Download Capacitance

Capacitance
PHY 2049 Chapter 25
Chapter 25
Capacitance
In this chapter we will cover the following topics:
-Capacitance C of a system of two isolated conductors.
-Calculation of the capacitance for some simple geometries.
-Methods of connecting capacitors (in series , in parallel).
-Equivalent capacitance.
-Energy stored in a capacitor.
-Behavior of an insulator (a.k.a. dielectric) when placed in the
electric field created in the space between the plates of a
capacitor.
-Gauss’ law in the presence of dielectrics.
(25 - 1)
Capacitors
Capacitor
 Composed of two metal plates.
 Each plate is charged
 one positive
 one negative
 Stores energy
SYMBOL
A simple Capacitor
TWO PLATES
WIRES
Battery
INSIDE THE DEVICE
What is STORED in the capacitor?





An Electric Field
Energy
Charge
All three
None of these
Two Charged Plates
(Neglect Fringing Fields)
d
Air or Vacuum
E
-Q
Area A
V=Potential Difference
+Q
Symbol
ADDED CHARGE
Where is the charge?
- Q-
d
Air or Vacuum
E
Area A
V=Potential Difference
+
+
+
+
+
+ +Q
AREA=A
s=Q/A
One Way to Charge:
 Start with two isolated uncharged plates.
 Take electrons and move them from the +
to the – plate through the region between.
 As the charge builds up, an electric field
forms between the plates.
 You therefore have to do work against the
field as you continue to move charge from
one plate to another.
Capacitor
More on Capacitors
Gauss
d
q
 E  dA  
Air or Vacuum
-Q
E
+Q
Area A
V=Potential Difference
Gaussian
Surface
0
 EA  0 A   EA  
Q
0
Q   0 EA
Q
(Q / A) s
E


0 A
0
0
Same result from other plate!
DEFINITION - Capacity
 The Potential Difference is
APPLIED by a battery or a
circuit.
 The charge q on the
capacitor is found to be
proportional to the applied
voltage.
 The proportionality
constant is C and is
referred to as the
CAPACITANCE of the
device.
q
C
V
or
q  CV
UNITS
 A capacitor which
acquires a charge
of 1 coulomb on
each plate with the
application of one
volt is defined to
have a capacitance
of 1 FARAD
 One Farad is one
Coulomb/Volt
q
C
V
or
q  CV
Continuing…
q
C
V
q  sA   0 EA 
so
C
0 A
d
 0 AV
d
 The capacitance of a
parallel plate
capacitor depends
only on the Area and
separation between
the plates.
 C is dependent only
on the geometry of
the device!
After the switch is closed, how
much charge passed through the
capacitor?
V




C/V
V/C
CV
C+V
P
Capacitance of a parallel plate capacitor
The plates have area A and are separated
by a distance d . The upper plate has a
S
n̂
charge  q and the lower plate a charge - q
N
We apply Gauss' law using the Gaussian surface S shown in the figure.
The electric flux   EA cos 0  EA.
From Gauss' law we have:  
q
o
 EA 
q
o
E
q
A o
The potential difference V between the positive and the negative plate is




given by: V   Eds cos 0 E  ds Ed 
The capacitance C 
qd
A o
A
q
q

 o
V qd / A o
d
A o
C
d
(25 - 6)
Units of 0
Coulomb 2 Coulomb 2
 0  

2
m  Joule
Nm
Coulomb 2

m  Coulomb  Volt
Coulomb Farad


m
m  Volt
and
 0  8.85 10 12 F / m  8.85 pF / m
pico
Simple Capacitor Circuits
 Batteries
 Apply potential differences
 Capacitors
 Wires
 Wires are METALS.
 Continuous strands of wire are all at the same
potential.
 Separate strands of wire connected to circuit
elements may be at DIFFERENT potentials.
NOTE
 Work to move a charge from one side
of a capacitor to the other is = qEd.
 Work to move a charge from one side
of a capacitor to the other is qV
 Thus qV = qEd
 E=V/d As before
TWO Types of Connections
SERIES
PARALLEL
Parallel Connection
q1  C1V1  C1V
q2  C2V
q3  C3V
QE  q1  q2  q3
V
CEquivalent=CE
QE  V (C1  C2  C3 )
therefore
C E  C1  C2  C3
Series Connection
q
V
-q
C1
q
-q
C2
The charge on each
capacitor is the same !
Series Connection Continued
V  V1  V2
q
V
C1
-q
q
-q
C2
q
q
q


C C1 C 2
or
1
1
1


C C1 C 2
More General
Series
1
1

C
i Ci
Parallel
C   Ci
i
Example
C1
C2
(12+5.3)pf
V
C3
C1=12.0 uf
C2= 5.3 uf
C3= 4.5 ud
More on the Big C
E=0A/d
+dq
+q
-q
 We move a charge
dq from the (-)
plate to the (+)
one.
 The (-) plate
becomes more (-)
 The (+) plate
becomes more (+).
 dW=Fd=dq x E x d
dW  dq  Ed
Gauss
s
q 1
E

0 A 0
So….
q 1
dW 
d  dq
A 0
Q
W U  
0
d
q2 d
q2 1
qdq 

A 0
2 A 0
2 ( A 0 )
d
or
Q 2 C 2V 2 1
U

 CV 2
2C
2C
2
Not All Capacitors are Created
Equal
 Parallel
Plate
 Cylindrical
 Spherical
Spherical Capacitor
Gauss
q
 E  dA  
4r E 
2
0
q
0
q
E (r ) 
2
4r  0
Calculate Potential Difference V
positive. plate
Eds

V
negative. plate
q 1
V  
 2 dr
40  r 
b
a
(-) sign because E and ds are in OPPOSITE directions.
Continuing…
q
b
dr
q
1
V

( )
2

40 a r
40 r
q  1 1
q ba
V
  


40  a b  40  ab 
q
ab
C   40
V
ba
Lost (-) sign due to switch of limits.