3.1 Laplaces Equation Common situation: Conductors in the system,
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Transcript of 3.1 Laplaces Equation Common situation: Conductors in the system,
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.
.0 scoordinate Cartesian
0: '
2
2
2
2
2
2
2
zV
yV
xV
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
One dimension
bmxxVdx
Vd
)(
02
2
Boundary conditions:
0)5( and 4)1( VV
.1)1(' and 4)1(or VV
.any for
)],()([21)( theis )(
a
axVaxVxVaveragexV
V has no local minima or maxima.
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
yV
xV
V has no local minima or maxima inside the boundary.
A ball will roll to the boundary and out.
Three Dimensions
0 2
2
2
2
2
2
zV
yV
xV
Partial 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.
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.
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.
Image Charges
What is V above the plane?
02 /)()()( dzyxqV
Boundary conditions:
0,
,,for 0
zz
yxV
There is only one solution.
0 ],[4
1),,( 222222 )()(0
zzyxVdzyx
q
dzyx
q
The region z<0 does not matter. There, V=0.
Image charge
Induced surface charge:
nV
0 2/3222 )(2),(
zyxqdyx
qdQ
Force on q: zF ˆ)2(4
12
0 dq
Force exerted bythe image charge
Energy: d
qdEW44
12
2
0
20
Different
from W of2 charges!!
222
1222
0 )2()1(
)2()1(1
4)(
dkzyxdkzyxrqrV
k
k
k
Example 3.2
Find the potential outside the conducting grounded sphere.