#2. Two parallel plates capacitors C1, and C2 are connected in parallel and then
ID: 1297935 • Letter: #
Question
#2. Two parallel plates capacitors C1, and C2 are connected in parallel and then to a 24.0 volt battery. Each plate has area 80.0cm^2 and the plate separation is 3.00mm. Capacitor C1, is filled with air but capacitor C2 is filled with dielectric constant of k =2.60. a) Find the capacitance of each capacitor b) Find the charge stored in each capacitor c) Find the total energy stored in the capacitors battery ?s disconnected and then the dielectric is removed from capacitor C2. d) Find the new charge stored in each capacitor e) Find the new voltage across each capacitor 1) Find the new stored energy in the capacitorsExplanation / Answer
a)
Capacitance
C=KeoA/d
C1=1*(8.85*10-12)(80*10-4)/(3*10-3) =2.36*10-11 F or 23.6 pF
C2=2.6*(8.85*10-12)(80*10-4)/(3*10-3) =6.136*10-11 F or 61.36 pF
b)
Charge on each capacitor
Q1=C1V=23.6*10-12*24 =5.664*10-10 C or 0.5664 nC or 566.4 pC
Q2=C2V=6.136*10-11*24 =1.4726*10-9 C or 1.4726 nC or 1472.6 pC
c)
equivalent capacitance
Ceq=23.6+61.36 =84.96 pF
Total energy
E=(1/2)CeqV2=(1/2)(84.96*10-12)*242
E=2.4468*10-8 J or 24.468 nJ
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a)
Total Charge
Q=5.664*10-10+1.4726*10-9=2.04*10-9 C or 2.04 nC
Without dielectric charge on each capacitors is
Q1=Q2=Q/2=1.02*10-9 C or 1.02 nC
b)
Now
C1=C2=23.6*10-12 F
Voltage across each capcitor
V1=V2=Q1/C1=1.02*10-9/23.6*10-12
V1=V2=43.22 Volts
c)
equivalent capacitance
Ceq=C1+C2= 4.72*10-11 F
Energy stored in capacitors
E=(1/2)CeqV2 =(1/2)(4.72*10-11)*43.222
E=4.41*10-8 J or 44.1 nJ
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