A parallel plate capacitor is made from circular plates of Aluminum foil each wi
ID: 1278506 • Letter: A
Question
A parallel plate capacitor is made from circular plates of Aluminum foil each with a radius of 3 cm.
The distance between the plates is 8 mm. A 120 V emf is connected to the plates.
(A) Find the charge stored in this capacitor.
(B) Calculate the amount of energy stored in the capacitor.
(C) If one were to keep the 120 V battery connected while increasing the separation
between the plates to 12 mm, what would be the final energy stored in the capacitor?
(D) If instead one disconnects the battery and increases the distance between the plates
to 12 mm, what would be the final energy stored in the capacitor?
Explanation / Answer
a)
Area
A=pi*r2=pi*0.032=2.83*10-3 m2
Capacitance
C=eoA/d =(8.85*10-12)(2.83*10-3)/(8*10-3)
C=3.13*10-12 F
Charge stored in capacitor is
Q=CV =3.13*10-12*120
Q=3.753*10-10 Coloumbs or 0.3753 nC or 375.3 pC
b)
Energy stored in capacitor is
U=(1/2)CV2=(1/2)*(3.13*10-12)*1202
U=2.25*10-8 J or 22.5 nJ
c)
Capacitance at d=12 mm
C=(8.85*10-12)(2.83*10-3)/(12*10-3)
C=2.085*10-12
Since battery remains connected Voltage remains same ,charge changes
U=(1/2)CV2=(1/2)(2.085*10-12)*1202
U=1.5*10-8 J or 15 nJ
d)
Since battery is disconnected,charge remains constant
U=(1/2)(Q2/C) =(1/2)(3.753*10-10)2/(2.085*10-12)
U=3.38*10-8 J or 33.8 nJ