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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)
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
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 :
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) :
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 :