8) A parallel plate capacitor consists of two plates, each having an area of 154
ID: 1427271 • Letter: 8
Question
8) A parallel plate capacitor consists of two plates, each having an area of 154.7 cm2 , separated by a distance of 0.0075 cm. How much charge accumulates on a plate if you hook this capacitor across a 12 Volt battery? What is the electric field between the capacitor plates? You connect the capacitors discussed in problem (8), initially uncharged, to a 5.0 V battery in a parallel configuration. You then disconnect them from the battery and from each other. You then connect the positive side of one capacitor to the negative side of the other and vice-versa. What is the final voltage across each capacitor?
Explanation / Answer
given that
A=154.7cm2
D=0.0075cm
V=12
C=q/v
C=0A/d
=[( 8.85x10-12)x(154.7x10-4)]/(0.0075x10-2)
=(1.3690x10-2)/(0.0075x10-2)
c=1.825x10-9
charge q=cv
=(1.825x10-9)x12
=2.19x10-8
electric field E=q/0
=(2.19x10-8)/(8.85x10-12)
e=2475.2
C2 and C3 in parallel.
capacitance = (0.01 + 0.02)F
= 0.03 F
C1 in series with this combination.
Total capacitance = 1 / [ 1/C1 + 1/0.03F]
= 1 / [ 1 / 0.02F + 1 / 0.03F]
= 1 / [100/2 + 100/3]
= 1 / [500/6]
Total capacitance = 6/500 F = 0.012 F
Total charge Q = CV = 0.012 X 10^-6 X 100
Q = 1.2 coulombs.
For capacitors in series the charge is the same on each capacitor. For parallel capacitors the charge is shared between them.
Charge of 1.2 C is on C1.
This charge is shared between C2 and C3 in the ratio
Q3 / Q2 = C3 / C2 = 2 : 1
Charge on C2 = 1/3 (1.2 C) = 0.4C
Charge on C3 = 2/3 (1.2C) = 0.8C
Voltage across each is found from V = Q / C
For C1
V1 = 1.2C / 0.02F = 60 volts
C2 and C3 have same voltage across them since they are in parallel.
Voltage across C2 and C3 = 100 - 60 = 40 volts.
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