Electrostatics. Boundary Value Problems

The Electrostatic Boundary value Problems.
Laplace’s and Poisson’s equations



1 Electrostatic Boundary value Problems
In numerous practical scenarios, the charge or potential distribution remains unknown. Conse-
quently, the application of Coulomb’s law or Gauss’s law is not feasible. An efficient approach
for addressing these situations involves utilizing boundary values. These types of problems can
be reformulated into either Laplace’s equation or Poisson’s equation.

We know that (from Gauss’s law),

∇.D = ρ (1)
where D is the electric flux density and ρ is the volume charge density.
This can be written in terms of the electric field as,

∇.(εE) = ρ (2)

since D = εE.
i.e.,
ρ
∇.E = (3)
ε
Now, the electric field can be written as the negative gradient of a potential

E = −∇V (4)

So we can write
ρ
∇.(−∇V) = (5)
ε
or
ρ
∇2 V = − (6)
ε
This is called Poisson’s equation. The operator ∇2 is called the Lapalacian operator.
If there are no volume charges , the RHS of the above equation becomes zero, and we obtain

∇2 V = 0 (7)

This is called Laplace’s equation.

1

, The expressions for Laplacian operator in Cartesian , Cylindrical and Spherical polar co-
ordinates system are given below.


In Cartesian co-ordinates,

∂ 2V ∂ 2V ∂ 2V
∇2 V = + 2+ 2
∂ x2 ∂y ∂z
In cylindrial co-ordinates,

1 ∂ 2V ∂ 2V
 
2 1 ∂ ∂V
∇ V= ρ + 2 +
ρ ∂ρ ∂ρ ρ ∂ φ 2 ∂ z2

In spherical polar co-ordinates,

1 ∂ 2V
   
2 1 ∂ 2∂V 1 ∂ ∂V
∇ V= 2 r + 2 sinθ + 2 2
r ∂r ∂r r sinθ ∂ θ ∂θ r sin θ ∂ φ 2



2 The uniqueness of the solution obtained by the Boundary
value method
It is worth considering whether the solution derived from solving Laplace’s or Poisson’s equa-
tions under specified boundary conditions is unique. The answer to this inquiry is affirmative.
This principle is commonly referred to as the uniqueness theorem. A straightforward proof
can be established using the method of contradiction. To illustrate this, let us assume the ex-
istence of two solutions, V1 and V2 , exist for the Lapalce’s equation subjected to the boundary
conditions.
Then,
∇2 V1 = 0 (8)
and
∇2 V2 = 0 (9)
If we define Vd = V2 −V1 , then evidently,

∇2 Vd = ∇2 V2 − ∇2 V1 = 0 (10)

Since the potential cannot have two values on the boundary, we have,

Vd = 0 (11)

on the boundary.
Now we know that Z I
∇.Adv = A.dS (12)
v S

2

, Let A = Vd ∇Vd .
Then,
∇.A = ∇.(Vd ∇Vd ) = Vd ∇2 Vd + ∇Vd ∇Vd (13)
The first term on RHS vanishes, since ∇2Vd = 0.
Thus,
∇.A = ∇Vd . ∇Vd (14)
Therefore, Z Z I
∇.Adv = ∇Vd . ∇Vd dv = (Vd ∇Vd .dS) (15)
v v S
The RHS vanishes, since Vd = 0 on the boundary.
Thus, we have Z Z
∇Vd . ∇Vd dv = |∇Vd |2 dv = 0 (16)
v v
i.e.,
∇Vd = 0 (17)
or,
Vd = V2 − V1 (18)
is constant everywhere in v. But since Vd = 0, on the boundary we should have V2 = V1 every-
where.
Thus, we obtained the result, V is unique.



3 Steps to solve Poisson’s or Lapalace’s equation
The following outlines the fundamental steps necessary for solving Poisson’s or Laplace’s
equation:

• In the case where ρv = 0, it is necessary to solve Laplace’s equation. Conversely, if
ρv 6= 0, one must address Poisson’s equation. When V is a function of a single variable, it
can be resolved through direct integration. However, if V depends on multiple variables,
the separation of variables technique may be employed. It is important to note that the
resulting solution is not unique, as the integration constants must be determined from
the boundary conditions.

• Apply the required boundary conditions. This makes the solution unique.

• Once V is obtained, we can calculate E using E = ∇V and D using D = εE.
R
• We can find the charges induced on the conductor using A = ρS dS. Note that ρS = Dn ,
S
the normal component of D.



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