A 5.20uF parallel-plate air capacitor has a plate separation of 5.50mm and is charged to a potential difference of 400 V.
2 Answers
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Energy density, D, is the energy of the charged capacitor, E, divided by the volume enclosed:
D = 1/2 (CV^2) / (AS)
where:
A = area of a plate
S = separation between plates = 5.50E-3 m
But for a parallel plate capacitor,
C = ϵ0 A/S
So:
A = CS/ϵ0
AS = CS^2/ϵ0
Then,
D = 1/2 (CV^2) / (CS^2/ϵ0)
= 1/2 (V^2) / (S^2/ϵ0)
= 1/2 (V^2)ϵ0 / S^2
= 1/2 (400)^2 (8.85E-12)/(5.50E-3)^2
= 2.34E-2 J/m^3
Observe that the energy density depends on only the voltage, the separation between the plates and the dielectric constant. In other words, we didn’t need to know the value of the capacitance because it was determined by the other parameters.
Let’s try it again using the given value of the capacitor:
E = 1/2 CV^2 = 1/2 5.20E-6 (400)^2
= 0.416 J
A = CS/ϵ0 = 5.20E-6 (5.50E-3)/(8.85E-12)
= 3.23E3 m^2
D = E/AS = 0.416 J / (3.23E3 m^2)(5.50E-3 m)
= 0.0234
= 2.34E-2 J/m^3
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find magnitude of electric field
E= V/d
d is the distance between the plates, in meters
E= 400/(5.5*10^-3)
The dielectric constant for air is
k=1.00059 ; so ~1 = no change
then to find the energy density:
u= 1/2 (ϵ(E^2))
ϵ= k*ϵ0 = 1* (8.85* 10^-12)
u= 1/2 ((8.85* 10^-12)* (0.0234 ^2)
