laplace’s equation, poisson’s equation and uniqueness theorem
DESCRIPTION
CHAPTER 6. LAPLACE’S EQUATION, POISSON’S EQUATION AND UNIQUENESS THEOREM. 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. - PowerPoint PPT PresentationTRANSCRIPT
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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
VE Use gradient concept :
v
v
V
V
2Operator :
Hence :
(1)
(2)
(3)
(4)
(5) => Poisson’s equation
is called Poisson’s equation applies to a homogeneous media.
22 / mVV v
0v 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.
02 0v
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 s
Solution :
0vSince 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
(c)
)/(2
)ˆ(2
ˆˆ
2
ˆ2
ˆˆ
200
00
00
000
mCd
V
zd
VznD
d
V
zd
VznD
dzs
zs
(d) Surface charge :
0
/
V
ds
VQC
s
ab
)/(/2
/
2
0
00
0
2
mFdV
dV
VmC s
(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.
03
Solution : Use Laplace’s equation in cylindrical coordinate :
and V = f(r) only :
011
2
2
2
2
22
z
VV
rr
Vr
rrV
BrAVr
A
r
V
Ar
Vr
r
Vr
r
r
Vr
rrV
ln
0
012
and V = f(r) only :
(1)
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
ab
rbVV
/ln
/ln0
abb
VrD
aba
VrD
brs
ars
/lnˆ
/lnˆ
0
0
Surface charge densities:
abV
b
ab
Va
brsbr
arsar
/ln
22
/ln
22
0
0
Line charge densities :
oab V
d
V
QC
Capacitance per unit length:
)/(/ln
2/
0
mFabV
mC
6/ and 0 0
VV 100 6/ 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 = 0
= /6
V = 100V
Solution : V = f ( ) in cylindrical coordinate :
01
2
2
2
2 dVd
rV
BAV
Ad
dV
d
Vd
0
2
2
/600
)6/(100
0
6/
0
A
AV
BV
Boundary condition :
Hence :
ˆ600
ˆ1
r
d
dV
rVE
600V
6/0 for region :
BAV
A
d
dV
Ad
dV
d
dV
d
d
d
dV
d
d
rV
2/tanlnsin
sin
0sin
0sinsin
12
2
= /10
= /6
V = 50 V
xy
z
6/ and 10/ E
Ex.6.4: Two infinite concentric conducting cone located at
10/ . The potential V = 0 V at
6/ and V = 50 V at . Find V and between the two conductors.
Solution : V = f ( ) in spherical coordinate :
2/tanlnsin
d
Using :
BAV 2/tanln BAV
BAV
12/tanln50
20/tanln0
6/
10/
Boundary condition :
Solving for A and B :
20/tan12/tan
ln
20/tanln50 ;
20/tan12/tan
ln
50
BA
1584.0
2/tanln1.95
20/tan
2/tanln
20/tan
12/tanln
50
V
ˆ
sin1.95
ˆ1
r
ddVr
VE
6/10/ Hence at region :
and
6. 4 SOLUTION FOR POISSON’S EQUATION
0v 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.
04 0v
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
V = f(x) :
2
2
0
00
2
00
0
d
d
VA
Add
VV
BV
dx
x
Boundary condition :
BAxx
V 2
20
dx 0In region :
xd
Vxd
xV 00
2
xxd
d
V
xdx
dVE
ˆ2
ˆ
00
;
xrv 1 and 0 Ex.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
0
Solution :
)1ln(
)1ln(
0
0
0
0
d
VA
dAVV
BV
dx
x
Boundary condition :
xdx
Vx
dx
dVE
d
xVV
ˆ)1ln()1(
ˆ
)1ln(
)1ln(
0
0
dx 0In region :BxAV )1ln(