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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 :
( )[ ]
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
−=∇
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 :
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.
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)
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)
)(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)
)/(
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
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
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)
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 <<;
( )
( )
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 :
oab V
d
V
Q
C
ρ
==
Capacitance per unit length:
( )
)/(
/ln
2
/
0
mF
abV
mC
περ
== 
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
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 :
( ) 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 :
( ) 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
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) :
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
;
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 :
)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(

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Chap6 laplaces and-poissons-equations

  • 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
  • 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 :