Finding capacitance, excess charge, stored energy, electric field, and energy density between plates...?

Question

A parallel-plate air-filled capacitor having area 40 cm^2 and plate spacing 1.0 mm is charged to a potential difference of 600 V.

Find (a) the capacitance, (b) the amount of excess charge on each plate, (c) the stored energy, (d) the electric field between the plates, and finally (e) the energy density between the plates.

What equations did you use? Can you explain your work? Thanks and good luck! :)

Answer 1

a.) C= E*A/d where:
e = relative permitivity = er*e0 where er = 1.0006 for air, and e0 = 8.854 * 10 ^-12 farads/meter. So e = (1.0006*8.854e-12 = 8.859e-12)
A = area = (40cm)(1e-4 m^2) = .004m^2
d = length = (1mm)= .001 m
So... C=(8.859e-12)(.004)/(.001) = 3.544e-11 F or 35.44 pF

b.) Q=C*V = (3.544e-11)(600) = 2.1e-8 coulombs. So +2.1e-8 c will be on one plate, and -2.1e-8 c will be on the other. I don't know what you mean by "excess charge"

c.) U = stored energy = 1/2Q*V = 1/2 C*V^2 = 1/2(3.544e-11)(600)^2 = 6.379e-6 Joules

d.) E=Q/(e*A) = (2.1e-8)/(8.859e-12*.004) = 5.926e5 Newtons/coulomb

e.) u = energy per unit volume = U/A*d or u = 1/2 e*E^2
u = U/A*d = (6.379e-6)/(.004*.001) = 1.595 Joules/m^3

All these equations can be obtained from a basic college physics text. Also, please be carefull not to confuse the e's above for permitivity with the E's for electric field with the e's for powers of ten