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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
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.
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 :
( )[ ]
ερ
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
ερ−=∇
0=vρ When the free charge density
=> Laplace’s equation(6)22 / 0 mVV =∇
2
2
2
2
2
22
z
V
y
V
x
VV
∂∂+
∂∂+
∂∂=∇
In rectangular coordinate :
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.
6.3 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.
02εε =0=vρ
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=vρSince and the problem is in rectangular form, thus
02
2
2
2
2
22 =
∂∂+
∂∂+
∂∂=∇
z
V
y
V
x
VV (1)
(a)
0
0
2
2
2
2
22
=
=
=∂∂=∇
dz
V
dz
d
dz
Vd
z
VV
We note that V will be a function of z only V = V(z) ; thus :
BAzV
Adz
dV
+=
=Integrating twice :
where A and B are constants and must be evaluated using given potential values at the boundaries :
00
===
BVz
dVA
VAdVdz
/0
0
=→==
=
(2)
(3)
(4)
(5)
(6)
(7)
)(0 Vzd
VV =∴
Substitute (6) and (7) into general equation (5) :
dz <<0
)/(ˆˆ
ˆˆˆ
0 mVd
Vz
z
Vz
z
Vz
y
Vy
x
VxVE
−=∂∂−=
∂∂+
∂∂+
∂∂−=−∇=(b)
)/(2
ˆ 200 mCd
VzED
εε −==(c)
)/(2
)ˆ(2
ˆˆ
2
ˆ2
ˆˆ
200
00
00
000
mCd
V
zd
VznD
d
V
zd
VznD
dzs
zs
ε
ερ
ε
ερ
+=
−•−=•=
−=
•−=•=
=
=
(d) Surface charge :
0
/
V
ds
VQC
s
ab
ρ=
=
)/(/2
/
2
0
00
0
2
mFdV
dV
VmC s
εε
ρ
==
=∴
(e) Capacitance :
z = 0
z = d
V = 0 V
V = V0 V
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 :
011
2
2
2
2
22 =
∂∂+
∂∂+
∂∂
∂∂=∇
z
VV
rr
Vr
rrV
φ
BrAV += ln
BbAV
BaAVV
br
ar
+==
+==
=
=
ln0
ln0
Boundary condition :
( ) ( )babV
Bba
VA
/ln
ln;
/ln00 −
==
Solving for A and B :
( )( )ab
rbVV
/ln
/ln0=∴
Substitute A and B in (1) :
(1)
bra <<;
( )
( ) rabr
VED
rabr
V
r
VrVE
ˆ/ln
ˆ/ln
ˆ
0
0
εε ==
=∂∂−=−∇=
( )( )ab
rbVV
/ln
/ln0=∴
( )
( )abb
VrD
aba
VrD
brs
ars
/lnˆ
/lnˆ
0
0
ερ
ερ
−=−⋅=
=⋅=
=
=
Surface charge densities:
( ) ( )( ) ( )ab
Vb
ab
Va
brsbr
arsar
/ln
22
/ln
22
0
0
πεπρρ
πεπρρ
−==
==
==
==
Line charge densities :
6/ and 0 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
Solution : V = f ( φ ) in cylindrical coordinate :
01
2
2
2
2 ==∇φdVd
rV
BAV
Ad
dV
d
Vd
+=
=
=
φφ
φ0
2
2
π
ππφ
φ
/600
)6/(100
0
6/
0
=
==
==
=
=
A
AV
BV
Boundary condition :
Hence :
φπ
φφ
ˆ600
ˆ1
r
d
dV
rVE
−=
−=−∇=
φπ
600=V
6/0 πφ ≤≤for region :
( ) BAV
A
d
dV
Ad
dV
d
dV
d
d
d
dV
d
d
rV
+=
=
=
=
=
=∇
2/tanlnsin
sin
0sin
0sinsin
12
2
θθθ
θθ
θθ
θ
θθ
θθ
θ = π/10
θ = π/6
V = 50 V
xy
z
6/ and 10/ 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/tanlnsin
θθ
θ =∫d
Using :
( ) BAV += 2/tanln θ( )( ) BAV
BAV
+==
+==
=
=
12/tanln50
20/tanln0
6/
10/
π
π
πθ
πθ
Boundary condition :
Solving for A and B :
−=
=
20/tan12/tan
ln
20/tanln50 ;
20/tan12/tan
ln
50
ππ
π
ππ
BA
=
=
1584.0
2/tanln1.95
20/tan
2/tanln
20/tan
12/tanln
50
θ
πθ
ππ
V
θθ
θθˆ
sin1.95
ˆ1
r
ddVr
VE
−=
=−∇=
6/10/ θθθ ≤≤Hence at region :
and
6. 4 SOLUTION FOR POISSON’S EQUATION
0=vρ When 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=vρ
Find : (a) V in the range 0 < x < d ; (b) between the conductors E
Solution :
BAxx
V
Axdx
dVdx
Vd
V v
++−=
+−=
−=
−=∇
2
20
0
02
2
2
ερ
ερερερV = f(x) :
2
2
0
00
2
00
0
d
d
VA
Add
VV
BV
dx
x
ερερ
+=
+−==
==
=
=
Boundary condition :
BAxx
V ++−=2
20
ερ
dx ≤≤0In region :
( ) xd
Vxd
xV 00
2+−=
ερ
xxd
d
V
xdx
dVE
ˆ2
ˆ
00
−+−=
−=
ερ
;
xrv +== 1 and 0 ερEx.6.6: Repeat Ex.6.5 with
( )
( )( )
( )
BxAV
x
A
dx
dV
Adx
dVx
Vxdx
d
Exdx
d
E
D v
++−=+
=
−
=
−+
=∇−+
=+
=•∇=•∇
)1ln(
1
1
01
01
0
0ε
ερSolution :