please make sure you do all parts and you have the right answers. An air-gap par

Question

please make sure you do all parts and you have the right answers.

An air-gap parallel-plate capacitor that has a plate area of 2.60 m2 and a separation of 1.80 mm is charged to 90 V.

(a) What is the electric field between the plates?
..... kV/m

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

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

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

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

Explanation / Answer

Electric permitivity in air eo =8.85*10^-12 C^2/N^2.m^2

A =2.6 m^2 , d =1.8 mm , V =90V

(a) E = V/d = 90/(0.0018)

E = 50 kV/m

(b) u =(1/2)eoE^2 =(1/2)(8.85*10^-12)(50000*50000)

u = 11.0625 mJ/m^3

(c) U =uV = uAd = (11.0625*10^-3)(2.6)(1.8*10^-3)

U = 51.7725 µJ

(d) C =Aeo/d =(2.6*8.85*10^-12)/(1.8x10^-3)

C = 12.783 nF

(e) U =(1/2)CV^2 =(1/2)(12.783*10^-9)(90*90)

U =51.7725 µJ

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