# 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

(1)

where ρv is the charge density per unit volume in the enclosed volume.

In free space,

D =oE

(2)

Use eqn. (2) in eqn. (1) to get

E = v

(3)

o

But E can be expressed as potential gradient i.e. E = −V . Substituting

this value of E in eqn. (3) yields

(−V ) = v

o

Or

(V ) = – v

Or

(V ) = – v

Or

o

o

2V = –

v

o

(4)

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

(5)

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 = (

V

V

V

aˆ x + aˆ y + aˆ z ) (

aˆ x +

aˆ y +

aˆ z )

x

y

z

x

y

z

2V 2V 2V

= 2 + 2 + 2

y

z

x

(6)

In Cylindrical Coordinates:

2V =

1 V 1 2V 2V

+

+

2 2 z 2

(7)

In Spherical Coordinates (r, θ, ϕ)

2V =

2V

1 2 V

1

V

1

r

+

sin

+

r 2 sin 2 2

r 2 r r r 2 sin

(8)

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.

Example 1

Find the potential function for the region between the parallel circular disc shown in the

figure. Neglect fringing.

z

Solution

y

x

Since V is not a function of ρ or ϕ, Laplace’s equation reduces to

2V

= 0

z 2

V

V

=A

=0

z z

z

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

V

E=−

aˆ z = -A âz

z

Example 2

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.

Solution

V

=0

y y

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

and

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.

Example 3

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.

Solution

z

V=0

V = 150 V

Y

The potential is a constant with ϕ and z. Then Laplace’s equation reduces to

1 V

= 0

Integrating eqn. (1) once, we get

(1)

V

= A

(2)

Integrate (2) to get

V = Aln(ρ) + B.

(3)

Applying boundary conditions,

0 = Aln(0.001) + B

(4)

150 = Aln0.02 + B

which give

A = 50.1,

(5)

B = 345.9.

Substituting these values in eqn. (3) to get

V = 50.1ln ρ + 345.9 (V)

Now

V

50.1

E = â = −

â

Example 4

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.

Solution

Since V is not a function of or , Laplace’s equation reduces to

2 V

r

r r

(1/r2)(d/dr)(r2dV/dr) = 0

2 V

=A

=0r

r

V

A

= 2

r r

(r2dV/dr) = A

Integrating gives

and the second integration gives

V = (-A/r) + B

The boundary conditions give

0 = (-A/0.1) + B

and

100 = (-A/2) + B

Therefore

A = 10.53 Vm,

and

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)

(V/m)

Name:

Description:

…