Chapter 25 Capacitance-IIIn the last lecture:
we calculated the capacitance C of a system of two isolated conductors.
We also calculated the capacitance for some simple geometries.
In this chapter we will cover the following topics:
-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)
HITT
A. the work done by the field is positive and the potential energy of the electron-field system increases
B. the work done by the field is negative and the potential energy of the electron-field system increases
C. the work done by the field is positive and the potential energy of the electron-field system decreases
D. the work done by the field is negative and the potential energy of the electron-field system decreases
E. the work done by the field is positive and the potential energy of the electron-field system does not change
An electron moves from point i to point f, in the direction of a uniform electric field. During this displacement:
i j
This means that if we apply the same voltage across the capacitors in fig.a and fig.b (either right or left) by connectingto a battery, the same charge is provided by the battery. Alternatively,i
V
qf we place the same charge on plates of the capacitors in fig.a
and fig.b (either right or left), the voltage across them is identical. This can be stated in the following manner: If we place the
qV
capacitorcombination and the equivalent capacitor in separate black boxes, by doing electrical mesurements we cannot distinguish between the two.
Consider the combination of capacitorsshown in the figure to the left and to the right (upper part). We will substitutethese combinations of capacitor with a single capacitor eqC
Equivalent Capacitor
that is
"electrically equivalent" to the capacitor group it substitutes.
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The fig.a we show three capacitors connected in parallel. This means that the plate of each capacitor is connectedto the terminals of a battery of voltage . We will substitute
V
Capacitors in parallel
the parallel combination of fig.a with a single equivalent capacitor shown in fig.b which is also connectedto an identical battery
1 1 1 2 2 2
3 3 3 1 2 3 1
The three capacitors have the across their plates.The charge on is: . The charge on is: .The charge on is: . The net charge
C q C V C q C VC q C V q q q q C
same potential difference V
2 3
1 2 31 2 3
1 21
The equivalent capacitance
For a parallel combination of n capacitors is given by the expression:
...n
e n
eq
q jj
C C V
C C
C C C C C
C VqC C C CV V
1 2 3eqC C C C
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The fig.a we show three capacitors connected in series. This means that one capacitor is connected after the other.The combination is connected to the terminals of a battery of vol
Capacitors in series
tage . We will substitute the series combination of fig.a with a single equivalent capacitor shown in fig.b which is also connected to an identical battery.The three capacitors have the
V
same charg
1 1 1
2 2 2
3 3 3
1 2 3
on their plates.The voltage across is: / . The voltage across is: / . The voltage across is: / .The net voltage across the combina
tion
Thus we have:
C V q CC V q CC V q C
V V V V
V q
e q
1 2 3
1 2 3
1 1 1
The equivalent capacitance 1 1 1eq
C C Cq qCV
qC C C
1 2 3
1 1 1 1
eqC C C C (25 - 11)
In general a capacitor systemmay consist of smaller capacitorgroups that can be identified asconnected "in parallel" or "in series"
More complex capacitor systems
12 1 2
1 2
12 3
In the example of the figure and in fig.a are connected . They can be substituted by the equivalent capacitor as shown in fig.b. Cap
in parallel
acitors and in fig.b are cC C C
C C
C C
123
123123 12 3
onnected .They can be substituted by a single capacitor as shown in fig.c
is given by the equation:
in series
1 1 1
C
CC C C
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dAC o /
A questionWhat is the equivalent capacitance between the points A and B?
A.1 μFB. 2 μFC.4 μFD.10μFE.None of these
A B
What would a 10V battery do, i.e. how much charge will it provide, when it is connected across A and B? 40 μC
---
--
+++++
dq'
q'-q'
V'
q'q
V' V
Charge
VoltageO
A
B
Consider a capacitor which is has a charge .We can calculate the work required to chargethe capacitor by assuming that we transfer a charge
from the negative
C qW
dq
Energy stored an an electric field
plate to the positive plate.We assume that the capacitor charge is and thecorresponding voltage . The work required
for the charge transfer is given by:
We continue this proc
qV dW
qdW V dq dqC
0
ess till the capacitor charge is
1equal to . The total work
q
q W V dq q dqC
2 2
0
21 If we substitute we get: or 2
Work can also be calculated by determining the area of triangle OAB which is
equal to . Area
2 2
2
2
qq CVqW q CV
C
W AVqV d
q
q W
VW WC
(25 - 13)
2 2
The work spent to charge a capacitor is stored in the form of potential energy that can be retrieved when is capacitor is
discharged. Thus 2 2
WU W
q CV qVUC
Potential energy stored in a capacitor
2
We can ask the question: where is the potential energy of a chargedcapacitor stored? The answer is counter intuitive. The energy is stored in the space between the capacitor plate
Energy density
s where a uniform electric field / is generated by the capacitor charges.In other words the electric field can store energy in empty space!
E V d
2 2
2 2q CVUC
---
--
+++++
q-q
d
E
A A
2 2
We define as energy densiry (symbol ) the potential energy per unit volume.
The volume between the plates is: where is the plate area
Thus the energy density 2 2 2
o
Uu uV
V V Ad A
U CV V VuAd Ad Ad d
22
This result, derived for the parallel plate capacitor holds in general
2oE
2
2oEu
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q
-q
q'
q
q
-q'
-q
-q
V
V
V
V'
In 1837 Michael Faraday investigated what happens to thecapacitance of a capacitor when the gap between the plates is completely filled with an insulator (a.k.a. dielectri
C
Capacitor with a dielectric
c)Faraday discovered that the new capacitance is given by :
Here is the capacitance before the insertion of the dielectric between the plates. The factor is knownas the dielectric co
airairC CC
nstant of the material. Faraday's experiment can be carried out in two ways:
With the voltage across the plates remaining constantIn this case a battery remains connected to the plates . This is
V1.
shown in fig.a With the charge of the plates remaining constant.
In this case the plates are isolated from the batteryThis is shown in fig.b
q2.
airC C
(25 - 15)
q
-q
q'
q
q
-q'
-q
-q
V
V
V
V'
airC C
This is bacause the battery remains connected to the plates After the dielectric is inserted between the capacitor plates the plate charge changes from tq
Fig.a : Capacitor voltage V remains constant
o
The new capacitance airq κq q C κ κCV V V
q q
This is bacause the plates are isolated After the dielectric is inserted between the capacitor plates
the plate voltage changes from to
The ne
w capa
V VV
Fig.b : Capacitor charge q remains constant
citance / air
q q qC CV V V
(25 - 16)
conductor dielectric
qo
o o
In a region completely filled with an insulator ofdielectric constant , all electrostatic equations containing the constant are to be modified by replacing with
Electric field of
Example 1 :
2
a point charge inside
a dielectric:
The electric field outside an isolated conductor immersed in a dielectric becomes
14
:
o
o
qEr
E
Example 2 :
(25 - 17)
Dielectrics are classified into "polar" and "nonpolar"Polar dielectrics consist of molecules that have a non-zero electric dipole moment even at zero electric fielddue to
Dielectrics : An atomic view
2
the asymmetric distribution of charge within the molecule (e.g. H O ). At zero electric field(see fig.a) the electric dipole moments are randomly
oriented. When an external electric field is appoE
lied(see fig.b) the electric dipole moments tend to align
preferrentially along the direction of because thisconfiguration corresponds to a minimum of the potential energy and thus is a position
oE
of stable equilibrium. Thermal random motion opposes the alignment andthus ordering is incomplete. Even so, the partialalignment produced by the exterlal electric fieldgenerates an internal electric field that apposes
. Thus the net electric field is than
o oE E Eweaker
cosU pE
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A nonpolal dielectric on the other hand consists of molecules that in the absence ofan electric field have zero electric dipole moment (see fig.a) . If we place the dielectric
between the plates of a capacitor the external electric field induces an electric dipole
moment that becomes aligned with (see fig.b). The aligned molecules do not create any net charge inside the dielectric. A
o
o
E
p E
net charge appears at the left and right surfaces of the dielectric opposite to the capacitor plates. These charges come from negative and positive ends of the electric dipoles. These surfaceinduced charges have sign to that of of the opposing plate charges. Thus the induced charges create an electric field
which opposes the applied field (see fig.c). As a result is that theoE E
opposite
net electric field between the capacitor plates is weaker.
E
oEE
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S
S
n̂
n̂
In chapter 23 we formulated Gauss' law assuming that the
charges existed in vacuum. or
In this section we will write Gauss' law in a from which is suitab
o oE dA q q
Gauss' law and dielectrics
le for cases in which dielectrics are present. Consider first the parallel plate capacitor shown in fig.a.
We will the Gaussian surface S. The flux
Now we fill the space betwe
oo
oo
qE A
qEA
en the plates
with an insulator of dielectric constant (see fig.b) . We will apply Gauss' law for the same surface S. Inside S in addition to the plate charge
we also have the induced charge on the surface of the dielectric.
(eqs.1) Fro
o
o
qq qq EA
q qEA
m Faraday's experiments we have: (eqs.2)
If we compare eqs.1 with eqs.2 we have:
o
o
o
E qEA
qq A qq E d
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Gauss' law in the presence of dielectrics
o E dA q
Even though the equation above was derived for the parallel plate capacitoris true in general.
The flux integral now involves The charge that is used is the plate charge, also
Eq
Note 1 : Note 2 :
known as "free charge"Using the equation above we can ignore the induced charge
The dielectric constant is kept inside the integral to desrcibe themost general case in which is not con
q
Note 3 :stant over the Gaussian surface
(25 - 21)
HITT
A parallel-plate capacitor has a plate area of 0.2m2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 106 V/m between the plates, the magnitude of the charge on each plate should be:A. 8.9 x 10-7 C B. 1.8 x 10-6 C C. 3.5 x 10-6 C D. 7.1 x 10-6 C E. 1.4 x 10-5 C
A question
• Each of the four capacitors shown is 500 μF. The voltmeter reads 1000V. The magnitude of the charge, in coulombs, on each capacitor plate is:
A. 0.2 B. 0.5 C. 20 D. 50 E. none of these
Question
• A parallel-plate capacitor has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about:
• A. 0 B. 5N C. 9N D. 1 x104 N E. 9 x 105 N
Question• A parallel-plate capacitor
has a plate area of 0.3m2 and a plate separation of 0.1mm. If the charge on each plate has a magnitude of 5x10-6 C then the force exerted by one plate on the other has a magnitude of about:
A. 0B. 5NC. 0. 9ND. 1 x104 NE. 9 x 105 N
N
AqqEF
o
71.43.01085.82
105
2
12
26
2
The electric field = σ/2εo why?