Poission’s and Laplace’s equations
Lecture 4
1
Including the Electric Potential in Gauss’ Law
~ ·E
~ = ρ/ǫ. Insert the above expression for the electric
Gauss’ law in differential form is ∇
field in terms of the electric potential. This results in;
~ · ∇V
~ = −ρ/ǫ0
∇
This is a partial differential equation, Poisson’s equation, which we will solve in order
to obtain the electric potential, and from the potential, the electric field. In the case where
ρ = 0 the equation is called Laplace’s equation and the operator, ∇2 , is the Laplacian. These
equations are 2nd order, linear, partial differential equations. We will later look at methods
for their solution.
2
Total energy of a charge distribution
We assume a set of positive charges placed at positions in space. An energy is required to
assemble this distribution which can be calculated by moving each charge from far away
to its position. As each charge is assembled, it experiences an potential due to the other
charges previously moved into position, and there an energy given by the values of the charge
multiplied by the electric potential due to each assembled charge. We write for this energy;
W = κ [q1 q2 /r12 + q1 q3 /r13 + · · ·
q2 q3 /R23 + q2 q4 /r24 + · · · ]
This is written as ;
W = κ
P
qi qj /rij = (κ/2)
i>j
P
qi qj /rij
i6=j
Note that this sum can be rearranged to be written in the form;
W = (κ/2) [q1
P
qj /r1j + q2
16=j
P
qj /r2j + · · · ]
26=j
Here the electric potential is just;
Vi = κ
P
qi /rij
i6=j
1
Which can be used above to get;
W = (1/2)
P
qi Vi
i
Now change the discrete charge distribution to a continuous one so that the sum is
changed to an integral.
W = (1/2)
R
dτ ρ V
~ ·E
~ = ρ/ǫ0
Then apply Poisson’s equation ∇
W = (ǫ0 /2)
R
~ ·E
~
dτ ρ ∇
Here;
~ ·E
~ = ∂Ex + ∂Ey + ∂Ez
∇
∂x
∂y
∂z
~ · (V E)
~ =
Substitute into the above equation and integrate by parts or use the identity ∇
~ )·E
~ + V (∇
~ · E).
~ If the surface terms in the integration by parts vanish (ie taken at
(∇V
large distances) we obtain;
W =
R
~ · E)
~
dτ (ǫ0 /2)(E
~ 2 to the electric field.
From the above we assign an energy per unit volume of (ǫ0 /2)E
This is the energy that was needed to assemble the charge distribution to create the field.
3
Example
To illustrate that we should assign the energy required to create the field, to the field, we
consider the field of a parallel plate capacitor. We separately obtain this field by superposition of the field from each plate of the capacitor. The field of an infinite plate with uniform
charge density can be obtained using Gauss’ law. Consider figure 1. The flux out of the
cylindrical volume that encloses a surface charge, σ da, penetrates only the cylindrical end
caps of the cylinder, since by symmetry the field can only be perpendicular to the plate.
Thus we write;
H
~ · dA
~ = 2E(A) = Q/ǫ0 = σ(A)/ǫ0
E
The factor of 2 comes from the integration over each end cap, and note that the field
~ is uniform
points in oposite directions above and below the plane. Then the magnitude of E
2
E1
E2
E2
σ da
Plane 1
E1
E2
d
E2
σ
Plane 2
E2
Field generated in this volume
E1
E2
Figure 1: The field of a parallel plate capacitor used to find the energy stored in the electric
field
over the end cap area so it factors from the integral. The resulting field is E = σ/(2ǫ0 ),
pointed away from the plate, both above and below the plate. Note the field strength is
independent of the distance away from the plate. This is due to the fact that the plate is
infinite in extent, which is an approximation for a large plate with distances not far from
the surface. Now a parallel plate capacitor has two plates of opposite charge, that are parallel and closely spaced. This is shown in figure 1. The field strength due to each plate
is the same as calculated above but the sign of the field is reversed due to the opposite
charges. We see that the fields between the plates add, but cancel outside the capacitor.
The total field between the plates is E = σ/ǫ, and E = 0 outside the capacitor. Now we
find the force per unit area on plate 2 due to plate 1. The field of plate 1 at the position
of plate 2 is E1 = σ/(2ǫ). The force on an area of plate 2 containing a charge (σ Area) is then;
F2 = (σ Area)E1 = σ 2 Area/(2ǫ0 )
F/(Area) = σ 2 /(2ǫ0 )
Then substitute for σ the value of the total field between the plates, ET = σ/ǫ0 ;
F/Area = ǫ0 E 2 /2
We are now in a position to look at the energy that is put into the field by the geometry
of the charge distribution. We have the field, E = σ/ǫ0 , between the plates and no field
outside the plates. Then we uniformly separate the plates a distance, d. Because there is
an attractive force between the plates we put work (potential energy) into the system. This
energy is equal to the force times the distance the plates are moved. Thus there is an energy;
3
Q1
Q2
R1
R2
Figure 2: The simplified geometry used to examone the field strength near a sharp edge of
a conductor
W = (F/Area) × Area × d = [ǫ0 E 2 /2] Volume
The Volume is the additional volume created as the plates are moved apart, and in this
volume we have created an electric field where there was none originally. If we let the plates
collapse they will release the energy that was put into the system by pulling the plates apart
and at the same time, destroying the field. This energy is then presumed to be stored in the
field. Therefore we assign an energy per unit volume , ǫ0 E 2 /2, to the field.
4
Conductors
If an electric field is applied, there will be a force on a charge which may cause the charge
to move. In fact, a conducting material is one in which charges are free to move. At the
moment we are considering only static charge. If a conducting material is inserted into an
electric field, charges will move until they generate their own field which cancells the applied
one. Thus a static condition will eventually be reached, and for a static charge distribution
there is no electric field in a conducting medium or when the conducting medium is a perfect
conductor (ie has no resistance to the flow of charge). An electric field equal to zero implies
a constant potential. This has the consequence that all points of a continuous, conducting
medium have the same value of electric potential.
As an illustration we consider the electric field near a sharp edge of a conductor. We
approximate this geometry by two conducting spheres connected by a thin, perfectly conducting wire, as shown in figure 2. The wire allows charge to move between the spheres so
that they have the same potential. On the other hand it is thin, and short so its contribution
to the field is ignored.
The field about a conducting sphere must be radial (from the argument above there can
be no tangential compoment of E field) and must have the same magnitude at all points on
a spherical surface. As we have previously seen from Gauss’ law, the field is ;
4
~ = κ Q2 R̂
E
R
where Q is the charge on a sphere and R is its radius. The electric potential of a sphere is ;
Q
V = κR
When the two spheres are connected by a conducting wire they have the same potential.
We write;
Q
Q
V = κ R1 = κ R2
1
2
From this we find that the charge on the spheres is related by;
Q1 = Q2 (R1 /R2 )
The ratio of the electric field strength is;
E1 /E2 = R2 /R1
Then as R1 ≫ R2 we have that E2 ≫ E1 . Applying this simple solution to the more
complicated problem of a conductor with a sharp edge, we have that the electric field parallel to the conducting surface is zero and the potential is constant, but the perpendicular
component of E is much larger at the edge than on the smooth surface. This is because the
charge per unit area is larger at a sharp edge.
If this conductor is placed in air, the strong field can strip electrons from the air molecules
and attract free charge that might be in the vacinity of the point. These charges are accelerated in the strong field, colliding with other molecules and producing an avalanche of charge
that can create a spark of electricity. If the conductor is in vacuum, the strong field can pull
electrons from the conducting surface, which is called field emission.
5
Charge on the Earth’s surface
The earth has a strong electric field at its surface of about 100 volt per m pointed inward to
the surface (negative charge on the surface). Assuming a uniform distribution over a spherσ with σ the surface charge density. Substitution
ical surface the field is approximately 2ǫ
0
for E gives σ = 1.8 × 10−9 c/m2 or approximately 1010 excess electrons per square meter.
Presumably the earth remains charged due to lightning. Currents in a lightning stroke can
be 104 amperes.
5
6
Examples
We now determine the energy required to assemble a uniformly charged spherical shell of
charge. The electric field of a uniformly charged shell of charge is obtained by Gauss’ law.
The field at a radius greater than the radius of the spherical surface, R, is the same as if all
the charge were placed at the center of the charge distribution. The field within the shell
vanishes as there is no enclosed charge and the field is symmetric about the center of the
sphere. Thus for r > R;
2
σ
σR2 r̂
~ = κ Q2 r̂ = κ (4πR
r̂
=
E
r
r2
ǫ0 r 2
We can get the energy using the energy density;
W =
R
dτ (ǫ0 /2) E 2
R
R∞
2 4
4
r 2 dr σ R
2 r
ǫ
0
R
2
2 4
Q
W = 2πσǫ0 R = 8πǫ R
0
W = (ǫ0 /2)
dΩ
This is just Q times the electric potential as expected for a spherical shell of charge.
We could immediately obtain this result from the expression for the energy in terms of the
charge density.
W = (1/2)
R
dτ ρ V
This result is obvious because the charge density is confined to a spherical shell of radius,
R. Finally we could also obtain the result by finding the energy required to bring a small
charge, δQ, from infinity to the shell, and uniformly distributing this charge over the shell
(distribution requires no energy). The field remains that of a uniformly charged spherical
shell as we continue to bring in charge to the shell. Each increment adds a charge, δQ, to
the charge, Q. Therefore the energy is;
W = κ
RQ
0
Q
dQ R
This gives the same result above which was calculated by two different ways ways.
Next suppose a uniform spherical charge distribution of radius, R . We calculate the
electric potential by finding the potential at a point r > R and for 0 < r < R. The potential
Q
is obtained from the electric field, which itself is obtained from Gauss’ law as E = κ enclosed
r2
R
~ · d~r
V = E
6
r>R
V = −κ
Rr
dr
RR
Rr
Q
Q
ρr
dr r − dr 3ǫ ] = 8πǫ R [3 − (r/R)2 ]
0
0
R
∞
Q
Q
2 = κ r
r
0<r<R
V = −κ [
∞
~ = ∇V
~ .
From this we obtain the field both inside and outside the sphere using, E
Then to get the energy to assemble a uniformly charged sphere, we can use any of the above
ρR3
and for
methods. Suppose we use the energy density. The field strength for r > R is
3ǫ0 r 2
ρr
0 < r < R is 3ǫ . The energy is;
0
RR
R∞ ρ2 R6 2
ρ2 r 4
0 [
dr
+
dr
r ]
W = 4πǫ
2 0
9ǫ20
9ǫ20
R
3Q2
W = 20πǫ R
0
Finally can a solid conductor have “free” charge in its interior ? Clearly a conductor can
have enclosed charge due to atomic structure, but these charges are bound and the positive
charge equals the negative charge inside the conductor. If there were free charge inside the
conductor, there would be an electric field which would cause these charge to move, violating
the static assumption. In static equilibrum the free charge on a conductor can only reside
on its surface. Although not a proof, one sees above that the energy of the spherical charge
distribution is greater than that of the spherical shell. A system will always move from a
higher to a lower energy state.
7
Energy minimization
The forces on charges in a charge distribution of a set of conductors will move the charge
so that a minimum in the energy of the field is produced. This generates a technique to
determine the field by approximation, although we do not show this in detail here.
8
Capacitors
If a potential difference is placed on a set of conductors, they will have excess charge on their
surface, and this charge acts as the source of the required electric field which is produced.
As our previous discussion demonstrates, the creation of this charge distribution not only
7
creates an electric field but also involves a change in the system potential energy. The geometrical arrangement of conductors has a property called capacitance, which is the ability
to store electric charge and thus energy. Capacitance depends on the system geometry as
can be seen from a calculation of the system electric potential at the position of conductor
1.
V1 = κ
R
dτ ′
ρ
|~r1 − ~r′ |
Use the mean value theorem of calculus to extract the average value of
1
from
|~r1 − ~r′ |
the integral. This results in an expression;
1/C1 = Z V1
dτ ρ
which is independent of the charge distribution. Here C is the capacitance of the system
and mathematically is expressed as
1/C = κh
1
i
|~r1 − ~r′ |
This expression is not useful to calculate capacitance, but does show that there is a
linear relation between 1/C, the potential, and the charge. Thus suppose that there are N
conductors that have charges Qi and potentials, Vi . The development above demonstrates
that we can write;
Vi =
N
P
sij Qj
j=1
The terms sij are elements of a matrix which can be inverted to obtain the capacitance
matrix.
Qi =
P
Cij Vj
The inversion element, Cii , is the coefficient of capacitance, and the elements, Cij , are
the coefficients of mutual capacitance. Then the capacitance is defined as the charge to
potential ratio with all other conductors grounded. It is always greater than zero.
9
Energy in terms of capacitance
The energy of a configuration of charge, as given above is;
8
σ
V = V0
d
E
V=0
σ
Area = A
E
Figure 3: The geometry of a parallel plate capacitor
W = (1/2)
P
qi Vi
i
Then we can substitute for either Q or V written in terms of the capacitance.
W = (1/2)
P
Cij Vi Vj = (1/2)CiiVi2
ij
W = (1/2)Q2i /Cii
The above expressions are obtained by setting Vij = 0 when i 6= j from the definition of
capacitance.
10
Examples
We look first at the simple example of a parallel plate capacitor with large plate area and
small spacing between the plates. The geometry is shown in figure 3. We have seen that
the electric field between the plates is E = σ/ǫ and vanishes outside the capacitor. The
potential between the plates is;
V =
Ra
~ · d~l
E
b
~ is perpendicular to the plates and the potential on one plate is set equal to zero
Then E
to satisfy the requirements of measuring capacitance. The other plate is held at a potential,
V0 which is the same everywhere on the plate.
V = Ed
Substitution for the value of E gives;
V = (σ/ǫ)d = (Q/Area ǫ)d
From the definition of capacitance ;
9
1
V = V0
E1
2
E2
V=0
Figure 4: The geometry used to calculate total capacitance for 2 capacitors joined in parallel
or series
C = Q/V = (ǫ0 Area)/d
The energy per unit area stored in this capacitor is then;
W Volume = (1/2)C V 2 = (1/2)
(ǫ0 Area)E 2 d2
= (ǫ/2)E 2
d
as is expected. Now consider an arrangement of capacitors either in parallel or in series
as is showm in figure 4. In the case of the capacitor array the voltage across each of the capacitors is the same. The total charge is the sum of the charges on each capacitor. Therefore;
QT = (Q1 + Q2 ) = V0 (C1 + C2 )
CT = C1 + C2
In the case of capacitors in series, the charges on the plates connected by conductors
must be equal but opposite as shown in figure 4.
V1 = Q/C1 and V2 = Q/C2
Then since V0 = V1 + V2 Q(1/C1 + 1/C2 )
1/CT = (1/C1 + 1/C2 )
The general technique for calculating capacitance is to find the potential, perhaps by
integration of the field and divide this into the charge that created the field. As another
example of practical interest, we calculate the capacitance per unit length of a coaxial cable.
10
V = V0
L
V= 0
E
a
b
Figure 5: The geometry of a coaxial cable used to find the capacitance per unit length
The geometry is shown in figure 5. The cable is assumed to be infinite in length so by
symmetry the electric field is radial in cylindrcal geometry, ( ie independent of both z and
φ). We use a Gaussian surface about the inner wire of the cable that carries a charge per
unit length, λ. We have previously found that the field is;
~ = λ
E
2πǫ0 r
This is the field between the two concentric cylinders of the coaxial cable. There is no
field inside the inner conductor and outside the outer conductor. The potential between the
conductors is then;
V =
Rb
~ · d~r
E
a
λ ln(b/a)
V = 2πǫ
0
The capacitance, since λ = Q/L is then;
C/L = Q/V =
2πǫ0
ln(b/a)
We now find the stored energy per unit length in the capacitor. This is given by;
λ2 ln(b/a)
λ2 L2 ln(b/a)
Q2
=
W/L = 2 C =
2
4πǫ0
2 2πǫ0 L
We could also get this result by integrating the field energy per unit volume.
W = (ǫ0 /2)
R
dl rdr dφ E 2
W = (ǫ0 /2)(2πL)
Rb
a
rdr
λ2
4π 2 ǫ20 r 2
11
W/L =
λ2 ln(b/a)
4πǫ0
12