A 200 µF capacitor is constructed out of two parallel plates of very large area which are separated by 1 mm. A battery is used to charge the capacitor to a potential of 100 V .
(a) If the capacitor is disconnected from the battery, how much work (in Joules) must be done to pull the plates apart to a separation of 3 mm?
A 200 µF capacitor is constructed out of two parallel plates of very large area which are separated by 1 mm.?
2018-10-28 3:13 am
回答 (1)
2018-10-28 3:34 am
Initial energy is E = ½C₀V² = ½(200µ)(100)² = 1.00 J
Q = C₀V = (200µ)(100) = 20000 µC = 20 mC
C changes when you move the plates, Q is unchanged
New C₁ = C₀/3 = 66.67 µF
new E₁ = ½Q²/C = ½(20 mC)²/(66.67 µF) = 200µ/66.67µ = 3 J
work done is 2 J, positive.
Parallel plate cap
C = ε₀εᵣ(A/d) in Farads
ε₀ is vacuum permittivity, 8.854e-12 F/m
εᵣ is dielectric constant or relative permittivity
of the material (vacuum = 1)
A and d are area of plate in m² and separation in m
Energy in a Capacitor in Joules
E = ½CV² = ½QV = ½Q²/C
Q = CV
Q is charge in coulombs
C is capacitance in Farads
V is voltage in volts
E is energy in Joules
Q = C₀V = (200µ)(100) = 20000 µC = 20 mC
C changes when you move the plates, Q is unchanged
New C₁ = C₀/3 = 66.67 µF
new E₁ = ½Q²/C = ½(20 mC)²/(66.67 µF) = 200µ/66.67µ = 3 J
work done is 2 J, positive.
Parallel plate cap
C = ε₀εᵣ(A/d) in Farads
ε₀ is vacuum permittivity, 8.854e-12 F/m
εᵣ is dielectric constant or relative permittivity
of the material (vacuum = 1)
A and d are area of plate in m² and separation in m
Energy in a Capacitor in Joules
E = ½CV² = ½QV = ½Q²/C
Q = CV
Q is charge in coulombs
C is capacitance in Farads
V is voltage in volts
E is energy in Joules
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