A parallel plate capacitor has an area of 5.00 cm2 and a capacitance of 3.50 pF. The capacitor is connected to a 12.0 V battery. After the capacitor is completely charged up, the battery is removed
a. What must be the separation of the plates and how much energy is stored in the electric field between the plates?
b. If a sheet of Mica with εr=3.5 is placed between the plates, how will the voltage across the plates and the charge density on the plates change?
c. If the sheet of Mica has an area of 3.00 cm 2 so that it does not cover the whole area of the plates, what would be the effective capacitance of this arrangement?
d. If the Mica sheet is removed and the plates are moved 1.00 mm farther from each other, what will be the voltage between the plates, and how much work will it take to move the plates apart?
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Expert's answer
2021-02-28T07:37:42-0500
Area of plates, A = 5 cm2 = 5 x 10-4 m2
Capacitance, c = 3.50 pF = 3.50 x 10-12 F
Voltage, V = 12V
(a) c=dϵοA
⇒d=cϵοA
⇒d=3.50×10−128.85×10−12×5×10−4
⇒d=12.64×10−4m
Energy store, E=21cV2
⇒E=21×3.50×10−12×(12)2
⇒E=242×10−12J
(b) A dielectric medium, mica with relative permittivity ϵr=3.5 is placed between the plates
c=dϵA
⇒c=dϵοϵrA
⇒c=12.64×10−48.85×10−12×3.5×5×10−4
⇒c=12.25×10−12F
Energy stored, E=21cV2
⇒V2=c2E
⇒V=c2E
⇒V=12.25×10−122×252×10−12
⇒V=6.414V
Voltage of the battery decreases to 6.414V after placing mica sheet
ϵrEο=Eο−Ei
⇒ϵοϵrσο=ϵοσο−ϵοσi
⇒σi=σ(1−k1)
From above equation it is clear that k=ϵr
When mica sheeet is placed between the plates, the charge density decreases.
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