An air-gap parallel-plate capacitor that has a plate area of2.60 m 2 and a separ

Question

An air-gap parallel-plate capacitor that has a plate area of2.60 m2 and a separation of1.80 mm is charged to 90 V. (a) What is the electric field between theplates?
1 kV/m

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

given area of plates = A = 2.60m2 distance between the plates = d = 1.80mm =1.80*10-3m voltage acroos plates = v = 90v capacitence = c = 0A/d                         = 8.9*10-12*2.60/1.80*10-3 =12.85nF total energy = U = 1/2cv2=1/2*12.85*10-9 *(90)2                                       = 52.04J elecrtic field bettween plates E = q/0A    wher q = cv = 12.85*10-9F *90 =1.15c plug values in the above eq                                       = 52.04J elecrtic field bettween plates E = q/0A    wher q = cv = 12.85*10-9F *90 =1.15c plug values in the above eq

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