Chapter 25
Capacitance-II
In 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
An electron moves from point i to point f, in the direction of a
uniform electric field. During this displacement:
i
j
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
Equivalent Capacitor
Consider the combination of capacitors
shown in the figure to the left and to the
right (upper part). We will substitute
these combinations of capacitor with a
single capacitor Ceq that is
"electrically equivalent"
to the capacitor group it substitutes.
This means that if we apply the same voltage V across the
capacitors in fig.a and fig.b (either right or left) by connecting
to a battery, the same charge q is provided by the battery. Alternatively,
if we place the same charge q on plates of the capacitors in fig.a
and fig.b (either right or left), the voltage V across them is identical.
This can be stated in the following manner: If we place the capacitor
combination and the equivalent capacitor in separate black boxes,
by doing electrical mesurements we cannot distinguish between the two.
(25 - 9)
Capacitors in parallel
The fig.a we show three capacitors
connected in parallel. This means that
the plate of each capacitor is connected
Ceq  C1  C2  C3
to the terminals of a battery of voltage V .
We will substitute the parallel 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 same potential difference V across their plates.
The charge on C1 is: q1  C1V . The charge on C2 is: q2  C2V .
The charge on C3 is: q3  C3V . The net charge q  q1  q2  q3   C1  C2  C3  V
q  C1  C2  C3  V

 C1  C2  C3
V
V
For a parallel combination of n capacitors is given by the expression:
The equivalent capacitance Ceq 
n
Ceq  C1  C2  ...  Cn   C j
j 1
(25 - 10)
Capacitors in series
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 voltage V . 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 same charge q on their plates.
The voltage across C1 is: V1  q / C1.
The voltage across C2 is: V2  q / C2 .
The voltage across C3 is: V3  q / C3 .
The net voltage across the combination V  V1  V2  V3
(25 - 11)
 1
1
1 
Thus we have: V  q  
 
 C1 C2 C3 
q
The equivalent capacitance Ceq  
V
1
1
1
1
 

Ceq C1 C2 C3
q

 1
1
1 
q 
 
C
C
C3 
2
 1
More complex capacitor systems
In general a capacitor system
may consist of smaller capacitor
groups that can be identified as
connected "in parallel" or
"in series"
In the example of the figure C1 and C2 in fig.a are connected in parallel.
They can be substituted by the equivalent capacitor C12  C1  C2 as
shown in fig.b. Capacitors C12 and C3 in fig.b are connected in series.
They can be substituted by a single capacitor C123 as shown in fig.c
C123 is given by the equation:
(25 - 12)
1
1
1


C123
C12 C3
C  o A/ d
A question
What is the equivalent
capacitance between the
points A and B?
A. 1 μF
B. 2 μF
C. 4 μF
D. 10μF
E. None of these
A
What would a 10V battery do, i.e. how much charge will it provide,
when it is connected across A and B? 40 μC
B
-q'
-
q'
Energy stored an an electric field
+
+
+
+
+
dq'
Consider a capacitor C which is has a charge q.
We can calculate the work W required to charge
the capacitor by assuming that we transfer a charge
dq from the negative plate to the positive plate.
We assume that the capacitor charge is q and the
corresponding voltage V . The work dW required
V'
q
dq
C
We continue this process till the capacitor charge is
Charge
for the charge transfer is given by: dW  V dq 
A
q
q'
q
B Voltage
O
1
equal to q. The total work W   V dq   qdq
C0
V' V
q
1  q2 
q2
W   
C  2  0 2C
CV 2
If we substitute q  CV we get: W 
2
or W 
qV
2
Work W can also be calculated by determining the area A of triangle OAB which is
equal to  V dq. Area  W 
Vq
2
(25 - 13)
E
-q
q
-A
-
+
+
A +
+
+
d
Potential energy stored in a capacitor
The work W spent to charge a capacitor is stored in the form of
potential energy U  W that can be retrieved when is capacitor is
q 2 CV 2 qV
discharged. Thus U 


2C
2
2
Energy density
We can ask the question: where is the potential energy of a charged
q 2 CV 2
U

2C
2
u
oE
2
capacitor stored? The answer is counter intuitive. The energy is
stored in the space between the capacitor plates where a uniform
electric field E  V / d is generated by the capacitor charges.
In other words the electric field can store energy in empty space!
2
We define as energy densiry (symbol u ) the potential energy per unit volume. u 
The volume V between the plates is: V  Ad where A is the plate area
o  V  oE2
U
CV 2
V2
Thus the energy density u 


   
Ad 2 Ad 2 Ad 2  d 
2
This result, derived for the parallel plate capacitor holds in general
2
(25 - 14)
U
V
C   Cair
Capacitor with a dielectric
q
In 1837 Michael Faraday investigated what happens to the
V
-q
capacitance C of a capacitor when the gap between the plates
is completely filled with an insulator (a.k.a. dielectric)
Faraday discovered that the new capacitance is given by :
q'
V
-q'
C   Cair Here Cair is the capacitance before the insertion
of the dielectric between the plates. The factor  is known
as the dielectric constant of the material.
Faraday's experiment can be carried out in two ways:
q
V
In this case a battery remains connected to the plates .
-q
q
-q
1. With the voltage V across the plates remaining constant
This is shown in fig.a
V'
2. With the charge q of the plates remaining constant.
In this case the plates are isolated from the battery
This is shown in fig.b
(25 - 15)
C   Cair
q
V
Fig.a : Capacitor voltage V remains constant
-q
This is bacause the battery remains connected to the plates
q'
After the dielectric is inserted between the capacitor plates
the plate charge changes from q to q   q
V
-q'
q
V
q κq
q
The new capacitance C  
 κ  κCair
V
V
V
Fig.b : Capacitor charge q remains constant
This is bacause the plates are isolated
-q
After the dielectric is inserted between the capacitor plates
q
-q
V'
the plate voltage changes from V to V  
The new capacitance C 
V

q
q
q

    Cair
V V /
V
(25 - 16)
In a region completely filled with an insulator of
dielectric constant  , all electrostatic equations
q
containing the constant  o are to be modified by
replacing  o with  o
Example 1 : Electric field of a point charge inside
1
q
a dielectric: E 
4 o r 2
conductor
dielectric
Example 2 :
The electric field outside an isolated conductor
immersed in a dielectric becomes:
E

 o
(25 - 17)
Dielectrics : An atomic view
Dielectrics are classified into "polar" and "nonpolar"
Polar dielectrics consist of molecules that have a non-zero
electric dipole moment even at zero electric field
due to the asymmetric distribution of charge
within the molecule (e.g. H 2O ). At zero electric field
(see fig.a) the electric dipole moments are randomly
oriented. When an external electric field Eo is applied
(see fig.b) the electric dipole moments tend to align
preferrentially along the direction of Eo because this
configuration corresponds to a minimum of the potential
energy and thus is a position of stable equilibrium.
Thermal random motion opposes the alignment and
U   pE cos 
thus ordering is incomplete. Even so, the partial
alignment produced by the exterlal electric field
generates an internal electric field that apposes
(25 - 18)
Eo . Thus the net electric field E is weaker than Eo
A nonpolal dielectric on the other hand consists of molecules that in the absence of
an 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 Eo induces an electric dipole
moment p that becomes aligned with Eo (see fig.b). The aligned molecules do not create
any net charge inside the dielectric. A 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 induced surface charges have sign opposite
to that of of the opposing plate charges. Thus the induced charges create an electric field
E  which opposes the applied field Eo (see fig.c). As a result is that the
net electric field E between the capacitor plates is weaker.
E
Eo

(25 - 19)
Gauss' law and dielectrics
In chapter 23 we formulated Gauss' law assuming that the
S
charges existed in vacuum.  o  E  dA  q or  o   q
n̂
In this section we will write Gauss' law in a from which
is suitable 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   Eo A 
S
n̂
 Eo 
q
o A
q
o
Now we fill the space between 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 q
q  q
we also have the induced charge q on the surface of the dielectric.   EA 

o
E
q  q
q
E
(eqs.1) From Faraday's experiments we have: E  o 
(eqs.2)
A o
  A o
If we compare eqs.1 with eqs.2 we have: q  q 
q

  o   E  dA  q
(25 - 20)
Gauss' law in the presence of dielectrics
 o   E  dA  q
Even though the equation above was derived for the parallel plate capacitor
is true in general.
Note 1 : The flux integral now involves  E
Note 2 : The charge q that is used is the plate charge, also known as "free charge"
Using the equation above we can ignore the induced charge q
Note 3 : The dielectric constant  is kept inside the integral to desrcibe the
most general case in which  is not constant 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
none of these
C. 20
D. 50
E.
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. 0
B. 5N
C. 0. 9N
D. 1 x104 N
E. 9 x 105 N
F  qE 

q2
2 o A
5  10 
6 2
2  8.85  10
12
 0.3
 4.71N
The electric field = σ/2εo why?