L8a - Laplace’s Equation: Potentials and Fields
Rev 12/21/2014
`And how exactly like an egg he is!' she said
aloud, standing with her hands ready to catch
him, for she was every moment expecting him
to fall.
`It's very provoking,' Humpty Dumpty said after
a long silence, looking away from Alice as he
spoke, `to be called an egg -- very!'
`I said you looked like an egg, Sir,' Alice gently
explained. `And some eggs are very pretty, you
know,' she added, hoping to turn her remark
into a sort of compliment.
`Some people,' said Humpty Dumpty, looking
away from her as usual, `have no more sense
than a baby!'
For a thorough discussion of the Laplace Equation, refer to
Dawkins PDE, pages 45-55. In particular, work through the two
solutions to the steady-state heat flow problem presented in
Examples 1 and 2. Then complete the other two solution cases to
form the full solution. Those results will not be reproduced here.
Note this requires eigenfunctions and Fourier Series Solutions.
Functions that solve the Laplace equation are often referred to as
harmonic functions.
Example: Consider the function u(r) = 1/r, where r  ( x  x0 ) 2  ( y  y0 ) 2  ( z  z0 ) 2 in
R3. Show that u is harmonic on a domain that does not include the point (x0, y0, z0).
In many electric potential problems, we want to know the distribution of surfaces of
equal potentials outside of a closed charge-bearing surface. So more work is needed…
In general, Laplace’s equation yields well to separation of variables.
Let V ( x, y )  F ( x)G ( y )
1
F '' G  G '' F  0
F ''
G ''

 2
F
G
However, we do not have enough information to specify the sign that should be
associated with the separation constant, 2.
Case 1: Try  = 0
F’’(x) = 0 = G’’(y) so that V ( x, y )  (C1  C2 x)(C3  C4 y )  A  Bx  Cy  Dxy .
Case 2: keep it positive,  > 0
This yields the familiar G ( y )  C cos  y  D sin  y . But the F equation, F ''  2 F  0 ,
has solutions F ( x)  C1e x  C2 e  x . These are usually written in terms of the hyperbolic
sine and hyperbolic cosine functions ( sinh x 
e x  e x
e x  e x
, cosh x 
).
2
2
In fact, we can say that F ( x)  C1 cosh  x  C2 sinh  x is the general solution to the F(x)
ODE. Thus V ( x, y )  (C1 cosh  x  C2 sinh  x)( A cos  y  B sin  y ) .
As usual, we need BCs, so let V(x,0) = 0 = V(0,y):
V (0, y )  C1G ( y )  0  C1  0
V ( x, 0)  C2 (sin  x) A  A  0
 V ( x, y )  BC2 (sinh  x)(sin  y )
We now need to find the eigenvalues and corresponding eigenfunctions using sin y.
Letting V(x, L) = 0,
sin  L  0;   n

L
.
Taking the coefficients and L to be
1, here is a representative plot for
n = 1.
2
Looking along the x axis, we see V(x,-1) = V(x,0) = V(x,1) = 0; these are the zeros of sin
y. At y =+/- .5, we see sinh x.
Forces and fields: Don’t give me any static (electricity)
Any electric charge (usually represented with the symbols
+Q or –Q) in space gives rise to a vector electric field E.
The field points away from a positive charge and towards a
negative charge. The field strength E is a measure of the
attractive or repulsive force on an external positive “test
charge.”
We use Coulomb’s Law to describe the force between two

Qq
charges: F COULOMB  k 2 r , where k is a positive constant,
r
Q is the field charge, q is the test charge, r is the distance
between Q and q and the unit vector r is the direction from Q to q. A positive force is
repulsive (the resulting acceleration would increase the distance between the charges);
a negative force is attractive.

 F
Since the magnitude of the test charge q might not be known, the electric field E 
q
is a more useful concept: E is in units of force per unit charge.
Since charges may be distributed in a volume of space, we often refer to charge density
(charge Q per unit volume of space). Coulomb’s Law is one of Maxwell’s Equations,
rewritten to state that the divergence of the electric field is proportional to the charge
density:
,
where permittivity is a fundamental constant describing how electric fields
propagate through space (0 = 8.85x10-12 volts/meter).
In three dimensions, we can write the electric field as the vector sum of its components:

E  Ex i  E y j  Ez k and therefore:
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Ex E y Ez  ( x, y, z )
. Subscripts are components, not partial derivatives.



0
x
y
z
Note that in two dimensions, we usually use surface charge density  in place of ; in one
dimension, charge per unit length  (another use for ).
Making a vector from a scalar

The electric field E is further related to a scalar quantity electric potential, which
expresses the energy per unit of charge stored by the field. We know this electric
potential as voltage and note that the field is the negative gradient of potential:
(Recall: force is the negative derivative of potential energy and field is force per
unit charge). We arbitrarily define the potential to be 0 at an infinite distance
from all charge.
Putting these together, the scalar potential V is determined by the charge density
according to a second-order Poisson PDE
If there are no charges in a region of space,  = 0, this reduces to the
homogeneous case (Laplace's equation)
2
2
2
The divergence of the gradient is called the Laplacian:     2  2  2 .
x y z
2
To keep things (somewhat) manageable, we will only look at the Laplace case: What is
the potential at any point in a charge-free region of space? What is the field in response
to nearby charges? Our solutions will depend on the boundary condition set up by the
charge distribution. Once the potential function has been determined by solving the
Laplace equation, the electric field can be computed by taking the gradient of the
potential.
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The figure at right shows a conductive
object with a positive charge density 
(Coulombs/meter2) on its outer surface.
At equilibrium, the entire conductive surface has the same potential – and that requires
that the field inside the surface is 0. Outside the object, the electric field is oriented away
from the positive charges; field lines are perpendicular to the surface of the object. Since
the field is the negative gradient of potential, lines of equal potential (equipotentials)
must be perpendicular to the field lines. Close to the surface of the object, the
equipotentials are parallel to the surface. Our problem is to calculate the potential and
field at locations outside of this object; the edges of this object form our boundary
conditions.
Example: Ideal parallel plate capacitor
Two very large metallic plates are oriented so that they are
parallel to the yz plane; the one located at x = -d/2 is given a
negative charge -Q, the other, located at x = d/2, a positive
charge +Q. We can arbitrarily define the potential at the
V
negative plate to be  0 and the potential at the positive plate
2
V
to be 0 . The potential difference between the two plates is
2
V0; there will be a 0 equipotential midway between the two
plates.
d
d
 x  , the potential only varies in x;
2
2
d 2V
Laplace’s equation reduces to
 0 which is very
dx 2
easily solved by V ( x)  C1  C2 x .
For 
5
Using the boundary conditions,
V
d
d
V ( )   0  C1  C2
2
2
2
V
d
d
V ( )  0  C1  C2
2
2
2
so that C1 = 0 and C2 
V0
.
d
Thus the potential between the plates is given by V ( x) 
V0
x ; the field
d
V
dV
  0 is a constant. The minus sign signifies that the field points to the left
dx
d
(the negative charge is at –d/2).
E ( x)  
This device is called a parallel plate capacitor; it stores charge and energy in the
electric field that is concentrated between the plates.
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