An air-gap parallel-plate capacitor that has a plate area of3.20 m 2 and a separ

Question

An air-gap parallel-plate capacitor that has a plate area of3.20 m2 and a separation of2.80 mm is charged to 110 V. (a) What is the electric field between theplates?
1 kV/m

(b) What is the electric energy density between the plates?
2mJ/m3

(c) Find the total energy by multiplying your answer from Part (b)by the volume between the plates.
3 µJ

(d) Determine the capacitance of this arrangement.
4 nF

(e) Calculate the total energy from U =½CV2, and compare your answer with yourresult from Part (c).
5 µJ (a) What is the electric field between theplates?
1 kV/m

(b) What is the electric energy density between the plates?
2mJ/m3

(c) Find the total energy by multiplying your answer from Part (b)by the volume between the plates.
3 µJ

(d) Determine the capacitance of this arrangement.
4 nF

(e) Calculate the total energy from U =½CV2, and compare your answer with yourresult from Part (c).
5 µJ

Explanation / Answer

a) E = V/d = 110V/2.80*10-3 m = 3.93*104V/m = 39.3 kV/m b) the capacitance C = 0 A/d =8.85x10-12 C 2/N.m2*3.20m2 / 2.80*10-3 m = 1.01*10-8F electric energy density between the plates u = Energy / A*d = (1/2)CV2 /(3.20m2*2.80*10-3 m ) = 6.82*10-3J/m3 = 6.82 mJ / m3 c) V = A*d total energy E = u*A*d = 6.82mJ/m2*3.20m2*2.80*10-3 m = 61.1 J d) the capacitance C = 0 A/d =8.85x10-12 C 2/N.m2*3.20m2 / 2.80*10-3 m = 1.01*10-8F = 10.1 nF e) U=(1/2)CV2 = 0.5*10.1*10-9 F*(110V)2 = 61.1J c) V = A*d total energy E = u*A*d = 6.82mJ/m2*3.20m2*2.80*10-3 m = 61.1 J d) the capacitance C = 0 A/d =8.85x10-12 C 2/N.m2*3.20m2 / 2.80*10-3 m = 1.01*10-8F = 10.1 nF e) U=(1/2)CV2 = 0.5*10.1*10-9 F*(110V)2 = 61.1J

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