3.1 Laplace’s Equation
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Transcript of 3.1 Laplace’s Equation
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3.1 Laplace’s EquationCommon situation: Conductors in the system, which are a at given potential V orwhich carry a fixed amount of charge Q.
The surface charge distribution is not known.
We want to know the field in regions, where there is no charge.
Reformulate the problem.
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.0 scoordinate Cartesian
0: '
2
2
2
2
2
2
2
z
V
y
V
x
V
VequationsLaplace
+ Boundary conditions. (e.g. over a surface V=const.)
Important in various branches of physics:gravitation,magnetism,heat transportation,soap bubbles (surface tension) …fluid dynamics
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One dimension
bmxxVdx
Vd
)(
02
2
Boundary conditions:
0)5( and 4)1( VV
.1)1(' and 4)1(or VV
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.any for
)],()([2
1)( theis )(
a
axVaxVxVaveragexV
V has no local minima or maxima.
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Two Dimensions
Partial differential equation.To determine the solution you must fix V on the boundary – boundary condition.
Rubber membraneSoap film
0 2
2
2
2
y
V
x
V
V has no local minima or maxima inside the boundary.
A ball will roll to the boundary and out.
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Three Dimensions
0 2
2
2
2
2
2
z
V
y
V
x
VPartial differential equation.To determine the solution you must fix V on the boundary, which is a surface, – boundary condition.
V has no local minima or maxima inside the boundary.
Earnshaw’s Theorem: A charged particle cannot be held in a stable equilibrium by electrostatic forces alone.
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First Uniqueness Theorem
The solution to Laplace’s equation in some volume V is uniquely determined if V is specified on the boundary surface S.
The potential in a volume V is uniquely determined if a) the charge density in the region, andb) the values of the potential on all boundaries are specified.
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Second Uniqueness Theorem
In a volume surrounded by conductors and containing a specified charge density, the electrical field is uniquely determined if the charge on each conductor is given.
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Image Charges
What is V above the plane?
02 /)()()( dzyxqV
Boundary conditions:
0,
,,for 0
zz
yxV
There is only one solution.
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0 ],[4
1),,(
222222 )()(0
zzyxVdzyx
q
dzyx
q
The region z<0 does not matter. There, V=0.
Image charge
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Induced surface charge:
n
V
0 2/3222 )(2),(
zyx
qdyx
qdQ
Force on q: zF ˆ)2(4
12
0 d
q
Force exerted bythe image charge
Energy: d
qdEW
44
1
2
2
0
20
Different
from W of2 charges!!
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222
1222
0 )2(
)1(
)2(
)1(1
4)(
dkzyxdkzyxr
qrV
k
k
k
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Example 3.2
Find the potential outside the conducting grounded sphere.
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