A parallel plate capacitor has plates of area A=250cm^2 and separation d=2.00cm.

Question

A parallel plate capacitor has plates of area A=250cm^2 and separation d=2.00cm. The capacitor is charged to a potential difference Vo=120V. Then the battery is disconnected (the charge Q on the plates won't change), and a dielectric sheet (k=3.25) of the same area A but thickness l=5.00mm is placed between the plates.

determine:

a. The initial capacitance of the air filled capacitor.

b. The charge on each plate before the dielectric is inserted.

c. The charge induced on each face of the dielectric after it is inserted.

d. The electric field in the space between each plate and the dielectric.

e. The electric field in the dielectric.

f. The potential difference between the plates after the dielectric is added

g. The capacitance after the dielectric is in place.

Explanation / Answer

a) c = Ae0/d

c = 250*10^-4*8.85*10^-12/0.02 = 11.06*10^-12 F

B) Q = CV = 11.06*10^-12*120 = 1.33*10^-9 C

C) C' = Ae0/d-t(1-1/k) = 13.38*10^-12 F

v' = v/k = 120/3.25 = 36.92 volts

q = c'v' = 13.38*10^-12*36.92 = 0.49*10^-9 C

d) E = V'/d = 36.92/0.02 = 1846 V/m

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