Note on poisson and laplace equation
Poisson’s and Laplace’s Equations
The solution to an electrostatic problem is straightforward for cases in which the charge
distribution is everywhere specified. The Coulomb law or Gauss’s law can be used to determine
the potential and electric field.
However, many of the problems encountered in practice are not of this type.
The Poisson’s and Laplace’s equations offer an alternative method in obtaining the potential and
electric field given the boundary or initial conditions.
Consider the Gauss’s law in differential form which is expressed as
D = v
where ρv is the charge density per unit volume in the enclosed volume.
In free space,
Use eqn. (2) in eqn. (1) to get
E = v
But E can be expressed as potential gradient i.e. E = −V . Substituting
this value of E in eqn. (3) yields
(−V ) = v
(V ) = – v
(V ) = – v
2V = –
Eqn. (4) is called the Poisson’s equation. If the charge configuration is known,
then the equation can be solved to obtain the potential function and then
electric field. If the region is charge free then eqn. (4) reduces to
2V = 0
Eqn. (5) is known as the Laplace equation. The LHS is the divergence of
Gradient of V.
The operator 2 is called the Laplacian operator. It takes different forms for
different coordinate systems as listed below.
In Cartesian Coordinates:
2V = (
aˆ x + aˆ y + aˆ z ) (
aˆ x +
aˆ y +
aˆ z )
2V 2V 2V
= 2 + 2 + 2
In Cylindrical Coordinates:
1 V 1 2V 2V
2 2 z 2
In Spherical Coordinates (r, θ, ϕ)
1 2 V
r 2 sin 2 2
r 2 r r r 2 sin
The Laplace equation can be solved for appropriate coordinate system to
obtain the potential V and the electric field E.
The Uniqueness Theorem
This states that if a region of space is bounded by surfaces at known potentials and
contains given charges, there is only one solution for the potential and electric field
within the region. i.e. if V1 is a solution and V2 is also a solution, then V1 = V2
because there is only one solution.
Find the potential function for the region between the parallel circular disc shown in the
figure. Neglect fringing.
Since V is not a function of ρ or ϕ, Laplace’s equation reduces to
And the solution is V = Az + B. The parallel circular discs have a potential function
identical to that for any pair of parallel planes. For another choice of axes, the linear
potential function might be V = Ay + B or V = Ax + B
aˆ z = -A âz
Two parallel conducting planes in free space are at y = 0 and y = 0.02 m, and the zero
voltage reference is at y = 0.01 m. If D = 253ay nC/m2 between the conductors, determine
the conducting voltages.
From Example 4.1, V = Ay + B. Then
E = D/o = -V = -Aay = [(253 x 10-9)/(8.854 x 10-12)]ay
whence A = -2.86 x 104 V/m. Then,
0 = (-2.86 x 104)(0.01) + B or
B = 2.86 x 102 V
V = -2.86 x 104y + 2.86 X 102 V
Then for y = 0, V = 286 V and for y = 0.02, V = -286 V.
Find the potential function and the electric field intensity for the region between two
concentric right circular cylinders shown in Fig. 4.3, where V = 0 at ρ = 1 mm and
V = 150 V at ρ = 20 mm. Neglect fringing.
V = 150 V
The potential is a constant with ϕ and z. Then Laplace’s equation reduces to
Integrating eqn. (1) once, we get
Integrate (2) to get
V = Aln(ρ) + B.
Applying boundary conditions,
0 = Aln(0.001) + B
150 = Aln0.02 + B
A = 50.1,
B = 345.9.
Substituting these values in eqn. (3) to get
V = 50.1ln ρ + 345.9 (V)
E = â = −
In spherical coordinates, V = 0 for r = 0.10 m and V = 100 for r = 2.0 m. Assuming free
space between these concentric spherical shells, find E and D.
Since V is not a function of or , Laplace’s equation reduces to
(1/r2)(d/dr)(r2dV/dr) = 0
(r2dV/dr) = A
and the second integration gives
V = (-A/r) + B
The boundary conditions give
0 = (-A/0.1) + B
100 = (-A/2) + B
A = 10.53 Vm,
B = 105.3 V. Then
V = (-10.53/r) + 105.3,
E = – V = -(dV/dr)ar = -(10.53/r2)ar
D = oE = (-9.32 x 10-11/r2)ar (C/m2)