Download PChapter 14 Capacitors and Capacitance

Chapter 14
Capacitors and Capacitance
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Capacitor and Capacitance
Capacitor:
A device comprising of two conductors separated by an
insulator or a vacuum is used for storing electricity. The
insulation is called “dielectric”.
or
symbol of a constant capacitor (battery
disconnected after charging)
or
A charging process
using a battery is
required to make the
2 conductors have
charges (equal
magnitude but
opposite sign). The
net charge on the
capacitor remains
zero.
zero
symbol of a variable capacitor
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Capacitance: A measure of the ability of a capacitor to store energy and
is defined as:
Q
C=
V12
Where:
C = capacitance [unit: farad (Michael Faraday) or 1 F]
Q = magnitude of charge on each conductor [Coulomb or C]
V12 = potential difference (voltage) between the conductors (equal to
the voltage of the battery which provides the charge) [Volt or V]
+Q
1
1 microfarad = 1 µF = 10-6 F
1 picofarad = 1 pF = 10-12 F
-Q
2
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Capacitance of a parallel-plate capacitor in vacuum:
Q
A
C=
= ε0
V12
h
+Q
V12
h
Where:
C = capacitance [F]
Q = magnitude of charge on each conductor [C]
V12 = potential difference between the conductors [V]
ε0 = a universal constant [= 8.85 × 10-12 F/m or C2/Nm2]
A = area of each plate [m2]
h = separation between two plates [m]
-Q
1 F = 1 C2/Nm = 1 C2/J
Electric field (E) and potential difference (V):
σ
Q
E= =
ε0 ε0 A
1 Qh
V12 = Eh =
ε0 A
(uniform field)
σ : is the magnitude of the surface charge
density on each plate
V = ∫ E ⋅ ds
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Capacitance of an isolated spherical conductor:
+
+
R
Q
C = = 4πε 0 R
V
+
+
+
+
ε0
Total electric flux through any closed surface ~
total electric charge inside the surface
Q
Aε 0
∞
∞ Q
∞
Q
Q
Q
V = = ∫ Edr = ∫
dr = ∫
dr = −
R Aε
R 4πr 2ε
C R
4πε 0 r
0
0
V=
+
+
r r Q
Gauss ' s Law : ∫ E • dA =
E=
Q
∞
R
Q
Q
=
C 4πε 0 R
C = 4πε 0 R
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Example 14.1: The plates of a parallel-plate capacitor are 3.28mm apart, and
each has an area of 12.2 cm2. Each plate carries a charge of magnitude 4.35 x
10-8 C. The plates are in vacuum. a) What is the capacitance? b) What is the
potential difference between the plates? c) What is the magnitude of the electric
field between the plates?
Solution:
A
0.00122 m 2
(a) C = ε 0 = ε 0
= 3.29 pF .
d
0.00328 m
Q
A
C=
= ε0
V12
h
Q 4.35 ×10 −8 C
(b)
=
= 13.2 kV .
−12
C 3.29 ×10 F
Q
C = = 4πε 0 R
V
V 13.2 × 103V
(c ) E = =
= 4.02 ×106 V / m.
d 0.00328 m
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Capacitors in series:
V1
+Q
V2
-Q
+Q
-Q +Q
-Q
A
C1
V4
V3
C2
C3
…
+Q
-Q
B
Cn
V = V1 + V2 + V3 + ... + Vn
Q
V=
C
1 1
1
1
1
=
+
+
+ ... +
C C1 C2 C3
Cn
For conservation of charge, the
charge stored in each capacitor
must be the same.
Charge in capacitor 1 = charge in capacitor 2 = …
= charge in capacitor n
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C1
Capacitors in parallel:
+Q1
Q = CV ⇒ Q1 = C1V
Q = Q1 + Q2 + Q3 + ... + Qn
-Q1
C2
C = C1 + C2 + C3 + ... + Cn
+Q2
-Q2
B
A
…
Cn
+Qn
-Qn
Potential difference (voltage) across each capacitor is the same
(v = constant for these capacitors in parallel)
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Example 14.2: In the circuit shown in the figure, C1 = 3 µF, C2 = 5 µF, and C3 =
6 µF. The applied potential is Vab = +24 V. Calculate a) the charge on each
capacitor; b) the potential difference across each capacitor; and c) the potential
differential between points a and d.
C1
Solution:
(a)
1
1
1
1
1
a
=
+
=
+
−6
−6
Ceq C1 + C2 C3 ((3.0 + 5.0) ×10 F ) (6.0 ×10 F )
⇒ Ceq = 3.42 ×10 −6 F .
C2
d
The magnitude of the charge for capacitors in series is equal,
while the charge is distributed for capacitors in parallel. Therefore, b
Q3 = Q1 + Q2 = VCeq = (24.0 V)(3.42 x 10-6 F) = 8.21 x 10-5 C.
C3
Since C1 and C2 are at the same potential,
Q1 Q2
C
5
=
⇒ Q2 = 2 Q1 = Q1 ,
C1 C2
C1
3
5
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Q3 = Q1 + Q2 = Q1 + Q1 = Q1 = 8.21×10 −5 C
3
3
⇒ Q1 = 3.08 ×10 −5 C , and Q2 = 5.13 ×10 −5 C .
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(b) V1 = Q1 / C1 = (3.08 x 10-5 C)/(3.00 x 10-6 F) = 10.3 V.
Similarly: V2 = 10.3 V ; V3 = 13.7 V.
V1 = V2
(c) The potential difference between a and d: Vad = V1 = V2 = 10.3 V.
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Energy stored in capacitors:
Energy stored in a capacitor = total work done in transferring the charge Q
from one plate to the other
The work to charge a capacitor from zero charge to final charge Q:
W
Q
Q
0
0
0
W = ∫ dW = ∫ vdq = ∫
Where
q
1 Q2
dq =
C
2 C
∆W
∆Q →0 ∆Q
V12 = lim
q is the charge in intermediate state; and
v is the potential difference in intermediate state.
-Q
+Q
0
initial
final
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The potential energy (P) stored in a charged capacitor:
1 Q2 1
1
2
= CV = QV
P=
2 C 2
2
Units:
C=
Q
V
Q [Coulomb]
C [farads or coulomb per volt]
V [volt or joules per coulomb]
P [joules]
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