capacitance the ability to store charge. common types electrolytic tanalumsilver mica
TRANSCRIPT
Capacitance Depends on… Separation of the plates (d) [d in m]
d
1C
AC
C
Plate Area (A) [A in m2]
Dielectric Constant () [ in C2/N-m2]
Capacitance
d
AC
farad J
C
Nm
C
Nm
mC
m
mNmC
:Units
22
3
22
22
2
Capacitance depends on the construction of the capacitor, not on the circuit it is used in.
Example
What is the capacitance of a capacitor if the charge is 0.075 C held at V = 400 V
C = Q/V =0.075 C/400 V = 1.875 X 10-4 f = 187.5 f
Storage of Charge Charges flow from the wire to one plate of
the capacitor. The insulator (dielectric) prevents charges from flowing to the other plate. Charge accumulates.
The accumulated charge attracts opposite charges on the other plate.
The second plate charge pushes opposite charges down the other wire.
+
+++
++
――――――
Capacitors in DC Circuits
Capacitor in a dc circuit Capacitor will begin charging and current
will flow When fully charged (Q= CV), current
stops Electric field between plates cancels the
field produced by the battery voltage
Dielectric Constant
Many texts give in terms of a fundamental constant (o) and a number that depends on the material (K)
= Ko
o = permittivity of free space
= 8.854 X 10-12 C2/Nm2
K = dielectric constant
Typical Values of KMaterial K
Air (1 atm) 1.00059
Vacuum 1.00000
Ammonia 22
Glass 5-10
Mica 3-6
Paraffin Wax 2.1-2.5
Porcelain 6.0-8.0
Rubber 2.5-3.0
Example Two rectangular sheets of copper foil 16 X 20 cm are
separated by a thin layer of paraffin wax 0.2 mm thick. Calculate the capacitance if the dielectric constant for the wax is 2.4.
pf3400f10X4.3C
m10X2
)m032.0)(4.2)(Nm/C10X854.8(C
m10X2mm2.0d
m032.0cm320)cm20)(cm16(A
d
AKC
9
4
22212
4
22
o
Combining Capacitance
Parallel – plates of the capacitors are at the same voltage as the poles of the battery.
Voltage across AA’ = BB’ = E
Parallel Capacitance
Q1 = C1E Q2 = C2E Q3 = C3E
Total Charge stored is…
QT = Q1 + Q2 + Q3
QT = E(C1 + C2 + C3) = CTE
CT = C1 + C2 + C3
Series Capacitance
The same current flows through each capacitor
Current is charge/time, so Q1 = Q2 = Q3 = QT
Series Capacitance Kirchhoff’s Voltage Law
E = V1 + V2 + V3
321T
3
T
2
T
1
T
T
T
3
3
2
2
1
1
T
T
C
1
C
1
C
1
C
1
C
Q
C
Q
C
Q
C
Q
C
Q
C
Q
C
Q
C
Q
Example
Three capacitors of 2.0, 3.0, and 5.0 f are connected in parallel to a 12 V source.1.Find the charge on each capacitor.2.Find the total charge of the combination.3.Find the total charge if the same three are connected in series to the 12 V source.
Solution (Parallel connection)
1. Q1 = VC1 = (12 V)(2.0X 10-6f) =24 C
Q2 = VC2 = (12 V)(3.0X 10-6f) =36 C
Q3 = VC3 = (12 V)(5.0X 10-6f) =60 C
2. QT = Q1 + Q2 + Q3 = (24+36+60)C
= 120 C
Solution (part 3) Series connection
f31
30C
30
31
30
61015
f0.5
1
f0.3
1
f0.2
1
C
1
C
1
C
1
C
1
T
321T
C6.11Q
)f10X31
30)(V12(VCQ
T
6TT
Charge Form For a dc voltage flowing for a time t
I = Q/t E = Q/C + QR/t = Q(1/C + R/t)
RCt
ECtQ
CtRCt
E
tR
C
EQ
1
Time dependent forms Notice that RC must have units of time –
most books call this = RC Current
/teR
Ei
)e1(EV /t Voltage