Chapter 24: Capacitance and Dielectrics
Capacitors and Capacitance
A capacitor (“cap”, for short) is a device for storing charge. A capacitor
consists of two conductors, called the plates, separated by an insulator.
The insulator is called the dielectric. Common materials used for the
diectric include air, mica, polystyrene, polyester, and ceramic materials.
Capacitors are used in camera flashes, DC power supplies, amplifiers,
filters, and in tuning applications, to name just a few.
1
Ch. 24: Capacitance and Dielectrics
The Defining Equation for the Capacitance
We “charge up” a capacitor by connecting it to a voltage source (battery,
e.g.). When this is done, the charges on the plates are the same, but
opposite in sign. We call the magnitude of charge on either plate “ Q ”.
It turns out that Q is just proportional to the magnitude of the voltage
between the plates. This happens because the magnitude of the electric
field E between the plates is proportional to Q and the voltage ΔV is
proportional to E .
The constant of proportionality between Q and ΔV is called the
capacitance, C . So:
Q = C ΔV
It is customary to call the absolute value of the potential difference
between the plates just “V ”, so from now on, I will write the above
equation:
Q = CV
(1)
This is the “defining equation” for the capacitance, C . (That is, Eq. (1)
defines what C means.)
2
Ch. 24: Capacitance and Dielectrics
Notes:
C
≡ “Farad”, F (after Michael Faraday)
V
• C just depends on geometry (area of plates, distance of separation)
and on the material used for the dielectric.
• Unit for C :
3
Ch. 24: Capacitance and Dielectrics
Calculating Capacitances
Three special cases for which it’s “easy” to write down C :
1. Parallel-plate capacitor
2. Cylindrical capacitor
3. Spherical capacitor
General Procedure:
G
• Write down E between the plates from Gauss’s law.
rf
G G
• Get ΔV between the plates from ΔV = − ∫ E ⋅ dr
ri
• Get C from C =
Q
Q
=
V ΔV
4
Ch. 24: Capacitance and Dielectrics
Case 1: Parallel-plate capacitor
In Chapter 22, found E due to a single “infinite” sheet:
σ
E=
2ε 0
G
G
Let E+ be the electric field due to the positive plate and E− be the field
due to the negative plate. Then consider three regions:
• Region I: above the positive plate
• Region II: between the plates
• Region III: below the negative plate
Everywhere in Region I:
and:
G
σ ˆ
E+ =
j
2ε 0
G
σ ˆ
E− = −
j,
2ε 0
so the net field in Region I is:
5
Ch. 24: Capacitance and Dielectrics
Similarly, in Region III:
and:
G G
so E = 0 .
G G
G G
E = E+ + E− = 0
G
σ ˆ
E+ = −
j
2ε 0
G
σ ˆ
E− =
j,
2ε 0
In Region II, however, we have:
G
σ ˆ
E+ = −
j
2ε 0
and:
G
σ ˆ
E− = −
j,
2ε 0
so:
6
Ch. 24: Capacitance and Dielectrics
G
σ
E = − ˆj
ε0
Now to get ΔV between the plates, imagine moving along a path from a
point a somewhere on the bottom plate (at yi ) to a point b directly above
a on the top plate (at y f ). Then:
yf
G G
⎛ σ
ΔV = − ∫ E ⋅ dr = − ∫ ⎜ −
ε0
ri
yi ⎝
Finally, the capacitance is:
rf
ˆj ⎞⎟ ⋅ dy ˆj = σ
ε0
⎠
( )
yf
∫ dy =
yi
σ
σ d Qd
y
y
−
=
=
(
f
i)
ε0
ε 0 Aε 0
Q
Q
=
ΔV ( Qd Aε 0 )
A
C = ε0
(2)
d
Notice that I didn’t say anything about what the dielectric was made of.
This means that I assumed that the dielectric was vacuum (literally
nothing at all). If there is a dielectric between the plates, then Eq. (2)
must be modified slightly. I’ll discuss this in more detail later.
C=
7
Ch. 24: Capacitance and Dielectrics
Note:
• The capacitance does just depend on geometrical factors, as we
said. Specifically, C is large if:
o A is large (big plates)
o d is small (plates close together)
8
Ch. 24: Capacitance and Dielectrics
Case 2: Cylindrical capacitor
From Gauss’s law, the electric field between the plates is:
G
λ
E=
rˆ
2πε 0 r
Now consider moving along a radial path from a point a on the inner
cylinder (at r = ra ) to a point b on the outer cylinder (at r = rb ). Then:
rf
rb
r
G G
⎛ λ
⎞
⎛ rb ⎞
λ b1
λ
ΔV = − ∫ E ⋅ dr = − ∫ ⎜
ln ⎜ ⎟
rˆ ⎟ ⋅ ( drrˆ ) = −
dr = −
∫
2πε 0 r ⎠
2πε 0 ra r
2πε 0 ⎝ ra ⎠
ri
ra ⎝
The V in Q = CV is the absolute value of this potential difference:
⎛ rb ⎞
λ
ln ⎜ ⎟
V=
2πε 0 ⎝ ra ⎠
And the capacitance is:
2πε 0Q
Q
C= =
V
⎛r ⎞
λ ln ⎜ b ⎟
⎝ ra ⎠
The Q and the λ are related, however:
9
Ch. 24: Capacitance and Dielectrics
λ=
So:
Q
L
⇒ Q = λL
2πε 0 λ L
⎛ rb ⎞
λ ln ⎜ ⎟
⎝ ra ⎠
2πε 0 L
C=
⎛r ⎞
ln ⎜ b ⎟
⎝ ra ⎠
C=
(3)
G
In using Gauss’s law to get E , we assumed that the capacitor was very
long (i.e., essentially infinitely long) compared with the spacing between
the plates. If L were literally infinite, the capacitance would be infinite,
from (3). No real cylindrical cap has an infinite capacitance, however, so
it’s common to speak of the capacitance per unit length instead:
2πε 0
C
(4)
=
L
⎛r ⎞
ln ⎜ b ⎟
⎝ ra ⎠
10
Ch. 24: Capacitance and Dielectrics
Case 3: Spherical Capacitor
From Gauss’s law, the electric field between the plates is:
G
Q
E=
rˆ
2
4πε 0 r
Now consider moving along a radial path from a point b on the outer
sphere (at r = rb ) to a point a on the inner sphere (at r = ra ). Then:
rf
ra
r
G G
⎛ Q
⎞
Q a⎛ 1 ⎞
Q ⎛1 1⎞
ΔV = − ∫ E ⋅ dr = − ∫ ⎜
rˆ ⋅ drrˆ ) = −
⎜ − ⎟
⎜ 2 ⎟ dr =
2 ⎟ (
∫
4πε 0 r ⎠
4πε 0 rb ⎝ r ⎠
4πε 0 ⎝ ra rb ⎠
ri
rb ⎝
Or:
Q ⎛ rb − ra ⎞
ΔV =
⎜
⎟
4πε 0 ⎝ ra rb ⎠
So the capacitance is:
⎛ ra rb ⎞
(5)
C = 4πε 0 ⎜
⎟
⎝ rb − ra ⎠
11
Ch. 24: Capacitance and Dielectrics
Capacitors in Series and Parallel
Key Terms
• A component of a circuit is a part of the circuit (capacitor, battery,
resistor, etc.)
• A terminal of a component is a conducting lead to which other
components can be connected.
• A node is the junction of two or more terminals of components.
• Two components are said to be in series if they share exactly one
node, with nothing else connected to that node.
• Two components are said to be in parallel if each terminal of one
component is connected to a unique terminal of the other
component.
Note:
• It is possible for components to be connected so that they are neither
in series nor in parallel with one another.
12
Ch. 24: Capacitance and Dielectrics
Caps in Parallel
Consider two capacitors, C1 and C2 , connected in parallel across a
battery that supplies voltage V . The two caps have the same voltage
across them, namely, the battery voltage, V . However, they store
different charges, Q1 and Q2 , because the capacitances are different.
The equivalent capacitance, Ceq , of any network of capacitors is the
single capacitor that could replace the entire network and store the same
total charge. What is Ceq for the parallel combination of C1 and C2 ?
13
Ch. 24: Capacitance and Dielectrics
To answer this question, consider the “defining equation” for the
capacitance, Q = CV , applied to C1 and C2 .
For C1 :
and for C2 :
Q1 = C1V1 = C1V
Q2 = C2V2 = C2V
Now add these two:
Q1 + Q2 = ( C1 + C2 )V
Qtot = ( C1 + C2 )V
This looks like Qtot = CeqV , with:
Ceq = C1 + C2
This means that if the parallel combination of C1 and C2 were replaced by
a single capacitor Ceq connected across the same battery voltage V , this
single cap would store the same total charge as the parallel combination
of C1 and C2 if Ceq were chosen to be simply the sum of C1 and C2 .
14
Ch. 24: Capacitance and Dielectrics
In general, for N caps in parallel across a battery supplying voltage V ,
we would have:
Q1 = C1V
Q2 = C2V
#
QN = C NV
Adding these, as before, I get:
Q1 + Q2 + " + QN = ( C1 + C2 + " + CN )V
Qtot = ( C1 + C2 + " + CN )V
This looks like
Qtot = CeqV ,
with:
Ceq = C1 + C2 + " + CN
So, caps in parallel just add.
For N equal capacitors, C , in parallel:
Ceq = NC
(6)
(7)
15
Ch. 24: Capacitance and Dielectrics
Caps in Series
Consider two caps, C1 and C2 , initially uncharged, connected in series.
Now imagine connecting C1 and C2 in series with a battery (supplying
voltage V ) and a switch, S , that is open (not conducting) until time t = 0 ,
then closed (conducting) at t = 0 .
The bottom plate of C1 , the top plate of C2 , and the wire connecting them
constitute one “hunk” of conducting stuff that is electrically neutral.
This conducting “stuff” must stay electrically neutral since there is no
way for charge to cross the insulating gap between the plates of C1 and
C2 (well, unless you do something rather extreme to the caps, that is).
16
Ch. 24: Capacitance and Dielectrics
When the switch S is closed, positive charge +Q flows onto the top plate
of C1 . This positive charge draws free electrons to the bottom plate of C1 .
This migration of free electrons to the bottom plate of C1 stops when the
charge on the bottom plate of C1 is −Q . However, this charge −Q has to
come at the expense of charge someplace else. This means that there
must be +Q on the top plate of C2 . And this draws −Q to the bottom
plate of C2 .
The upshot: caps in series store the same charge.
17
Ch. 24: Capacitance and Dielectrics
What equivalent capacitance Ceq could replace the series combination of
C1 and C2 and store charge Q ?
To answer this question, apply Q = CV to C1 and C2 :
Q
Q = C1V1 ⇒ V1 =
C1
Q
Q = C2V2 ⇒ V2 =
C2
Now add these:
⎛ 1
1 ⎞
V1 + V2 = Q ⎜ + ⎟
⎝ C1 C2 ⎠
But V1 + V2 = V , the battery voltage, so:
⎛ 1
1 ⎞
V = Q⎜ + ⎟
⎝ C1 C2 ⎠
or:
⎛ 1 ⎞
Q = ⎜ 1 1 ⎟V
⎜C +C ⎟
⎝ 1 2⎠
18
Ch. 24: Capacitance and Dielectrics
This looks like Q = CeqV , with the equivalent capacitance being:
1
Ceq = 1 1
C1 + C2
For this special case of two caps in series, we can rewrite Ceq in an
especially simple form:
CC
Ceq = 1 2
C1 + C2
(8)
19
Ch. 24: Capacitance and Dielectrics
For the general case of N caps in series:
Q ⎫
Q = C1V1 ⇒ V1 =
C1 ⎪
⎪
Q ⎪
Q = C2V2 ⇒ V2 =
⎛ 1
1
1 ⎞
⎪
C2 ⎬ ⇒ V = V1 + V2 + " + VN = Q ⎜ +
+"+
⎟
C
C
C
N ⎠
2
⎝ 1
⎪
#
⎪
Q ⎪
Q = CNVN ⇒ VN =
CN ⎪⎭
⎛
⎞
1
Q=⎜ 1 1
V
⎜ C + C + " + C1 ⎟⎟
N ⎠
⎝ 1 2
This looks like Q = CeqV for a single cap, Ceq , given by:
1
Ceq = 1 1
1
C1 + C2 + " + C N
This is usually written in the somewhat simpler form:
1
1
1
1
= +
+" +
Ceq C1 C2
CN
(9)
20
Ch. 24: Capacitance and Dielectrics
Finally, for N equal caps, C , in series, (9) gives:
1
1 1
1
⎛1⎞
= + +"+ = N ⎜ ⎟
Ceq C C C
⎝C ⎠
N terms
C
N
This says that if we put N equal caps in series, the equivalent
capacitance is one- N th of any one of them.
Ceq =
(10)
21
Ch. 24: Capacitance and Dielectrics
Energy Stored in a Capacitor
Because there’s some electric potential energy U elec associated with any
configuration of charges, caps store energy as well as charge. But how
do we write down an expression for this energy in terms of Q , C , and V ?
To do this, imagine a capacitor C that’s initially uncharged. Now imagine
charging up the cap by “grabbing” an infinitesimal bit of charge dq and
lifting it from the lower plate to the upper plate. For the first little bit of
charge moved, no work is required. (There is not yet any electric field
that we have to “fight against” to lift dq .) But for each subseqent dq , we
will have to do some work. If we imagine lifting dq quasistatically, then
the infinitesimal amount of work, dW , required is equal to the
infinitesimal amount of potential energy, dU elec , gained by dq . But
U elec = Vq , so:
dU elec = V dq
Let q be the magnitude of the charge on either plate of the cap at any
time when the voltage between the plates is V . These are related by the
defining equation for the capacitance:
22
Ch. 24: Capacitance and Dielectrics
q = CV
⇒ V=
q
C
So:
q
dq
C
To get the total energy U elec stored in the cap after a total charge Q is
moved from the lower plate to the upper plate, we need to integrate:
Q
q
U elec = ∫ dU elec = ∫ dq
C
0
Q2
(11)
U elec =
2C
Eq. (11) can be written in two alternative forms using Q = CV :
1
U elec = CV 2
(12)
2
and:
1
U elec = QV
(13)
2
dU elec =
23
Ch. 24: Capacitance and Dielectrics
Energy Density Stored in Electric Field
The energy density, u , stored in a capacitor is the energy per unit
volume:
U elec
(14)
u≡
Volume
The volume bounded by the capacitor plates is:
Volume = Ad ,
in which A is the area of the plates and d is the distance of separation.
So:
U
(∗ )
u = elec
Ad
24
Ch. 24: Capacitance and Dielectrics
It is possible to express the energy density u only in terms of the
magnitude E of the electric field between the plates. To do this, recall
that, for a parallel-plate cap, we found:
σ
Q
E= =
ε 0 Aε 0
Rearranging this for Q gives:
Q = ε 0 AE
And the capacitance of a parallel-plate cap was found to be:
A
C = ε0
d
Plugging these into (11) gives:
2
ε 0 AE )
(
1
U=
= ε 0 ( Ad ) E 2 ,
2 (ε 0 A / d ) 2
and plugging this into ( ∗ ) above gives the energy density in terms of E :
1 / 2 ) ε 0 ( Ad ) E 2
(
u=
Ad
1
u = ε0E2
2
25
Ch. 24: Capacitance and Dielectrics
Notice that this result doesn’t depend on any other properties of the cap
except the electric field between the plates. This suggests that the
energy stored by a capacitor can be thought of as energy stored in the
electric field between the capacitor plates.
In fact, it turns out (won’t show this here) that the expression for the
energy density just derived is correct no matter how the electric field
gets there. Whenever there is an electric field in any region of space,
there is some energy per unit volume stored in the field, and this energy
density is given by the formula above. As a reminder that this energy
density is stored in the electric field E , I will call this energy density “ uE ”:
1
uE = ε 0 E 2
(15)
2
26
Ch. 24: Capacitance and Dielectrics
Capacitors With Dielectrics
In deriving all of the results so far for capacitances of the parallel-plate,
cylindrical, and spherical caps, as well as the energy stored in a
capacitor, I haven’t said anything about the material used for the
dielectric. This means that I assumed the dielectric was just a vacuum –
literally nothing at all.
But what happens if there is a dielectric present?
To answer this question, imagine a parallel-plate cap – initially with no
dielectric between the plates – that has been charged up and then
disconnected from any other components (batteries, etc.). Let Q0 be the
magnitude of the charge on either plate of the cap and E0 be the
magnitude of the electric field between the plates.
Now suppose you insert a dielectric between the plates. When you do
this, atoms and molecules within the dielectric become polarized. This
polarization gives rise to an additional electric field within the dielectric –
and induced electric field, Eind , that opposes the field E0 due to the
charges on the plates, partially cancelling E0 .
27
Ch. 24: Capacitance and Dielectrics
The net electric field E within the dielectric is, therefore, some fraction of
E0 . We express this fact mathematically as follows:
E
(16)
E= 0
K
The factor K is called the dielectric constant of the material used for the
dielectric.
Note:
• K is a property of the material; it has to do with the extent to which
the material becomes polarized when placed in the field E0 . That is,
K depends on how tightly bound the electrons in the material are to
their nuclei.
• K ≥1
28
Ch. 24: Capacitance and Dielectrics
Effect of Dielectric on Capacitance
How does the presence of a dielectric between the plates affect our
earlier formulas for the capacitance C of the parallel-plate, cylindrical,
and spherical capacitors?
Consider the parallel-plate cap just discussed. We’ve seen that inserting
a dielectric reduces the field between the plates by the factor K . But E
is related to the magnitude of the voltage between the plates by:
V = Ed
So if E is reduced by the factor K , then V is reduced by the same factor:
V
V = 0,
(17)
K
in which V0 means the voltage between the plates before the dielectric
was inserted.
29
Ch. 24: Capacitance and Dielectrics
From the defining equation for the capacitance, the capacitance after the
dielectric is inserted is:
Q
C= ,
V
in which Q means the magnitude of the charge on either plate with the
dielectric inserted.
Because the cap was assumed to be disconnected from all other
components before the dielectric was inserted, there is no way – if the
dielectric does not conduct – for charge to get off of one plate and onto
the other plate. Therefore:
Q = Q0 ,
and the capacitance with the dielectric inserted is:
⎛Q ⎞
Q
Q0
C= 0 =
= K⎜ 0 ⎟
V (V0 K )
⎝ V0 ⎠
C = KC0
(18)
This says that inserting the dielectric boosts the capacitance by the
factor K .
30
Ch. 24: Capacitance and Dielectrics
For the parallel-plate cap, recall that we found the capcitance without a
dielectric to be:
A
C0 = ε 0
d
With a dielectric present, then, the capacitance is:
A
C = Kε 0
(19)
d
The product K ε 0 is called the permittivity, ε , of the material used for the
dielectric:
ε ≡ Kε 0
(20)
With this, we can write C as:
A
C =ε
(21)
d
For any capacitor, then, we can just use this same trick: to modify the
formula to reflect the presence of a dielectric, just replace ε 0 with ε .
31
Ch. 24: Capacitance and Dielectrics
Effect of Dielectric on Energy Density
Recall that the energy density stored in the electric field in a region of
space where there is no dielectric present was:
1
uE = ε 0 E 2
2
By a procedure similar to what we just did for the capacitance, we can
derive the expression for the energy density stored in the electric field in
a region of space where there is a dielectric. I won’t do this here, but the
result turns out to be that we just replace ε 0 with ε :
1 2
uE = ε E
(22)
2
32
Ch. 24: Capacitance and Dielectrics
Induced Surface Charge Density
We’ve seen that when a dielectric is inserted into a capacitor, the atoms
and molecules that make up the dielectric are polarized. This gives rise
to some induced charge per unit area on the surfaces of the dielectric
that are facing the plates. Let the induced surface charge densities on
the surfaces of the dielectric be called σ ind and −σ ind .
How are these related to σ and −σ , the surface charge densities on the
capacitor plates?
Consider a Gaussian “pillbox” that has two sides parallel to the plates –
one buried in the conducting, positive plate of the capacitor, the opposite
one buried in the dielectric. The only non-zero contribution to the flux
through this pillbox is from the side that is buried in the dielectric. The
flux through this side is EA . Therefore, Gauss’s law says:
Qencl (σ − σ ind ) A
EA =
=
ε0
E=
ε0
σ − σ ind
ε0
33
Ch. 24: Capacitance and Dielectrics
But we also know that
E=
So:
E0
σ
=
K Kε 0
σ − σ ind
σ
=
ε0
Kε 0
which gives (after a little algebra):
⎛
⎝
1⎞
⎠
σ ind = σ ⎜1 − ⎟
K
(23)
Notes:
• K ≥ 1, so σ ind is always some fraction of σ .
• For materials that are easily polarizable, K is large. For materials
with larger K , σ ind is closer to σ than for materials with smaller K .
34