Forces and Energy functions for dielectric systems
Prof. R.S. Sorbello, UWM
March, 2008
Consider a dielectric slab partly inside and partly outside a parallel plate capacitor
as shown, for example, in Fig. 4.30 on page 194 of Griffiths. The capacitor plates have
length l , width w, and are separated by a distance d. The slab has thickness d and the
coordinate of the left-hand edge of the slab is x. (If x= 0, the slab would be occupying the
entire space between the plates.)
The energy stored in the electric field is
W   12 (r) E (r)2 d ,
(1)
where (r) is the permittivity at position r and E(r) is the electric field at position r.
The volume integral is over all space, and this reduces to the space between the plates if
we ignore the energy associated with the fringing fields near the edge of the capacitor and
beyond. This means that the volume integral in Eq.(1) is taken only over the space
between the plates, and in this region, (r)   in the sub-region that is occupied by the
dielectric and (r)  0 in the sub-region between the plates where there is no dielectric.
Therefore,
W 

vacu um
1
2 0
E 2 d 

d ielectric
1
2
E 2 d
 12  0 E 2 xwd  12  E 2 (l  x)wd
(2)
where we have used the fact that E is uniform throughout the region between the plates,
and hence is the same in both the vacuum and dielectric regions inside the capacitor.
Let F be the mechanical force that you need to supply in the x-direction to hold
the dielectric slab in place. Since in an infinitesimal displacement dx, you do work Fdx
and this goes into increasing the energy stored in the electric field, the force that you
exert is given by
F
dW
.
dx
(3)
It will turn out that F is positive, meaning that the dielectric is sucked into the region
between the plates by the action of electric field (or Coulomb forces) in the system.
(Recall that you have to exert a mechanical force F that is equal and opposite to the force
due to the electric field. Also note: Griffiths uses –F to denote the force that you exert
on the dielectric rather than +F as we use here.)
Now comes the subtle part: When we take the derivative on the RHS of Eq. (3),
what is to be treated as a constant? Can we treat E as a constant, which would seem to
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imply that a battery is somehow involved in keeping the voltage V fixed? Or should we
treat the free charge +Q and –Q on the plates as constant, which implies that the system
is electrically isolated, meaning that no additional free charge can flow onto either plate?
The answer is that we must treat Q as fixed and not E as fixed. This is true even if a
battery is connected to the capacitor, in which case V and E are both kept fixed as you
move the dielectric by an amount dx! How can this be?
The key point is that it is only for an isolated system (fixed Q) that you can say
that when you move the dielectric by dx , the change in the stored electrical energy dW
comes entirely from the mechanical work Fdx that you do. Thus Eq. (3) is correct only
for such a fixed-Q process. To be more precise, we must therefore write the derivative
on the RHS of Eq.(3) as a derivative in which the condition to be obeyed is that Q is held
constant. In the language of partial derivatives, this is
 W 
F
.
 x  Q
(4)
On the other hand, if you say that there is an attached battery that keeps the
voltage (and therefore E-field) constant as you move the dielectric by dx , then the change
in the stored electrical energy dW comes partly from the mechanical work Fdx that you
do and partly from additional electrical work dWb attery done by the battery. But the latter
is given by dWba ttery  VdQ , where dQ is the extra charge that goes from the battery into
the capacitor to maintain the capacitor at a fixed voltage during the displacement dx.
Therefore,
dW  Fdx  dWb attery
 Fdx  VdQ
.
(5)
Dividing by dx and re-arranging terms, we get from the last equation the result
dW
dQ
F
V
, where in this case we are holding V fixed in taking these derivatives.
dx
dx
In the language of partial derivatives, this result is
 W 
 Q 
.
F
V

 x  V
 x  V
(6)
We emphasize that Eq.(4) and Eq.(6) must both give exactly the same result for
F! This follows from the fact that the electric force on the dielectric when it is at any
given position x, depends only on the charge distribution in the system (both free and
bound charge densities). So when the dielectric is at any position x and the free charge
on the plates is specified, then there is a unique force on the dielectric whether or not the
capacitor is attached to a battery. Of course, a battery would have been needed to
originally charge-up the capacitor so that it has free charges Q on the plates, but once
this is done, the battery can be disconnected. The force on the dielectric after the battery
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is disconnected will be the same as it was with the battery connected provided that the
position x is held fixed.
Now to actually calculate F, we can use Eq.(4), with W given by Eq.(2). But in
using Eq.(2), we must be careful to write E in terms of Q so that we can perform the
indicated derivative at constant Q. Alternatively, we can use Eq.(6), in which case the
expression for W in Eq. (2) is easy to use because V being held constant as the condition
on the derivatives in Eq.(6), means that E is to be held constant when taking the
derivatives. However, there is the extra term V (Q / x)V that needs to be evaluated on
the RHS of Eq.(6), so we would have to express Q as a function of V and x so that we
could evaluate that partial derivative. This means that we have to calculate the
capacitance of the structure since Q =C(x)V. But if we are going to evaluate the
capacitance, there is a better way to proceed, as we now describe.
To determine F in terms of the capacitance of the system, we note that W is the
electrical energy stored in a capacitor, which can be expressed in terms of the capacitance
C of the structure in either of the two equivalent ways:
W 
or
Q2
2C
CV 2
W 
2
(7a)
.
(7b)
As discussed in Griffiths, W contains contributions from the distortion of "molecular
springs" in addition to the pure Coulomb potential energy of the free and induced charges
in the system. (This fact is irrelevant for the subsequent analysis.)
If we use Eq.(7a) into Eq.(4), we get
   Q2  
F 

 x  2C   Q
.
(8)
Since C is a function only of x, this gives
Q2 d 1
Q2 dC
.
F
( ) =– 2
2 dx C
2C dx
(9)
Upon using Q=CV in the last equality, we finally get
V 2  dC
F    
.
 2  dx
(10)
We remark that C(x) can be readily determined by equating the expressions for W given
by Eqs.(2) and (7b) and making the substitution E=V/d in Eq.(2).
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Now let's see if we get the result given by Eq.(10) by applying Eq.(6) rather than
Eq.(4). Substituting Eq.(7b) into Eq.(6), we find
 
F
 x
 CV 
2
1
2
 (CV ) 
,
V
 x  V
V
(11)
which gives
V 2  dC
F    
.
 2  dx
(12)
Comparing Eqs (10) and (12), we see that we do get the same result for F
whether we use Eq.(4) or Eq. (6). In the case of Eq. (4) we imagined the capacitor plates
to be isolated; in the case of Eq. (6) we imagined the capacitor plates to be hooked up to
a battery. Of course, in any specific process between initial and final values of V and Q,
the values of V and Q at any step in the process will depend on the details of the path in
the "QV-space" that the process follows. So for example, V will be constant in a process
where a battery is attached, and Q will be a constant in a process the plates are isolated
(no battery attached). This means that Wmech will depend on the specific path of the
process between initial values of Q,V and final values of Q,V. (In math language, this
means that dWmech is an "imperfect differential," and the notation dWmech is often used to
signify that property.)
There is an interesting thing to notice here. If you were to use Eq. (3) and
substituted expression (7b) for W in Eq. (3), you would get
F
dW
d CV 2 


.
dx
dx  2 
(13)
So far this is correct, but it must be understood that in the derivative, one must keep Q
constant, as pointed out in the discussion leading to Eq.(4), which is the exact, explicit
form of Eq.(3). Now if you made the mistake of taking the derivative on the RHS of
Eq.(13) by keeping V fixed instead of keeping Q fixed, you would get
 V 2  dC
F   
 2  dx
(WRONG!!!)
(14)
This last expression is wrong because it was obtained by incorrectly keeping V constant
rather than Q constant in taking the derivative. However, comparing the wrong result
(14) with the correct result (10) or (12), we notice that the wrong method turns out to give
the correct magnitude for the force but the wrong sign! Since C will decrease as x
increases, the quantity dC/dx is negative. So the wrong result would predict that the
force you would apply to keep the dielectric in place is negative, implying that if you
were to let go of the dielectric, the dielectric would be ejected from the capacitor, as if
fired from a gun! The correct result is that the dielectric is sucked into the region
between the plates rather than repelled from it.
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A caveat: It is tempting to say that one can always get the correct magnitude of F
by making the simple, though erroneous, analysis leading from Eq.(13) to Eq.(14), so
why bother about all the details in the correct analysis just to get the correct sign? After
all, we can always fudge our final result by tacking on the right sign at the end. It is just
an accident that an incorrect analysis turns out to give the correct magnitude of the force,
so don't expect to get credit for such an analysis. Furthermore, besides the fact that
Eq.(14) is wrong because the method leading up to it is wrong, there are cases where
Eq.(14) also gives the wrong magnitude for F. This happens if we are dealing with a
certain class of dielectric media, known as non-linear dielectrics. Although we normally
don't encounter such exotic media in an undergraduate course, such media do exist and
do have technological applications. A non-linear dielectric medium is characterized by a
permittivity that depends upon the strength of the electric field. This leads to a
capacitance that depends on voltage, which, of course, is never the case for simple
dielectric media. For any medium, exotic or not, Eqs (4) and (6) remain valid, since their
derivation was based on the work-energy principle and requires no explicit assumption
about the dielectric medium. On the other hand, assuming that C =C(x) independent of V
or Q is true only for linear dielectrics (and vacuum). In fact, the final expressions (10) or
(12) for F would have to be modified for nonlinear media, and it turns out that the wrong
result given in Eq. (14) is no longer just in error by a minus sign. (Incidentally, at
sufficiently high electric fields, all dielectric media become non-linear, so it is not really
accurate to refer to non-linear media as "exotic".)
Optional Material:
Connection with thermodynamic functions and Legendre Transformations:
Eq.(6) can be written in the following form:
 W  
,
F
 x  V
(15)
W   W  QV .
(16)
where
The function W  is known as the Legendre transform of W.
Eq. (16), we get
dW   dW  QdV  VdQ .
Taking the differential of
(17)
Using dW from Eq. (5), we find that Eq.(17) gives
dW   Fdx  QdV
(18)
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Note that Eq.(15) follows directly Eq.(18) by using the definition of partial
derivatives. The set of equations Eq.(5), (16), (18), and the associated expressions Eq.(4)
and (15) have the same structure as the equations for thermodynamic potentials. Recall,
for example, that the internal energy function U=U(S,V) satisfies the law of
thermodynamics given by
dU  TdS  PdV .
(19)
The Legendre transformation to a new function A, called the "Helmholtz free
energy," is defined by
(20)
A  U  TS ,
from which it follows that
dA  SdT  PdV .
(21)
 U 
P  
 V  S
(22)
 A 
.
P  
 V  T
(23)
From Eq (19) we get
and from Eq. (21) we get
Eqs (19)-(23) are analogous to Eq.(5), (16), (18), and the associated expressions Eq.(4)
and (15). (Note the minus signs on the RHS of Eqs (19), (22), and (23) occur because
the definition of pressure P in thermodynamics is that it is the force per unit area exerted
by the gas on the walls of the system, as opposed to the force per unit area exerted by the
walls on the gas, which would differ by a minus sign since the forces are equal and
opposite.)
Interpretation of Legendre Transformation as including the external source of energy:
The Legendre transformation given by Eq. (16) can be interpreted as follows: For
all intents and purposes, W  acts as the total energy for a system that includes the battery
as part of the system. This contrasts with the original view that the system consists of the
dielectric and charged capacitor plates, and that the system can exchange energy with the
external world by means of a battery and the external mechanical force F. The new
interpretation follows by noting that the work done by the battery equals VdQ, and that
for a battery of fixed voltage V, this equals d(QV). It then follows from Eq.(16) that
dW   dW  d(QV )
 dW  dWbattery  d(W  Wbattery )
(24)
Eq. (24) states that for purposes of keeping track of energy changes, W  acts like the
energy of the original system minus the work done by the battery. This is what is to be
expected if we imagine the battery to be a large reservoir of charge at fixed voltage V and
we incorporate that large reservoir of charge as part of the system. Then the change in
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the total energy of this new system is the sum of the change of W from the dielectric part
and a second contribution equal to the negative of the energy change of the battery or
charge reservoir. (The change in energy of the charge reservoir subsystem is dWbattery
when the battery does work dWbattery on the dielectric subsystem.) The same idea holds
for the thermodynamic analogy Eq.(23), where we can interpret A as being the energy for
a larger system that includes the thermal reservoirs at temperature T.
A subtle point is that the second line of Eq.(24) is valid only for processes at
constant V, yet we use W  for any kind of process or path in the QV-plane. To make
sense of the interpretation of incorporating the battery into the new system for processes
in which V undergoes a change, one needs to imagine a very large (actually, infinite) set
of batteries, each differing from the previous one by an infinitesimal amount dV. Then
we connect and disconnect these batteries to our system at each step in the path where our
process requires a change in V. (Note that in each infinitesimal step in the process where
the battery voltage goes from V to V+dV and the charge goes from Q to dQ, the work
done by the battery in each step is VdQ. Therefore, nothing gets lost in the dV jump that
accompanies the switch to a new battery or charge reservoir.) In this sense, the statement
that W  acts as the total energy for a system that includes the battery as part of the
system has to be generalized to include the whole set of batteries or charge reservoirs.
Again, a similar thing is to be imagined for the thermodynamic analogy Eq.(23), where
the new system includes an infinite set of thermal reservoirs at different temperatures T.