01
= 1.8 X 10^5 Nm^-1
3. The current follows the equation, I = Io e^(-t / RC)
As the term e^(-t / RC) is decreasing, so the current decreases with time.
OR:
As the capacitor is being discharged, the voltage across the two metal plates is decreasing. As a result, the current decreases with time.
4. Total energy dissipated
= ∫P dt (from t = 0 to t = ∞)
= ∫I^2R dt
= (Io)^2R ∫e^(-2t / RC) dt
= (Io)^2R (RC / -2) ∫e^(-2t / RC) d(-2t / RC)
= (IoR)^2C / (-2) [e^(-2t / RC)] (t = 0 to t = ∞)
= 1/2 (IoR)^2C
= 1/2 CV^2
= 1/2 (6.95 X 10^-12)(1800)^2
= 1.13 X 10^-5 J
參考: Physics king
收錄日期: 2021-04-19 13:40:07
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