A square PP-capacitor has sides of length 0.750 m and separation distance 1.00 m

Question

A square PP-capacitor has sides of length 0.750 m and separation distance 1.00 mm and is charged to 120 V . (a) What is E~ between the plates? (b) What is the energy per unit volume between the plates? (c) Using your answer from part (b), find the total electrostatic field energy between the plates (ignore fringing effects) (d) What is the capacitance of the PP-capacitor? (e) Using your answer from part (d), find the total energy stored in the capacitor. Compare this answer to you answer from part (b).

Explanation / Answer

a) Electric Field = E = V/d = 120/1*10^-3 = 120000 N/C

b) The energy per unit volume between the plates = u = 0.5e0*E^2 = 0.5*8.85*10^-12*(120000)^2

u = 0.06372 = 63.72 mJ/m^3

c) The total electrostatic field energy between the plates = U = uV = uAd

U = 63.72*10^-3 * (0.750)^2*1*10^-3 = 35.84 uJ

d) Capacitance = C = e0A/d = (8.85*10^-12*0.750^2)/10^-3 = 4.98 nF

e) The total energy stored in the capacitor = 0.5CV^2 = 0.5*4.98*10^-9*(120)^2 = 35.86 uJ which is well in agreement with (c).

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