Potential
The work done by the electric ¤eld on a particle with charge q that moves
from a point A to a second point P is
Z
W =
dW
where dW is the work done as the particle moves through a displacement d~s
~ ¢ d~s
dW = F~ ¢ d~s = qE
Thus
Z
W =
B
A
~ ¢ d~s
qE
If we are dealing with static electric ¤elds, the force is conservative, that is, the
work done is independent of the path we choose between points A and B: We
can check this assertion using one of the ¤elds we have already found: say, the
¤eld due to an in¤nite line charge.
W =q
Z
B
A
¸
^r ¢ d~s
2¼r"0
The dot product selects the radial component of d~s; leaving
Z B
¸
dr
¸
rB
W =q
=q
ln
2¼" 0 A r
2¼"0
rA
The result depends only on the radial coordinates of the points A and B; independent of the path taken from one to the other.
Thus we can express the work done as the change in potential energy of the
system. The work done by the system reduces the stored energy in the system.
Thus
¸
rB
W (A ! B) = U (A) ¡ U (B) = q
ln
2¼" 0
rA
where the ¤nal expression is for the particular system (line charge) that we
used above. Since the value of the test charge that we moved comes out of the
integrals, we can use a trick similar to what we did in de¤ning electric ¤eld. We
de¤ne a ¤eld quantity called potential as follows
Z B
1
~ ¢ d~s
V (A) ¡ V (B) = W (A ! B) =
E
(1)
q
A
~ and the two
The integral on the far right depends only on the electric ¤eld E;
points A and B:
Some people like to write the result with a minus sign, by ¢ipping the limits:
Z A
~ ¢ d~s
V (A) ¡ V (B) = ¡
E
B
1
It doesnzt matter which way you do it, so long as you remember that the integral
is a path integral along a path that you may choose from one point (the "lower
limit") to the other (the "upper limit"). A trick I use to get the signs right is
to remember that "Potential decreases along ¤eld lines". If we go along a ¤eld
~ ; we are going toward lower potental.
line in the direction of E
As with potential energy, there is an arbitrary constant in this de¤nition.
We can set this constant by choosing a reference point R at which we set V ´ 0:
Then we may write our line integral as follows:
Z B
~ ¢ d~s
V (A) ¡ V (B) =
E
V (A) ¡ 0 ¡ [V (B) ¡ 0]
Z
=
A
R
A
= ¡
and
V (P ) = ¡
Z
Z
~ ¢ d~s +
E
A
R
Z
B
R
~ ¢ d~s +
E
Z
~ ¢ d~s
E
B
R
~ ¢ d~s
E
P
R
~ ¢ d~s
E
(2)
If our system has a ¤nite amount of charge that does not extend out to in¤nity,
we usually choose the reference point to be at in¤nity. With this choice, the
potential at a distance r from a point charge q is
kq
(3)
r
However, our line charge extends to in¤nity, so that choice for R is not available
to us. We can see that by noticing that the log function is in¤nite as r ! 1
(and also negative in¤nite as r ! 0): Thus for this ¤eld we must choose our
reference point at some ¤nite distance from the line.
Because the potential integral (1) is path independent, we may evaluate the
potential di¤erence using two di¤erent paths C 1 and C 2 :
Z B
Z B
~ ¢ d~s =
~ ¢ d~s
V (A) ¡ V (B) =
E
E
V (r) =
A;C 1
2
A;C 2
and so
Z
B
A; C1
~ ¢ d~s ¡
E
Z
Z
B
A;C 2
C=C 1¡C 2
~ ¢ d~s
E
= 0
~ ¢ d~s
E
= 0
where C is the closed path formed by going from A to B along C 1 and back to
A along C 2 :
~ true for any closed
Again this is a global statement about the electric ¤eld E;
curve C: To get a local statement, we use Stokezs theorem (Gri¢ths 1.3.5)
Z
Z ³
´
~
~ £E
~ ¢n
E ¢ d~s = 0 =
r
^ dA
C
S
where S is any surface spanning the curve C: Since this result must be true for
any C (and thus any S); we may conclude that
~ £E
~ =0
r
(4)
This is our second Maxwell equation, but unlike Gaussz law, it is not yet complete. This result is true only for static ¤elds.
Now we have learned something very powerful about static electric ¤elds:
The curl of a gradient is always zero (G 1.2.7) so we may express our electric
¤eld as the gradient of a scalar function ©:
~ = ¡r©
~ )r
~ £E
~ =0
E
~ = ¡ r©;
~
The minus sign is conventional. Now if E
then we can compute the
path integral as follows:
Z
B
A
~ ¢ d~s =
E
Z
B
A
~ ¢ d~s = ¡
¡r©
Z
B
A
d© = © (A) ¡ © (B)
Comparing this result with (1), we can see that this © is exactly the potential
V that we have already de¤ned!
~ and V tells us that V will be an extremum
Note that the relation between E
~
~
( rV = 0) wherever E = 0; but V need not be, and in general is not, zero!
~ at
Conversely, if V = 0, that fact does not tell us anything about the value of E
that point. We are free to add any constant to V everywhere in space without
~ at all.
changing the value of E
Calculating V:
If we know the electric ¤eld, we can use the de¤ning integral (1) to calculate
V ; as we already did for the in¤nite line charge case. This method works well
when the ¤eld is easy to get and is mathematically simple to express (so that
~
the integral is not too hard.) Usually if you have enough symmetry to get E
using the integral form of Gaussz law, it is pretty easy to get V this way.
3
If the charge distribution is ¤nite, we may model it as a collection of di¤erential elements (point charges) and use the point charge potential (3). Remember:
when you use this method you are putting the reference point at in¤nity.
Letzs compute the potential on its axis due to a disk of radius a with uniform
charge density ¾:
MODEL We model the disk as a collection of di¤erential rings of width dr:
The potential due to one ring is easy to get, because every element of the ring is
at the same distance from our point P: Then we sum (integrate) the potentials
due to all the rings.
SETUP An element of the ring has length rdÁ; thickness
dr and charge
p
dq = ¾dA = ¾rdrdÁ: Its distance from point P is d = r2 + z 2 ; the same for
each element. The potential due to a ring of radius r is then
dV =
Z
2¼
0
¾rdÁ
k¾rdr
k
dr = p
d
r2 + z 2
Z
2¼
0
2¼ k¾r
dÁ = p
dr
r2 + z2
Then the potential at P due to all the rings is
V (P ) =
Z
a
0
2¼k¾r
p
dr = ¼k¾
r2 + z2
Z
z
a 2+z2
du
u 1=2
where u = r2 + z2 ; du = 2rdr
SOLVE The integral is easy, giving
¯a 2+z 2
³p
´
¯
V (P ) = ¼k¾ 2u 1=2 ¯
= ¼k¾2
a2 + z 2 ¡ z
z
ANALYZE The surface charge density has dimensions of charge/length2 ; so
our answer is k£charge/length, as required. As z ! 1; the answer goes to zero,
which is correct, but not very informative. The important thing is whether it
goes to zero in the right way. Wezd expect to get the result for the potential
due to a point charge Q = ¼ a2 ¾: (This is just RULE 1 again.) Letzs investigate:
à r
!
a2
V (P ) = ¼k¾2 z 1 + 2 ¡ z
z
4
Now expand the square root using the binomial expansion:
· µ
¶
¸
1 a2
V (P ) = ¼k¾ 2 z 1 +
+
¢
¢
¢
¡
z
2 z2
·
¸
1 a2
¼a2 ¾
= ¼k¾ 2 z 2 + ¢ ¢ ¢ ! k
as z ! 1
2z
z
so we get the correct result.
Now we express the integral in its most general form, as we did previously
~ : The potential at ~r due to a di¤erential element at ~r0 , with reference
with E
point at in¤nity, is
k½ (~r 0) d¿ 0
dV =
j~r ¡ ~r 0 j
and so
V (~r) = k
Z
½ (~r0 ) d¿ 0
j~r ¡ ~r0 j
(5)
The ¤nal method for getting V is to solve the di¤erential equation that we
~
can get from Gausszs Law using the relation of V to E.
~ ¢E
~
r
=
~ ¢ (¡rV ) =
r
r2 V
=
½
"0
½
"0
¡
½
"0
(6)
This equation is easy to write but is not always so easy to solve, as we shall see.
We can learn something interesting by looking at a point charge again. What
is its charge density? It is zero everywhere, except at the position of the charge.
For simplicity, letzs put the charge at the origin. Then when we integrate over
any volume including the origin, we must get
Z
q = ½dV
We can describe this behavior with a quantity called the delta function. (Thatzs
its name, but it is not a real function at all.) It has the properties
= 0 if x 6= 0
± (x)
Z
= 1 if x = 0
± (x)
+1
± (x) dx
= 1
¡1
It is actually de¤ned by a property called the sifting property:
Z +1
f (x) ± (x) dx = f (0)
¡1
5
Do you see how this follows from the previous properties of ± (x)? Actually all
we need is that the range of integration include the origin:
½
Z b
f (0) if
a<0<b
f (x) ± (x) dx =
0
otherwise
a
Now our charge is at x = 0 and y = 0 and z = 0; so we need three delta
functions:
½ (~r) = q± (x) ± (y) ± (z) = q± (~r)
The ¤nal expression is a sort of shorthand for the three delta functions in the
middle expression.
Now Poissonzs equation becomes
µ ¶
kq
q
r2
= ¡ ± (~r )
r
"0
and so it must be true that
r2
µ ¶
1
= ¡4¼ ± (~r )
r
(7)
This is a very useful result that we will want to use often.
Equipotentials
We can use potential to get a visual picture of electric ¤elds by drawing the
surfaces on which V is constantw the equipotential surfaces. Let d~s be a small
displacement along one of these surfaces. Then
~ ¢ d~s = 0
dV = E
~ at that point. Thus:
So d~s must be perpendicular to E
Equipotential surfaces are perpendicular to electric ¤eld lines.
Energy
The potential energy of any system equals the work done to assemble the
system. (LB Chapter 8) So if we want to ¤nd the energy stored in an electric
system, we just have to dream up a convenient procedure for putting the system
together. Letzs start with a system of point charges.
We bring in the ¤rst charge q1 : There are no pre-existing ¤elds, so no force
is needed and no work is done. We put the charge at position ~r1 :
Now we bring in the second charge q2 : At each point along our path, we have
to exert a force on q2 that is the exact opposite of the force exerted by q1 ; so
that the net force on q2 is zero.
F~e x er ted
by us
= ¡F~ e xe rte d
b y q1
~1
= ¡q2 E
Then we can move it at a constant, in¤nitesimally slow, speed, until we get to
the ¤nal position ~r2 : The work done is
Z P2
~ 1 ¢ d~s = q2 V1 (~r2 ) = q2 kq1
W2 =
¡q2E
(8)
j~r2 ¡ ~r1 j
1
6
Now we bring in the third charge. This time the force we need is the exact
opposite of the sum of the forces exerted by the ¤rst two charges:
"Z
#
Z P3
Z P3
³
´
P3
~1 + E
~ 2 ¢ d~s = ¡q3
~ 1 ¢ d~s +
~ 2 ¢ d~s
W3 =
¡q3 E
E
E
1
=
1
q3 q1
q3 q2
q3 [V 1 (P3 ) + V 2 (P3 )] = k
+k
j~r3 ¡ ~r1 j
j~r3 ¡ ~r2 j
1
Now we can see a pattern emerging.
W
=
U
=
W1 + W2 + ¢ ¢ ¢
q1
q3 q1
q3 q2
W = kq2
+k
+k
+ ¢¢¢
j~r2 ¡ ~r1 j
j~r3 ¡ ~r1 j
j~r3 ¡ ~r2 j
=
k
N X
X
i=1 j<i
N
N
qi qj
kX X
=
j~ri ¡ ~rj j
2
i=1 j=1;j6= i
qi qj
j~ri ¡ ~rj j
Another way to think of this is that the energy is stored in pairs of charges, and
we need to sum over all the distinct pairs.
We can also express the result in terms of potential: Let V (~ri ) denote the
potential at the position ~ri of charge qi due to all of the other charges, j = 1¡ N;
excluding charge qi : Then
U=
N
1X
qi V (~ri)
2
i=1
Now what if we have a continuous distribution of charge instead of a set of point
charges? Well, we model our distribution as a collection of di¤erential elements,
and treat each element as a point charge dq = ½d¿ . But now our sum becomes
an integral.
Z
1
U=
V (~r ) ½ (~r ) d¿
(9)
2
(Izm using Gri¢thzs notation ¿ for volume to avoid confusion with potential V :)
However therezs a catch here. Notice that we have no way to exclude i = j as
we did when summing over point charges. Wezll see what that does in a minute.
We are going to express the energy U in terms of the ¤eld alone, so we start
~ using Gaussz law,
by expressing ½ in terms of E
~ ¢E
~
½ = "0 r
Then
U=
"0
2
Z
~ ¢E
~ d¿
Vr
~ and V; E
~ = ¡rV
~ ; but how? We need
Wezd like to use the relation between E
to do an "integation by parts". Note that (G 1.2.6 iii)
~ ¢ (Á~u) = Ár
~ ¢ ~u + ~u ¢ rÁ
~
r
7
~ to get
We can use this result with Á = V and ~u = E
³
´
³
´
~ ¢E
~ =r
~ ¢ VE
~ ¡E
~ ¢ rV
~ =r
~ ¢ VE
~ +E
~ ¢E
~
Vr
Thus
Z h
³
´
i
"0
~ ¢ VE
~ +E
~ ¢E
~ d¿
r
2
Next we can use the divergence theorem to evaluate the ¤rst term:
Z
I
³
´
~ ¢ VE
~ d¿ =
~ ¢n
r
VE
^ dA
U=
S1
The original integral was over all space, so the bounding surface is "at in¤nity".
Now provided that the total charge in our system is ¤nite, we can use RULE 1
again to argue that
kQ
~ ' kQ r^
V '
;
E
r
r2
so that on our surface at in¤nity (which we might as well take to be a sphere
centered at the origin, so that ^r ¢ ^n = 1)
I
I
I
1 2
1
1
~ ¢ ^ndA = kQ2 lim
VE
r
d=
kQ
lim
d- = 4¼kQ lim
=0
r !1
r!1 r
r!1 r
r3
S1
Notice that idealizations like in¤nite line charges must be excluded here because
they extend all the way to in¤nity.
The remaining term in our energy expression is
Z
Z
"0
U=
E 2 d¿ =
u d¿
2
where the lower case u is the electric energy density
u=
"0 2
E
2
(10)
This expression is of the standard form for energies in physics: 1=2 times (constant) times (variable)2 : (KE = 12 mv 2 ; spring PE= 12 ks 2; etc), and, like these
energies, it is always positive. Letzs see what happens if we apply the result to
two point charges. For convenience put the origin on the ¤rst charge.
~ (~r ) = kq1 r^ + kq2 (~r ¡ ~r2)
E
3
r2
j~r ¡ ~r2 j
Thus
E
2
=
=
"
# "
#
kq1
(~r ¡ ~r2 )
kq1
(~r ¡ ~r2 )
^r + kq2
¢
r^ + kq2
r2
r2
j~r ¡ ~r2 j3
j~r ¡ ~r2 j3
Ã
!2
³ q ´2
kq2
q1 q2
(~r ¡ ~r2 )
1
k 2 +
+ 2k 2 2 ^r ¢
2
r
r
j~r ¡ ~r2 j
j~r ¡ ~r2 j3
8
The ¤rst term here depends only on q1 ; the second only on q2 ; and the third on
the product q1 q2 : The ¤rst term gives the self-energy of charge one, and it is
in¤nite if the charge is really a point.
Z
Z
Z Z
"0
"0 2¼ ¼ 1 ³ q1 ´2 2
U1 =
E 12 d¿ =
k 2 r sin µdµdÁdr
2
2 0
r
0
rmin
Z 2¼
Z ¼
Z 1
"0 2 2
1
=
k q1
dÁ
sin µdµ
dr
2
2
r
0
0
rmin
µ
¶¯ 1
1 ¯¯
= 2¼"0 k2 q12 ¡
r ¯
rmin
=
q12
k
! 1 as rmin ! 0
2 rmin
Similarly the second term is the self energy of charge 2. The third term is the
interaction energy that depends on both of the charges, and it is ¤nite. In fact,
it is identical to the result (8)
Z
q1 q2
(~r ¡ ~r2 )
q1 q2
U12 = 2k2 2 r^ ¢
d¿ = k
3
r
r12
j~r ¡ ~r2 j
that we obtained for two point charges. This is the only part of the total energy
that can ever be changed. The self energy is an intrinsic property of each charge
and we cannot get our hands on it. Thus we choose to ignore it, and look only
at the ¤nite, interaction energy. The self energy was introduced because we
could not exclude "i = j" from our integral (9).
This formulation tells us something about where the energy is stored in our
system: it is stored in the ¤eld itself and extends throughout the whole of space.
Notice that since the energy is quadratic in the ¤elds, we cannot just add
the energies of di¤erent parts of our system. The interaction terms are very
important. The superposition principle for ¤elds tells us that we can add the
¤elds due to di¤erent sources:
~ =E
~1 + E
~2 + E
~3 + ¢ ¢ ¢
E
but the corresponding energy density is proportional to
~1 ¢ E
~ 2 + 2E
~1 ¢ E
~3 + ¢ ¢ ¢
E 2 = E 12 + E 22 + ¢ ¢ ¢ + 2E
9