This document discusses Laplace's equation, Poisson's equation, and the uniqueness theorem. It begins by introducing Laplace's equation and Poisson's equation, which are derived from Gauss's law. Poisson's equation applies to problems with a non-zero charge density, while Laplace's equation applies when the charge density is zero. The uniqueness theorem states that for the potential solution to be unique, it must satisfy Laplace's equation and the known boundary conditions. Several examples are then provided to demonstrate solving Laplace's and Poisson's equations for different boundary value problems.

1. 1
LAPLACE’S EQUATION, POISSON’S
EQUATION AND UNIQUENESS
THEOREM
CHAPTER 6
6.1 LAPLACE’S AND POISSON’S EQUATIONS
6.2 UNIQUENESS THEOREM
6.3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE
6. 4 SOLUTION FOR POISSON’S EQUATION

2. 6.0 LAPLACE’S AND POISSON’S EQUATIONS
AND UNIQUENESS THEOREM
- In realistic electrostatic problems, one seldom knows the charge
distribution – thus all the solution methods introduced up to this
point have a limited use.
- These solution methods will not require the knowledge of the
distribution of charge.

3. 6.1 LAPLACE’S AND POISSON’S EQUATIONS
To derive Laplace’s and Poisson’s equations , we start with Gauss’s
law in point form :
vED ρε =•∇=•∇
VE −∇=Use gradient concept :
( )[ ]
ε
ρ
ρε
v
v
V
V
−=∇•∇
=∇−•∇
2
∇=∇•∇Operator :
Hence :
(1)
(2)
(3)
(4)
(5) => Poisson’s equation
is called Poisson’s equation applies to a homogeneous media.
22
/ mVV v
ε
ρ
−=∇

4. 0=vρWhen the free charge density
=> Laplace’s equation(6)
22
/0 mVV =∇
2
2
2
2
2
2
2
z
V
y
V
x
V
V
∂
∂
+
∂
∂
+
∂
∂
=∇
In rectangular coordinate :

5. 6.2 UNIQUENESS THEOREM
Uniqueness theorem states that for a V solution of a particular
electrostatic problem to be unique, it must satisfy two criterion :
(i) Laplace’s equation
(ii) Potential on the boundaries
Example : In a problem containing two infinite and parallel conductors,
one conductor in z = 0 plane at V = 0 Volt and the other in the z = d
plane at V = V0 Volt, we will see later that the V field solution between
the conductors is V = V0z / d Volt.
This solution will satisfy Laplace’s equation and the known boundary
potentials at z = 0 and z = d.
Now, the V field solution V = V0(z + 1) / d will satisfy Laplace’s equation
but will not give the known boundary potentials and thus is not a
solution of our particular electrostatic problem.
Thus, V = V0z / d Volt is the only solution (UNIQUE SOLUTION) of our
particular problem.

6. 6.3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE
Ex.6.1: Two infinite and parallel conducting planes are separated d
meter, with one of the conductor in the z = 0 plane at V = 0 Volt and the
other in the z = d plane at V = V0 Volt. Assume and
between the conductors.
0
2εε =0=v
ρ
Find : (a) V in the range 0 < z < d ; (b) between the conductors ;
(c) between the conductors ; (d) Dn on the conductors ; (e) on
the conductors ; (f) capacitance per square meter.
E
D sρ
Solution :
0=v
ρSince and the problem is in rectangular form, thus
02
2
2
2
2
2
2
=
∂
∂
+
∂
∂
+
∂
∂
=∇
z
V
y
V
x
V
V (1)
(a)

7. 0
0
2
2
2
2
2
2
=





=
=
∂
∂
=∇
dz
V
dz
d
dz
Vd
z
V
V
We note that V
will be a function
of z only V = V(z) ;
thus :
BAzV
A
dz
dV
+=
=
Integrating twice :
where A and B are constants
and must be evaluated using
given potential values at the
boundaries :
00
===
BV z
dVA
VAdV dz
/0
0
=→
===
(2)
(3)
(4)
(5)
(6)
(7)

8. )(0
Vz
d
V
V =∴
Substitute (6) and (7) into general equation (5) :
dz <<0
)/(ˆˆ
ˆˆˆ
0
mV
d
V
z
z
V
z
z
V
z
y
V
y
x
V
xVE
−=
∂
∂
−=






∂
∂
+
∂
∂
+
∂
∂
−=−∇=
(b)
)/(
2
ˆ 200
mC
d
V
zED
ε
ε −==
(c)

9. )/(
2
)ˆ(
2
ˆˆ
2
ˆ
2
ˆˆ
200
00
00
00
0
mC
d
V
z
d
V
znD
d
V
z
d
V
znD
dzs
zs
ε
ε
ρ
ε
ε
ρ
+=
−•−=•=
−=
•−=•=
=
=
(d) Surface charge :
0
/
V
ds
VQC
s
ab
ρ
=
=
)/(
/2
/
2
0
00
0
2
mF
dV
dV
V
mC s
εε
ρ
==
=∴
(e) Capacitance :
z = 0
z = d
V = 0 V
V = V0 V

10. Ex.6.2: Two infinite length, concentric and conducting cylinders of
radii a and b are located on the z axis. If the region between cylinders
are charged free and , V = V0 (V) at a, V = 0 (V) at b and b > a.
Find the capacitance per meter length.
03εε =
Solution : Use Laplace’s equation in cylindrical coordinate :
and V = f(r) only :
0
11
2
2
2
2
2
2
=
∂
∂
+
∂
∂
+





∂
∂
∂
∂
=∇
z
VV
rr
V
r
rr
V
φ

11. BrAV
r
A
r
V
A
r
V
r
r
V
r
r
r
V
r
rr
V
+=
=
∂
∂
=





∂
∂
=





∂
∂
∂
∂
=





∂
∂
∂
∂
=∇
ln
0
0
12
and V = f(r) only :
(1)

12. BrAV += ln
BbAV
BaAVV
br
ar
+==
+==
=
=
ln0
ln0
Boundary condition :
( ) ( )ba
bV
B
ba
V
A
/ln
ln
;
/ln
00 −
==
Solving for A and B :
( )
( )ab
rbV
V
/ln
/ln0
=∴
Substitute A and B in (1) :
(1)
bra <<;

13. ( )
( )
r
abr
V
ED
r
abr
V
r
V
rVE
ˆ
/ln
ˆ
/ln
ˆ
0
0
ε
ε ==
=
∂
∂
−=−∇=
( )
( )ab
rbV
V
/ln
/ln0
=∴
( )
( )abb
V
rD
aba
V
rD
brs
ars
/ln
ˆ
/ln
ˆ
0
0
ε
ρ
ε
ρ
−=−⋅=
=⋅=
=
=
Surface charge densities:
( )
( )
( )
( )ab
V
b
ab
V
a
brsbr
arsar
/ln
2
2
/ln
2
2
0
0
πε
πρρ
πε
πρρ
−==
==
==
==


Line charge densities :

14. oab V
d
V
Q
C
ρ
==
Capacitance per unit length:
( )
)/(
/ln
2
/
0
mF
abV
mC
περ
== 

15. 6/and0 πφφ ==
0=φ
VV 100= 6/πφ =
EV and
Ex.6.3: Two infinite conductors form a wedge located at
is as shown in the figure below. If this region is
characterized by charged free. Find . Assume V = 0 V at
and at .
z
x φ = 0
φ = π/6
V = 100V

16. Solution : V = f ( φ ) in cylindrical coordinate :
0
1
2
2
2
2
==∇
φd
Vd
r
V
BAV
A
d
dV
d
Vd
+=
=
=
φ
φ
φ
02
2
π
ππφ
φ
/600
)6/(100
0
6/
0
=
==
==
=
=
A
AV
BV
Boundary condition :
Hence :
φ
π
φ
φ
ˆ600
ˆ1
r
d
dV
r
VE
−=
−=−∇=
φ
π
600
=V
6/0 πφ ≤≤
for region :

17. ( ) BAV
A
d
dV
A
d
dV
d
dV
d
d
d
dV
d
d
r
V
+=
=
=
=



=



=∇
2/tanln
sin
sin
0sin
0sin
sin
1
2
2
θ
θθ
θ
θ
θ
θ
θ
θ
θ
θθ
θ = π/10
θ = π/6
V = 50 V
x
y
z
6/and10/ πθπθ ==
E
Ex.6.4: Two infinite concentric conducting cone located at
10/πθ =. The potential V = 0 V at
6/πθ =and V = 50 V at . Find V and between the two
conductors.
Solution : V = f ( ) in spherical coordinate :θ
( )2/tanln
sin
θ
θ
θ
=∫
d
Using :

18. ( ) BAV += 2/tanln θ
( )
( ) BAV
BAV
+==
+==
=
=
12/tanln50
20/tanln0
6/
10/
π
π
πθ
πθ
Boundary condition :
Solving for A and B :






−=






=
20/tan
12/tan
ln
20/tanln50
;
20/tan
12/tan
ln
50
π
π
π
π
π
BA






=






=
1584.0
2/tan
ln1.95
20/tan
2/tan
ln
20/tan
12/tan
ln
50
θ
π
θ
π
π
V
θ
θ
θ
θ
ˆ
sin
1.95
ˆ1
r
d
dV
r
VE
−=
=−∇=
6/10/ θθθ ≤≤Hence at region :
and

19. 6. 4 SOLUTION FOR POISSON’S EQUATION
0=vρWhen the free charge density
Ex.6.5: Two infinite and parallel conducting planes are separated d
meter, with one of the conductor in the x = 0 plane at V = 0 Volt and the
other in the x = d plane at V = V0 Volt. Assume and
between the conductors.
04εε =0=v
ρ
Find : (a) V in the range 0 < x < d ; (b) between the conductorsE
Solution :
BAx
x
V
Ax
dx
dV
dx
Vd
V v
++−=
+−=
−=
−=∇
2
2
0
0
0
2
2
2
ε
ρ
ε
ρ
ε
ρ
ε
ρV = f(x) :

20. 2
2
0
00
2
0
0
0
d
d
V
A
Ad
d
VV
BV
dx
x
ε
ρ
ε
ρ
+=
+−==
==
=
=
Boundary condition :
BAx
x
V ++−=
2
2
0
ε
ρ
dx ≤≤0In region :
( ) x
d
V
xd
x
V 00
2
+−=
ε
ρ
xx
d
d
V
x
dx
dV
E
ˆ
2
ˆ
00












−+−=
−=
ε
ρ
;

21. xrv +== 1and0 ερEx.6.6: Repeat Ex.6.5 with
( )
( )( )
( )
BxAV
x
A
dx
dV
A
dx
dV
x
Vx
dx
d
Ex
dx
d
E
D v
++−=
+
=





−
=





−+
=∇−+
=+
=•∇
=•∇
)1ln(
1
1
01
01
0
0ε
ε
ρSolution :

22. )1ln(
)1ln(
0
0
0
0
d
V
A
dAVV
BV
dx
x
+
−=→
+−==
==
=
=
Boundary condition :
x
dx
V
x
dx
dV
E
d
x
VV
ˆ
)1ln()1(
ˆ
)1ln(
)1ln(
0
0
++
−=−=
+
+
=
dx ≤≤0In region :
BxAV ++−= )1ln(