Lecture ( 11 )
Boundary Conditions , Poisson’s and Laplace’s Equation
3.7 Boundary Conditions
Boundary conditions is the condition that the field must satisfy at the interface separating the
media
•
The boundary conditions at an interface separating:
– Dielectric and dielectric
– Conductor and dielectric
– Conductor and free space
•
To determine the boundary conditions, we need to use Maxwell’s equation:
And
•
Decomposing the electric field intensity E into orthogonal components
where
and
are, respectively, the tangential and normal components of E to the interface of interest
1. Dielectric – dielectric boundary conditions

E1 and E2 in media 1 and 2 can be decomposed as

Applying Maxwell’s equation to the closed path ( abcda )
(1)
As ∆ℎ−> 0 , equation ( 1 ) becomes
(2)
is said to be continuous across the boundary
•
Since D =
=
+
, eq. (2) can be written as
Or
is said to be discontinuous across the interface
Applying the Gauss’s law , we have
Allowing ∆ℎ−> gives
Or
If no free charges exist at the interface , so
(1)
is continuous across the interface , since
=
,eq. ( 1 ) can be written as
The normal component of ( E ) is discontinuous at the boundary
2.Conductor – dielectric boundary conditions
Applying Maxwell’s equation to the closed path ( abcda )
As ∆ℎ -> 0,
Similarly, by applying the Gauss’s law to the pillbox and letting ∆ℎ → 0, we have
because D =
= 0 inside the conductor, so
Or
Thus under static conditions, the following conclusions can be made about a perfect conductor:
1. No electric field may exist within a conductor
2. Since E = -
= 0, there can be no potential difference any two points in the conductor
3. The electric field E can be external to the conductor and normal to its surface
3. Conductor – free space boundary conditions
This is a special case of the conductor – dielectric condition. Free space is a special dielectric for
which = 1
Thus the boundary conditions are
3.8 Poisson’s and Laplace’s Equation
A useful approach to the calculation of electric potential is to relate that potential to the charge +density which gives rise to it . The electric field is related to the charge density by the divergence
relationship
∇.
=
E : electric field , ρ : charge density , εo : permittivity
And the electric field is related to the electric potential by a gradient relationship
= −∇
Therefore the potential is related to the charge density by Poisson’s equation
∇ .∇
= ∇
=
−
In a charge – free region of space , this becomes Laplace’s equation
∇
=0
Tis mathematical operation , the divergence of the gradient of a function , is called ( The
Laplacian ) . Expressing the Laplacian in different coordinate systems to take advantage of the
symmetry of a charge distribution helps in the solution for the electric potential ( V ) . For
example , if the charge distribution has spherical symmetry, use the Laplacian in spherical polar
coordinates.
Potential of a Uniform Sphere of Charge
The use of Poisson’s and Laplace’s equation will be explored for a uniform sphere of charge.
In spherical polar coordinates , Poisson’s equation take the form
∇
=
+
1
+
1
+
2
+
cot
=
−
But since there is full spherical symmetry here , the derivatives with respect to ( θ ) and ( φ )must
be zero , leaving the form
+
2
=
−
Uniform
charge
density ρ
Total charge
=
4
3
R
Examining first the region outside the sphere , Laplace’s law applies.
+
Solution of form
=0
+
Since the zero of potential is arbitrary , it is reasonable to choose the zero of potential at infinity ,
the standard practice with localized charges. This gives the value ( b = 0). Since the sphere of
charge will look like a point charge at large distances , we may conclude that
=
=
4
So the solution to Laplace’s law outside the sphere is
=
4
Now examining the potential inside the sphere , the potential must have a term of order ( r 2 )to
give a constant on the left side of the equation, so the solution is of the form
=
+
Substituting in to Poisson’s equation gives
2 +4 =
=
giving
Now to meet the boundary conditions at the surface of the sphere , r = R
+
=
=
giving
+
The full solution for the potential inside the sphere from Poisson’s equation is
=
6
[
−
]+
4
=
6
[
−
]+
3