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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 : Use gradient concept : Operator : Hence : (1) (2) (3) (4) (5) => Poisson’s equation is called Poisson’s equation applies to a homogeneous media.
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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 EQUATIONSTo 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
zV
yV
xVV
In rectangular coordinate :
6.2 UNIQUENESS THEOREMUniqueness 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 VARIABLEEx.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.
ED s
Solution :0vSince and the problem is in rectangular form, thus
02
2
2
2
2
22
zV
yV
xVV (1)
(a)
0
0
2
2
2
2
22
dzV
dzd
dzVd
zVVWe note that V
will be a function of z only V = V(z) ; thus :
BAzV
AdzdV
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 VzdVV
Substitute (6) and (7) into general equation (5) :
dz 0
)/(ˆˆ
ˆˆˆ
0 mVdVz
zVz
zVz
yVy
xVxVE
(b)
)/(2ˆ 200 mCdVzED
(c)
)/(2
)ˆ(2ˆˆ
2
ˆ2ˆˆ
200
00
00
000
mCdV
zdVznD
dV
zdVznD
dzs
zs
(d) Surface charge :
0
/
Vds
VQC
s
ab
)/(/2
/
2
0
00
0
2
mFdV
dVV
mC 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 :
0112
2
2
2
22
zVV
rrVr
rrV
BrAVrA
rV
ArVr
rVr
r
rVr
rrV
ln
0
012
and V = f(r) only :
(1)
BrAV ln
BbAV
BaAVV
br
ar
ln0
ln0
Boundary condition :
babV
Bba
VA
/lnln
; /ln
00
Solving for A and B :
ab
rbVV/ln
/ln0
Substitute A and B in (1) :
(1)
bra ;
rabrVED
rabr
VrVrVE
ˆ/ln
ˆ/ln
ˆ
0
0
abrbVV
/ln/ln0
abbVrD
abaVrD
brs
ars
/lnˆ
/lnˆ
0
0
Surface charge densities:
abVb
abVa
brsbr
arsar
/ln22
/ln22
0
0
Line charge densities :
oab Vd
VQC
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 :
012
2
22
dVd
rV
BAV
AddVdVd
02
2
/600
)6/(100
0
6/
0
A
AV
BVBoundary condition :
Hence :
ˆ600
ˆ1
r
ddVr
VE
600V
6/0 for region :
BAV
AddV
AddVddV
dd
ddV
dd
rV
2/tanlnsin
sin
0sin
0sinsin1
22
= /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/tanln
20/tanln50 ;
20/tan12/tanln
50
BA
1584.02/tanln1.95
20/tan2/tanln
20/tan12/tanln
50
V
ˆsin
1.95
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 :
BAxxV
AxdxdVdxVd
V v
2
20
0
02
2
2
V = f(x) :
2
2
0
00
20
0
0
ddVA
AddVV
BV
dx
x
Boundary condition :
BAxxV 2
20
dx 0In region :
xdVxdxV 00
2
xxddV
xdxdVE
ˆ2
ˆ
00
;
xrv 1 and 0 Ex.6.6: Repeat Ex.6.5 with
BxAVxA
dxdV
AdxdVx
Vxdxd
Exdxd
E
D v
)1ln(1
1
01
01
0
0
Solution :
)1ln(
)1ln(
0
0
0
0
dVA
dAVV
BV
dx
x
Boundary condition :
xdx
VxdxdVE
dxVV
ˆ)1ln()1(
ˆ
)1ln()1ln(
0
0
dx 0In region : BxAV )1ln(
