WEBVTT  generated by VCS
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Good day, ladies and gentlemen.
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I warmly welcome you to another contribution
in the Electromagnetic Theory series.
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My name is Hans Georg Krauthäuser, I am the
holder of the professorship
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for Electromagnetic Theory and Compatibility
at TU Dresden.
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We've got electrostatics ourselves in the last
few events
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and are there until the formal solution of
the problem with the help of the
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Greenschen function, thus penetrating the fundamental
solutions.
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Today we are going to get to know a very extremely
important method for the
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Green's function can actually be found with
given boundary conditions .
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Because it depends of course, the formal solution
should only be used when
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you also know the Greensche function.
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The method we're going to go through today
is the socalled
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Mirroring method or image loading method.
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In principle, it is based directly on what
we used to derive the formal solution
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have already seen.
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The basic problem is therefore the solution
to the Poisson equation, but now with
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Boundary conditions in the finite.
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We saw in the employment of the Green function
that this
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Superposition principle is really fundamental.
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Because we can get out of the solution for
a point charge  and nothing else
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is ultimately the Greensche function  immediately
also solutions for any, not
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Open up deltashaped inhomogeneities.
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In the form we have already seen several times:
as an integral
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via the Greensche function multiplied by the
volume charge density rho_v.
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If we have this, then this part of the potential
is ultimately  as you can see here 
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completely determined by the solutions in the
solution volume.
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There are no interfaces yet.
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In order for the overall solution to meet
the intended boundary conditions, it must
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Green's function of open space  that is ultimately
nothing else than that
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Potential of a point charge, where we now form
the charge as one
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that is exactly what Green’s function is.
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We have to change that.
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An additional term is added here, the solution
of the Laplace equation
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so the associated homogeneous equation must
be.
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And it's clear when I find a solution to Laplace's
equation  the homogeneous equation
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add to G, then G will still be a solution
to Poisson's equation.
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This follows from the linearity of the Laplace
operator and only for linear ones
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Differential operators choose Green's formalism.
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The idea now is as follows: So if I use the
greens function of open space
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know and now introduce boundary conditions,
then these boundary conditions will be included
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lead to the fact that the field of point charges
 here in the case or this Greensche function

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is no longer the Green function of the changed
problem.
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Because the boundary conditions are no longer
met.
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We will actually see this again in more detail
using the example .
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Now of course I cannot place any additional
charges in the solution volume,
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because with that I would have a completely
different problem.
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But I have the option of modeling the influence
of the edge by using
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introduce charges outside of the solution volume
that are not there at all, I
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only take as model charges.
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And do it so that the original load plus the
model loads that
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associated field or the
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associated potential then
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the boundary conditions are met.
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If that's the case, then because of the uniqueness
, I'm actually done.
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I then found a potential that is the solution
in V and at the same time
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the specified boundary conditions are met.
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And then we already know: if we have found
that, then it is also clear,
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until it then goes to any additive constants
that give us
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no further interest.
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These charges, which I place outside of the
solution volume, have none
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direct influence, but of course they influence
the solution, otherwise we would
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don't do the whole thing.
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You can imagine it that way, that the surface
itself has an influence
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on the solution.
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Because, on a surface, charges are present
in the volume itself
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Surface charges influenced.
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And we try to investigate the influence of
these influenced surface charges
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simulate modeling by going outside the solution
volume instead
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Place charges.
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I think this becomes clear best if you look
at a specific example
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performs. Of course we start with a particularly
simple one, but also definitely
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with an important example.
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Namely, we consider a single point charge
Q that precedes a
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should be infinitely extensive plane.
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We assume that this level is also arbitrarily
good conductive, that is
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so a perfect ladder extended infinitely.
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Regarding the nomenclature: I already said
that we have charge Q, which is in one
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Distance a to the surface and if we now observe
this at a point r,
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then of course we have different sizes that
we look at.
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We have the position vector r, which is in
cylindrical coordinates and that is
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a cheap choice of coordinates, of course.
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In cylindrical coordinates it will have the
zcomponent on the one hand and on the other
hand
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a rho component, the phi component didn't
play any here at first
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Role, because the whole thing can also be
thought of as rotationally symmetrical
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the zaxis around. So, coordinate rho, coordinate
z are clear and as always at
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Of course, the distance between the cargo
and the load also plays a role in such arrangements
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Observation point a role.
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This is the vector R here and of course that
is nothing other than r minus r 'and
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Here r 'would be a times the unit vector in
the zdirection.
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The length of R which is obviously rho square
root of rho square
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plus z minus a square.
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z minus a is the piece in front here.
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Yes, a normal vector is already written on
it, that's because we
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now the boundary conditions will also be brought
into play.
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As always, we will arrange the normal vector
so that it comes out of the volume
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shows and we used to designate the materials
in such a way that one
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Side  this side here  is one and this side
into the n
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shows, then side is two.
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The boundary conditions for the electric field
or
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I wrote down the dielectric shift here again.
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It then applies to the normal components that
D_2 normal minus D_1 normal just that
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The surface charge density is rho_F and it
applies to the tangential component
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continuous E is tangential 2 minus E tangential
1 is equal to 0.
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If we now assume a vacuum here, then D 1 would
of course be epsilon_1 times E_1
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We will then use that right away and for
the metal, what is on this page
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is, we already know that there will be no
field on this side, that is
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both E_2 and also D_2 will be 0, so this equation
works accordingly here
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simplify. E tangentially equals 0, which is
what it says here, in particular, means that
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the Efield is perpendicular to this surface
of this perfect conductor.
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We have already derived that in the past and
have already done so
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used, so not surprising.
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Well, that is the arrangement and we can now
briefly recollect it
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shout what's going to happen now.
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If I didn't have this border here, then I
would have radial around Q here
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symmetrical electric field, with corresponding
circular potential lines.
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I won't have that here, but this electrical
charge that is here in front of the
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conductive wall is, of course, will have an
impact on the charge carriers, which yes
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are freely movable in the perfect ladder.
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Specifically, this means, let's assume that
there is a positive charge here, then
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negative charge will accumulate here and it
is also clear that
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it then drops and the further we are from
this point rho equal to 0,
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the less influence will of course be.
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This surface charge that results here is of
course on again
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Field  a potential  connected and what ultimately
interests us is that
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Overlay here in the solution volume.
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If we wanted to solve this directly, it would
lead to an integral equation for the
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Surface charge density and that is relatively
complicated.
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With such simple geometries, however, we can
very quickly consider how
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the problem is easier to solve.
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But we draw it a little differently.
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So on the one hand we still have our solution
volume with us
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our cargo. And I have now indicated what the
field would look like
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if this conductive plane weren't there.
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Then we would have this radially symmetrical
field and it is already
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of course, if you look at that, it would of
course be an Efield, the one up here
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and is of course not automatically perpendicular
to the surface down here .
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In the middle it cuts it, but outside the middle
we become
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always have a certain angle.
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But we are not allowed to have that.
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This tangential component here must disappear
because the tangential component
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is continuous, and we know it's 0 on one side,
so it must be on the
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on the other side also be 0.
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The tangential component would be a small
component here
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is directed from the bottom up.
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So we can only compensate for it by adding
another field
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have the one corresponding
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Has tangential component from top to bottom.
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As one can easily imagine, this will be fulfilled
when this is this one
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small, this blue arrow would be, then namely
the superposition of these two
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Field this brown arrow, which is now longer
overall, but vertical
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stands on the surface.
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It is easy to think that if I do it like that,
then it will also work in the
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Center. In the middle it will be twice as
strong, but also stand upright and
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down here, of course, it also works with the
corresponding construction instructions.
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Again with a larger resulting field, but above
all
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with vanishing tangential components in the
resulting field strength.
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Yes, and now of course it's easy to say: Okay,
where does this field come from?
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And if I trace that back, of course , I easily
come to the conclusion that I am here
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can introduce an image charge.
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And if I introduce this image load without
this layer being there now, then
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would be the tangential components in the
combination of the charge and the image charge
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along this surface  which is now for the
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Moment is no longer there  actually gone.
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For this it has to be a real image load and
that means the distance on the
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one side and the distance on the other side,
must be the same.
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And if I have a positive charge here, then
I need one on the side
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negative charge of the same size.
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Otherwise it will not work with the compensation
of the tangential component.
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By the way , we already know the picture that
comes out of it, of course, we have it
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Already looked at the dipole, here again called
to mind: If we
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draw a dipole, then exactly this condition
is here at the center line,
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that the Efield is always perpendicular to
the line is fulfilled.
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That is, the dipole does just that.
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From this point of view, for this simple example,
we can already say: One half
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this dipole field, that will also be the solution
and we calculate exactly that now
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but actually from one more time.
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Boundary conditions are very fundamental, I
have Etangential when z equals 0 is 0 resp.
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phi when z equals 0 is 0 because this is a
perfect conductor and it is grounded
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is, then I have potential 0.
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I specified that.
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I can now use the superposition of the two
potentials, i.e. one of them
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write down the original charge and once the
picture charge directly.
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Geometrically, it's easy to make clear: this
was the capital R that we did before
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had introduced.
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And the distance from the image charge to the
observation point is simply the root
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from rho square plus z plus a to the square.
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If I am here now, of course, the potential
in the appropriate way
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00:15:56.000 > 00:16:01.000
then of course I can immediately E equal minus
gradient phi that too
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Calculate the Efield. We're doing this in
cylinder coordinates.
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We don't have a phi dependency, so I need
an education for the gradient
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00:16:12.000 > 00:16:15.000
not to worry about it either.
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00:16:15.000 > 00:16:21.000
There is then only one rho and one z component
for the E field, which I now have
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00:16:21.000 > 00:16:23.000
just wrote down here.
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00:16:23.000 > 00:16:28.000
The interesting thing now is the level z equal
to 0, where we look at that again
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00:16:28.000 > 00:16:30.000
can see how is the Efield at z equal to 0.
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Now you simply set z equal to 0 everywhere,
and you find that the rho component
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disappears and actually just a zcomponent
remains, which is here
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00:16:43.000 > 00:16:46.000
is written accordingly.
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If I only have one z component, that's exactly
what I wanted.
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I obviously have no tangential component of
the electrical
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Field, but just a normal component.
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If we have the electric field, then we can
actually do a lot now
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00:17:05.020 > 00:17:09.020
quickly calculate the surface charge density
because we have that
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Continuity component for the normal component
of the dielectric displacement.
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And in that case I can just write it as epsilon_0
times E.
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That is only the one side that we called the
second side.
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On the second page is the E field or
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the dielectric shift is 0 anyway, so that
only this one term remains.
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The normal component  the normal vector was
directed outwards  so it is
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Normal component just the negative z component,
so this sign
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then just disappears here.
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And this discontinuity is just given by the
surface charge, so that we
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00:17:52.020 > 00:18:00.080
simply, when we have the field, we get the
rho_F by still having the field
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00:18:00.080 > 00:18:10.080
multiply once with ebsilon_0 at the place
of the surface of course and that gives
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00:18:10.080 > 00:18:16.080
then this expression, which we then wrote down
here again.
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Which, by the way, we have found so much easier
than when we found a corresponding one
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Integral equation should have solved.
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If we know the rho_F, then of course we can
also calculate the total charge of the area
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00:18:34.080 > 00:18:39.080
calculate. We do this to check the plausibility
of the whole thing again.
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For this we have to do the rho integration
from 0 to infinity and the
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phi integration from 0 to 2 pi.
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Since we have no pi dependency, there is of
course just a factor of 2 pi
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who hereby compensates.
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And the rest of the integral is then quickly
solved.
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In fact, if you do that, minus Q comes out
here .
226
00:19:05.040 > 00:19:13.040
That is very plausible, because the minus Q
is also exactly the charge that we call
227
00:19:13.040 > 00:19:18.040
Image charge have started and the overall
charge of the arrangement is allowed to increase
228
00:19:18.040 > 00:19:22.040
of course through this modeling step i did
that i instead of the
229
00:19:22.040 > 00:19:27.040
Surface charge density inserted an image charge
through this
230
00:19:27.040 > 00:19:31.040
Modulation step, of course, the overall charge
must not affect the arrangement
231
00:19:31.040 > 00:19:39.040
change. All in all, the arrangement has no
overall charge, and therefore none
232
00:19:39.040 > 00:19:42.040
Monopoly term in multipole expansion.
233
00:19:42.040 > 00:19:49.040
So this is such a pure dipole, as we already
suspected.
234
00:19:49.040 > 00:19:55.040
I had just shown the picture of the dipole
as the solution, which confirms that
235
00:19:55.040 > 00:19:56.040
just here again.
236
00:19:56.040 > 00:20:01.000
Please note: you always have to be careful
with the image loading method
237
00:20:01.000 > 00:20:06.000
The solution that we have now calculated are
still only solutions in this case
238
00:20:06.000 > 00:20:11.000
just for z greater than 0, i.e. in the original
solution volume.
239
00:20:11.000 > 00:20:17.000
On the other side of the conductive surface,
we also know the solution
240
00:20:17.000 > 00:20:23.000
beautiful. We know that E is equal to 0 and
phi is equal to 0, in other words, there we
have
241
00:20:23.000 > 00:20:26.000
absolutely no reason to calculate the Efield
and potential.
242
00:20:26.000 > 00:20:32.000
Only sometimes is it like this, if you calculate
this in the context of the image loading method,
243
00:20:32.000 > 00:20:36.000
you have almost forgotten the original problem
when you find the solution
244
00:20:36.000 > 00:20:42.000
found and then also writes this Efield for
z less than 0
245
00:20:42.000 > 00:20:46.000
as a solution. Of course you can't do that.
246
00:20:46.000 > 00:20:50.000
We found the potential.
247
00:20:50.000 > 00:20:53.000
I wrote it down here again.
248
00:20:53.000 > 00:20:59.000
And if we have the potential, we know we have
that too, in the end
249
00:20:59.000 > 00:21:01.060
Greensche function found.
250
00:21:01.060 > 00:21:07.060
That means that we can now use this potential
directly to do the Greensche function
251
00:21:07.060 > 00:21:14.060
the halfspace  the space with any highly
conductive surface
252
00:21:14.060 > 00:21:19.060
is complete calculate.
253
00:21:19.060 > 00:21:25.060
As if the halfspace z were larger than 0 and
I just need to write that down now
254
00:21:25.060 > 00:21:29.060
by simply setting Q = 1.
255
00:21:29.060 > 00:21:31.060
Then that is immediately the Green function.
256
00:21:31.060 > 00:21:36.060
And of course we wrote here without coordinates,
but the
257
00:21:36.060 > 00:21:40.060
Stepping from here to here shouldn't be too
difficult.
258
00:21:40.060 > 00:21:48.060
The key point is that I am using the Greenschen
feature that I am at
259
00:21:48.060 > 00:21:55.060
found an example, namely in which I said this
area is
260
00:21:55.060 > 00:22:02.020
grounded and has the potential 0 everywhere
I now also use these greens
261
00:22:02.020 > 00:22:09.020
can take to solve any other boundary value
problems of the same geometry.
262
00:22:09.020 > 00:22:14.020
The same geometry means that this conductive
plane must of course stay where it is.
263
00:22:14.020 > 00:22:17.020
Otherwise I would have a completely different
new problem.
264
00:22:17.020 > 00:22:22.020
We have seen that the boundary conditions
for the Efield are also very
265
00:22:22.020 > 00:22:27.020
were elementary. I took advantage of it, so
it has to continue indefinitely
266
00:22:27.020 > 00:22:36.020
be conductive so that E is perpendicular to
the surface , but we could too
267
00:22:36.020 > 00:22:42.020
any other values for the potential or
for the normal derivatives of the
268
00:22:42.020 > 00:22:52.020
Now accept potential on this surface and then
get out of the formal solution
269
00:22:52.020 > 00:22:56.020
immediately write down the solution for the
potential.
270
00:22:56.020 > 00:22:58.020
Because we now know the Greensche function.
271
00:22:58.020 > 00:23:00.080
And if we know the Green's function , we also
know the normal derivatives
272
00:23:00.080 > 00:23:07.080
the Greenschen feature, which I'll just repeat
here just in case
273
00:23:07.080 > 00:23:08.080
have written down.
274
00:23:08.080 > 00:23:15.080
That means, as soon as now at the places of
the surface, either phi or d phi after dn
275
00:23:15.080 > 00:23:22.080
is given, you can actually use the wellknown
Greensche function and the
276
00:23:22.080 > 00:23:27.080
known normal derivation of the Greenschen function
formally solve this expression here.
277
00:23:27.080 > 00:23:30.080
Formally solving means that everything that
is written here is known.
278
00:23:30.080 > 00:23:36.080
The integrals may then not be analytical in
the last step
279
00:23:36.080 > 00:23:40.080
but I have at least one closed term that I
am
280
00:23:40.080 > 00:23:42.080
in case of doubt can solve numerically.
281
00:23:42.080 > 00:23:48.080
We will be here in this case because that is
when you see this for the first time
282
00:23:48.080 > 00:23:54.080
It is amazing to actually do the math for
a simple case.
283
00:23:54.080 > 00:24:02.040
We now assume that the potential on the surface
is no longer 0
284
00:24:02.040 > 00:24:08.040
should be, as we have assumed so far, but
have a finite value phi_0.
285
00:24:08.040 > 00:24:15.040
What we expect is of course  since the potential
is only up to there anyway
286
00:24:15.040 > 00:24:22.040
a constant is determined  that I get the
same expression as for that
287
00:24:22.040 > 00:24:28.040
Problem where the potential at the surface
was 0 and just the phi_0 as additive
288
00:24:28.040 > 00:24:35.040
Constant added. And we are now trying to understand
that this is actually the case.
289
00:24:35.040 > 00:24:39.040
For the concrete calculation we are going
to use cylindrical coordinates again .
290
00:24:39.040 > 00:24:45.040
That is the familiar expression that we had
just seen and
291
00:24:45.040 > 00:24:52.040
in general  also wrote that down again 
I would have just that now
292
00:24:52.040 > 00:24:55.040
Volume integral with the charge distribution,
the Greens function and the two
293
00:24:55.040 > 00:25:07.000
Surface terms. Specifically, we have Dirichlet
boundary conditions here, that is, we
294
00:25:07.000 > 00:25:12.000
in this case would take a look, where we find
the normal derivatives of the Greens function
295
00:25:12.000 > 00:25:19.000
need. This term would not matter because we
have Dirichlet boundary conditions everywhere
296
00:25:19.000 > 00:25:25.000
have and yes we know that we do not have both
boundary conditions
297
00:25:25.000 > 00:25:26.000
must have a point.
298
00:25:26.000 > 00:25:32.000
That means, we can safely drop this term here,
we need it
299
00:25:32.000 > 00:25:41.000
not to be considered further. The surface integral
for the second term applies
300
00:25:41.000 > 00:25:43.000
ultimately to calculate.
301
00:25:43.000 > 00:25:47.000
But for this we first need the normal derivatives
at the location of the surface,
302
00:25:47.000 > 00:25:55.000
so d G after dn evaluated at the surface of
V, that is by definition
303
00:25:55.000 > 00:26:04.060
Gradient G times the normal vector and in the
special case  the normal vector is yes
304
00:26:04.060 > 00:26:12.060
the vector minus d z so the normal component
of the gradient would now be in that
305
00:26:12.060 > 00:26:19.060
Case the negative zcomponent, i.e. minus
d G evaluated according to dz when z equals
0.
306
00:26:19.060 > 00:26:25.060
The derivation of the Greens function here
with respect to z, of course, happened quickly.
307
00:26:25.060 > 00:26:30.060
I wrote down the intermediate step again ,
you are welcome to do it again
308
00:26:30.060 > 00:26:36.060
comprehend. And if you put in z equal to 0
here , you get first
309
00:26:36.060 > 00:26:44.060
this expression and then the final expression
as it is here in front.
310
00:26:44.060 > 00:26:54.060
Now I know the normal derivatives of the Greens
function.
311
00:26:54.060 > 00:26:58.060
Now all terms are known and I only need to
insert for them
312
00:26:58.060 > 00:27:00.020
Dirichlet boundary conditions.
313
00:27:00.020 > 00:27:06.020
So now the volume integral  that's highlighted
here as a green term  and
314
00:27:06.020 > 00:27:15.020
the surface term that's this integral here
in red and we need it now
315
00:27:15.020 > 00:27:18.020
of course only to use the concrete derivation.
316
00:27:18.020 > 00:27:20.020
We want to work out the red term.
317
00:27:20.020 > 00:27:26.020
Here in the first line I just started.
318
00:27:26.020 > 00:27:31.020
So then you pull the constants forward and
pull them together.
319
00:27:31.020 > 00:27:40.020
With the integral that is still here, you
can see very quickly that it is
320
00:27:40.020 > 00:27:46.020
is appropriate to substitute, for example,
as follows.
321
00:27:46.020 > 00:28:00.080
We put zeta equal to rho square plus a square,
then d zeta is equal to 2_rho d rho.
322
00:28:00.080 > 00:28:06.080
In principle, this is the correct expression
up here.
323
00:28:06.080 > 00:28:10.080
We just have to consider the 2.
324
00:28:10.080 > 00:28:12.080
That gives the factor half up here.
325
00:28:12.080 > 00:28:14.080
We extract the a from the integral.
326
00:28:14.080 > 00:28:20.080
And the integral, what is left  with which
of course the limits are adjusted
327
00:28:20.080 > 00:28:25.080
have to be  this is trivial to solve.
328
00:28:25.080 > 00:28:32.080
And if we do that now, it just ends up being
phi_0.
329
00:28:32.080 > 00:28:37.080
That's exactly what we were expecting.
330
00:28:37.080 > 00:28:44.080
That is, as expected, in that case the total
potential would simply be the potential
331
00:28:44.080 > 00:28:48.080
what we have calculated, in the event that
we have potential 0 on the surface
332
00:28:48.080 > 00:28:52.080
have, plus the potential that we are actually
in
333
00:28:52.080 > 00:28:54.080
the case on the surface.
334
00:28:54.080 > 00:28:56.080
That is very satisfying, the result.
335
00:28:56.080 > 00:29:01.040
This is exactly what we expected and shows
us again that we are
336
00:29:01.040 > 00:29:06.040
Didn't make any obvious mistakes in between
.
337
00:29:06.040 > 00:29:14.040
That was the first example, which is relatively
simple, but actually very
338
00:29:14.040 > 00:29:23.040
is important because conductive surfaces are
actually a very important part of electrical
engineering
339
00:29:23.040 > 00:29:25.040
play a fundamental role.
340
00:29:25.040 > 00:29:34.040
A second example where we can look at the
situation again is that
341
00:29:34.040 > 00:29:35.040
Reflection on the ball.
342
00:29:35.040 > 00:29:40.040
Spheres are also geometries that actually appear
again and again.
343
00:29:40.040 > 00:29:47.040
So now we consider the following geometry as
the second.
344
00:29:47.040 > 00:29:53.040
We have a sphere with a radius R also that
should again be arbitrarily conductive
345
00:29:53.040 > 00:29:56.040
and it too should first be grounded.
346
00:29:56.040 > 00:30:05.000
And at a distance s_1 from the center point
we have a charge Q.
347
00:30:05.000 > 00:30:13.000
We are interested in the potential or field
at observation point P.
348
00:30:13.000 > 00:30:25.000
And this point P is initially at a distance
R_1 from the original charge Q, the
349
00:30:25.000 > 00:30:30.000
we can of course simply specify.
350
00:30:30.000 > 00:30:39.000
One already suspects that one can cope with
reflections again here , but it is
351
00:30:39.000 > 00:30:45.000
not so obvious how big the picture load must
be now.
352
00:30:45.000 > 00:30:51.000
That is why we write something again here
in general that it is minus alpha times Q.
353
00:30:51.000 > 00:30:55.000
The alpha is still unknown and you do n't
know directly either
354
00:30:55.000 > 00:30:57.000
where to put them.
355
00:30:57.000 > 00:31:04.060
And so we say we put it at a distance s_2
from the center.
356
00:31:04.060 > 00:31:13.060
The potential for these two charges, we can
write that down again by
357
00:31:13.060 > 00:31:23.060
we make it clear what the corresponding distances
are.
358
00:31:23.060 > 00:31:33.060
The distance vector R_1 is simply actually
the vector R  and that is yes
359
00:31:33.060 > 00:31:42.060
nothing else than R times e_r minus the vector
r 'and that is the vector s_1 times
360
00:31:42.060 > 00:31:53.060
he'. For the other charge which has the charge
minus alpha times Q accordingly
361
00:31:53.060 > 00:32:02.020
written down. We are now demanding that the
potential be on the surface of the sphere
362
00:32:02.020 > 00:32:11.020
should vanish, i.e. phi of r equals R times
e_r, so when the point is on
363
00:32:11.020 > 00:32:15.020
the surface lies, that should be 0.
364
00:32:15.020 > 00:32:25.020
If you use that and demand that this is 0,
it leads to this relationship here.
365
00:32:25.020 > 00:32:31.020
This relationship we met, you can easily see
that this relationship
366
00:32:31.020 > 00:32:42.020
is fulfilled if s_2 the unknown s_2 is replaced
by the known R and
367
00:32:42.020 > 00:32:48.020
s_1 and if the alpha is again expressed by
the wellknown R and the
368
00:32:48.020 > 00:32:52.020
known s_1 is expressed in this way and whiteness.
369
00:32:52.020 > 00:32:57.020
If you use this accordingly, you will see
that the condition is met.
370
00:32:57.020 > 00:33:06.080
That is, we have determined the two unknown
quantities s_2 and alpha,
371
00:33:06.080 > 00:33:10.080
by considering the boundary conditions for
the potential the given
372
00:33:10.080 > 00:33:13.080
Boundary conditions for the potential  namely
the potential here
373
00:33:13.080 > 00:33:16.080
should be equal to 0.
374
00:33:16.080 > 00:33:23.080
In this way, of course, we can now also use
the Green's function of the sphere
375
00:33:23.080 > 00:33:26.080
determine. We know the potential.
376
00:33:26.080 > 00:33:33.080
And when we know the potential, we already
know that we only need the charge
377
00:33:33.080 > 00:33:42.080
formally to be set as 1 Q is now set to 1
the other sizes are used.
378
00:33:42.080 > 00:33:47.080
That's the way we get that expression for them
379
00:33:47.080 > 00:33:50.080
Green's function of the sphere.
380
00:33:50.080 > 00:33:55.080
You can also write it down differently by
clicking here accordingly
381
00:33:55.080 > 00:34:06.040
the amount and this is how it looks, not as
handy as in the example what
382
00:34:06.040 > 00:34:12.040
we just had, but still expressions that can
be handled.
383
00:34:12.040 > 00:34:16.040
That is obviously symmetrical, that’s another
check.
384
00:34:16.040 > 00:34:24.040
Because we know that the G of r and r 'must
be the same as G of r' and r, so
385
00:34:24.040 > 00:34:27.040
interchanging r and r 'must not play a role.
386
00:34:27.040 > 00:34:31.040
You quickly convince yourself that it is fulfilled
here.
387
00:34:31.040 > 00:34:34.040
The boundary conditions of the potential are
fulfilled.
388
00:34:34.040 > 00:34:41.040
The Poisson equation in the solution volume
is fulfilled and we then already know
389
00:34:41.040 > 00:34:45.040
that we have the uniqueness of the solution
and that means we know at that point
390
00:34:45.040 > 00:34:48.040
already, that we are already finished by then.
391
00:34:48.040 > 00:34:54.040
Again you can get here by solving this one
problem with a point charge
392
00:34:54.040 > 00:35:02.000
of a grounded sphere can actually solve a
whole class of problems by one
393
00:35:02.000 > 00:35:08.000
this Greens function that we have now found
, then just use and then
394
00:35:08.000 > 00:35:13.000
the potential  no matter how the boundary
conditions are given  whether as
395
00:35:13.000 > 00:35:18.000
von Neumann boundary condition or as Dirchlet
boundary condition or as
396
00:35:18.000 > 00:35:20.000
mixed boundary conditions do not matter.
397
00:35:20.000 > 00:35:24.000
Formally, we can calculate this because we
now know the Greens function.
398
00:35:24.000 > 00:35:26.000
Of course, we still have to know the normal
derivatives.
399
00:35:26.000 > 00:35:30.000
In that case it can be calculated quickly.
400
00:35:30.000 > 00:35:35.000
We don't have to go through this term by term
either , I just have it
401
00:35:35.000 > 00:35:36.000
written down again.
402
00:35:36.000 > 00:35:43.000
Well, now you can see how it works and honestly,
it is not
403
00:35:43.000 > 00:35:45.000
always easy to find the greens function.
404
00:35:45.000 > 00:35:52.000
But for certain, simple geometries, you can
actually use this
405
00:35:52.000 > 00:35:58.000
Image loading method  so with this mirroring
method  the Greens function
406
00:35:58.000 > 00:36:02.060
very easy and very elegant to find.
407
00:36:02.060 > 00:36:06.060
We have not yet considered an important case.
408
00:36:06.060 > 00:36:13.060
We have now only ever looked at one surface,
that for reflection
409
00:36:13.060 > 00:36:20.060
is used. But it is often the case that you
have several surfaces.
410
00:36:20.060 > 00:36:26.060
And even in this case you can actually sometimes
use the
411
00:36:26.060 > 00:36:31.060
Mirroring method to get ahead. We'll look
at an example very, very briefly
412
00:36:31.060 > 00:36:41.060
look at. Namely the case  we will see in
a moment  that we two parallel
413
00:36:41.060 > 00:36:45.060
have infinitely good governance levels.
414
00:36:45.060 > 00:36:53.060
Whenever you have several mirror planes or
have several edges, then you have to
415
00:36:53.060 > 00:37:00.020
note that you don't just have to mirror the
original charge on each of the edges,
416
00:37:00.020 > 00:37:05.020
but you actually have to reapply the mirror
charge at the edges
417
00:37:05.020 > 00:37:08.020
mirror on which they have not yet been mirrored.
418
00:37:08.020 > 00:37:14.020
So we have to take into account reflections
of the reflection.
419
00:37:14.020 > 00:37:18.020
It's really like standing between two mirrors
420
00:37:18.020 > 00:37:20.020
two real, optical mirrors stands.
421
00:37:20.020 > 00:37:25.020
Then it is also the case that you see very,
very many reflections and because of yourself
422
00:37:25.020 > 00:37:28.020
reflect the mirror images again.
423
00:37:28.020 > 00:37:29.020
This is actually the case here.
424
00:37:29.020 > 00:37:36.020
This is because we actually only solve this
for certain geometries
425
00:37:36.020 > 00:37:40.020
because that then automatically leads to a
series development.
426
00:37:40.020 > 00:37:46.020
Whether this series then converges or not
, that actually depends on the
427
00:37:46.020 > 00:37:48.020
concrete geometry.
428
00:37:48.020 > 00:37:53.020
A simple example: in which it will also converge,
but we will not
429
00:37:53.020 > 00:37:55.020
write down to the last point.
430
00:37:55.020 > 00:38:06.080
Let's assume that we actually have two conductive
planeparallel edges.
431
00:38:06.080 > 00:38:09.080
And in between we have a load distribution.
432
00:38:09.080 > 00:38:16.080
I've now written them down in triangular shapes,
that’s the same for them
433
00:38:16.080 > 00:38:18.080
To see reflections a little better.
434
00:38:18.080 > 00:38:24.080
What the mirror image of whom is seen a little
better with the triangle than when
435
00:38:24.080 > 00:38:26.080
I would only make one point.
436
00:38:26.080 > 00:38:33.080
What has to be done first, of course, is that
we use this charge distribution to the
437
00:38:33.080 > 00:38:36.080
Mirror the example at this first level.
438
00:38:36.080 > 00:38:44.080
That would result in a corresponding mirror
image, as I have drawn in here.
439
00:38:44.080 > 00:38:50.080
With a corresponding charge distribution with
the opposite sign.
440
00:38:50.080 > 00:38:57.080
Of course, I also have to mirror this at the
right interface and that gives it.
441
00:38:57.080 > 00:39:02.040
So now it goes on.
442
00:39:02.040 > 00:39:09.040
I need the charge that I have now mirrored
at this interface, this one
443
00:39:09.040 > 00:39:15.040
Image charge, I actually have to reflect it
on the other surface as well,
444
00:39:15.040 > 00:39:19.040
just as I still have to mirror this image
charge on the left surface.
445
00:39:19.040 > 00:39:21.040
Let's start with that.
446
00:39:21.040 > 00:39:26.040
This image charge, mirrored on the left surface,
obviously gives a charge
447
00:39:26.040 > 00:39:36.040
who sits here and which I could also imagine
as a mirror image of the first image charge,
448
00:39:36.040 > 00:39:41.040
if I simply put the structure in front of it
periodically, the red lines should indicate
that here.
449
00:39:41.040 > 00:39:44.040
It goes in the other direction accordingly.
450
00:39:44.040 > 00:39:53.040
This charge, now mirrored at the righthand
interface, is again a
451
00:39:53.040 > 00:40:00.000
Image charge  second order, so to speak 
and I could also imagine that as
452
00:40:00.000 > 00:40:06.000
the image of the first image mirrored on a
periodic continuation of the structure.
453
00:40:06.000 > 00:40:08.000
Yes, and now I think it's clear how this will
continue.
454
00:40:08.000 > 00:40:13.000
You would have to put this whole chain in
front of infinity, you have to
455
00:40:13.000 > 00:40:21.000
write down. Look at all the terms and then
just see that you
456
00:40:21.000 > 00:40:29.000
writes down the series expansion and shows
the convergence and the corresponding
457
00:40:29.000 > 00:40:35.000
Write down the end value of the series  the
sum of the series.
458
00:40:35.000 > 00:40:42.000
In this way you can also use the greens function
of this room in the
459
00:40:42.000 > 00:40:50.000
Calculate the gap between two infinitely extended
plates.
460
00:40:50.000 > 00:40:54.000
Once we have the Greens function, we know
that, then we can use it
461
00:40:54.000 > 00:40:57.000
Geometry solve all problems.
462
00:40:57.000 > 00:41:08.060
Well. Yes, it should actually have been at
the point and you can see how
463
00:41:08.060 > 00:41:16.060
The Greens functions are actually powerful
and how we actually do them for them
464
00:41:16.060 > 00:41:23.060
certain geometries are very easy with the help
of the mirroring method
465
00:41:23.060 > 00:41:32.060
can find. We will actually also be in the next
events
466
00:41:32.060 > 00:41:34.060
get to know other methods.
467
00:41:34.060 > 00:41:39.060
But the bottom line is that the electrostatics
468
00:41:39.060 > 00:41:46.060
The method of mirror charges is one of the
most important of all for the
469
00:41:46.060 > 00:41:49.060
Solve problems to find the Greenschen functions.
470
00:41:49.060 > 00:41:55.060
As always, thank you for your attention and
as usual you will find
471
00:41:55.060 > 00:42:01.020
there is also more information on the website.
Thank you, see you next time.