WEBVTT - generated by VCS
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Have a wonderful good day, ladies and gentlemen.
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I warmly welcome you to another contribution
in the Electromagnetic Theory
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series, whereby today it will be about "electrostatics"
.
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This is a topic we won't end today, but we
will definitely start with that
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Electrostatics here.
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My name is Hans Georg Krauthäuser, I am the
holder of the professorship for
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Electromagnetic Theory and Compatibility at
the TU Dresden.
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Electrostatics, as I already said - we will
start with it today - is concerned
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with the cases in which we have no changes
in time and in which we are only concerned
with
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take care of the electrical phenomena.
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That is, either we have no streams or we ignore
the what
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caused by currents. That is,
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a typical arrangement is that we have one somewhere
in the room
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Have charge carrier density distribution rho_V.
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And it will often be the case that we carry
out integrals over this charge density distribution
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have to. Then we run through the entire volume
with our starting point r ' and evaluate that
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at another point r out.
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Then it is clear that this connection vector
also plays an important role
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and the connection vector - that is clear -
that is then just r minus r '.
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We will start by actually looking at the whole
thing in a vacuum and only then
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sometimes derive important relationships.
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But even in a vacuum it will be the case that
there are also conductive bodies from time
to time
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- especially as limitations - and that's why
we say right at the beginning too
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something about electrostatics in or on conductive
bodies.
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Well, with that we can express the problem
again explicitly: Ie, we have in the
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Solution area as a source a charge carrier
density rho_V and apart from that,
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the space is either a vacuum, ie, epsilon is
equal to epsilon_0, or it can be called
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perfectly conductive, that is, the conductivity
tends towards infinity, to be viewed.
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We do electrostatics and in electrostatics
it's relatively easy to do with
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Dealing with these perfectly conductive objects
because a perfect conductor stands out
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from the fact that I have free charge carriers
and well , in the event of an external change,
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Of course, it takes until these free charge
carriers come back again if necessary
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have rearranged. But if I have - as in the
statics - no change, then I am
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I'm actually always in a state of equilibrium.
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And this relaxation of the free charge carriers,
if something were to change, that is what
it is
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in perfectly conductive areas so quickly that
we can assume that
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Compensate for electric fields practically
instantaneously by rearranging the free ones
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Charge carrier. This means that it is relatively
easy to come to the important statement that
within
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perfectly conductive areas the electric field
strength is zero.
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At the edge, of course, we have the transition
conditions of the field
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Pay attention to interfaces. There is already
another video about this and we will do that
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to be received again later.
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Of course, we are looking for the field outside
the perfect ladder.
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We don't need to look for it within, because
we know within the perfectly conductive areas
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we already have the solution.
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A small aside is allowed at this point .
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The question arises so automatically if we
are out here the
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Have dielectric constant epsilon equal to epsilon_0.
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How is it in the ladder?
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Can I also state something like the dielectric
constant?
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Yes, you can and we will actually look at
it again later , if
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we have a field excitation at higher frequencies
-or have any field excitation-
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then you will be able to develop these harmoniously.
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And then for such a harmonic component you
can use a complex one
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Specify dielectric constant.
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As I said, it is then complex and is written
as shown below, that is
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epsilon complex from Omega.
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Consists of a real part plus j times the imaginary
part.
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The real part is the normal epsilon as we
know it and the imaginary part results
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is the conductivity at this angular frequency
divided by the angular frequency itself.
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With the help of such a formula it will be
easy to do later , for one
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Decide on the specific material: do I now have
a good dielectric or do I have it
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now a good leader in front of me.
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With a good dielectric, the real part will
dominate over the
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Imaginary part. That is to say, the conductivity
is typically very low -or in particular
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the frequency is very high and with a good
conductor it is the other way round.
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Even at high frequencies, this imaginary part
becomes essential here
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be greater than the real part.
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But as I said, only as a side note to this
question that arises here
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urges not to just leave it standing in the
room.
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Well, let's get back to our problem.
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We can of course formulate basic equations
here - we have already done that before -
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so can repeat that here again.
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Basic equations of electrostatics now outside
of the perfectly conductive areas here
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written down again - so in a vacuum.
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Rotation E equals 0 or according to the integral
expression: integral around the border
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of an area E ds is equal to 0.
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And the second equation: Divergence D -what
then in a vacuum divergence epsilon_0 times
E is-
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is equal to rho_v or integrated accordingly
: Epsilon times the integral over the
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The surface of a volume VE dA is equal to
the volume integral rho_V d_V.
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And the latter here integral rho_v d_V that
is precisely the charge Q
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that I have in this volume.
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This equation back here - so the integral
form of the Coulomb-Gauss law in a vacuum-
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that can actually be done in simple cases
- and is called simple cases
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Use suitable symmetry to determine the E-field
directly from it .
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We will demonstrate this in an example here.
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And that's a simple example.
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Maybe even the simplest example ever.
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We assume a spherically symmetric charge distribution
at the origin.
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At the origin there is not really a restriction
because if it were somewhere else,
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we could of course place the coordinate system
so that this
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Charge distribution at the origin would be.
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The fact that it is spherically symmetrical
is, of course, a really tough one
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Symmetry requirement. We will see that she
really helps us here.
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Spherically symmetric charge distribution means
that our rho_v of vector r is in
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Reality just a rho_v from r.
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And if we have such a spherically symmetric
charge distribution, then it is clear
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-You know about the point charge - which will
also be directed radially in the E-field.
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That means, the E-field at the place r 'also
has an amount,
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which only depends on r 'amount.
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And the direction is the direction of r '.
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We now take as a volume a volume that is also
adapted to this symmetry.
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We are completely free to take any volume
.
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The equation that we just had - the Coulomb-Gauss
integral - is true for any
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Volumes. And if I have this freedom, I can
of course also have a volume
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Find one that is particularly appropriate to
the problem.
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And in that case I'll take a sphere with radius
R around the origin.
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That is, I take a volume whose symmetry is
also back to the symmetry of the
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corresponds to the original problem.
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And of course I can immediately specify the
UI element for such a volume .
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This is also directed outwards, so it has the
normal vector E_r´.
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And of course then the differential surface
dA.
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The nice thing is that vector E and vector
d_A are in the same direction here.
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In this way, the scalar product E d_A
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simple scalar to E of r d_A.
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This is a very important simplification, which
we of course put into our Coulomb-Gaussian
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Set can also be entered directly.
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And here a further step has already been taken,
namely E of r of R is before that
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Been drawn integrally. Why is that possible?
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On the surface of these spheres we have -
because we have a spherically symmetrical one
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Charge distributions have the same field strength
everywhere .
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Simply for reasons of symmetry.
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Ie, E of r is the same for all points on the
surface - so it is a constant -
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which we can draw in front of the integral
accordingly.
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Then only the surface integral over 1 d_A
remains here.
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And that's the surface itself.
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The surface of the sphere is just 4 pi_r square,
so here's just that
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Surface integral E d_A becomes 4 pi_r square
E of r.
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That's wonderful because the E of r is of course
exactly what I want it to be.
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If I have the amount E of r here, I also have
the entire E of r because it will be yes
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just multiplied the unit vector ran.
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That is, the fact that I can easily transform
this surface integral into
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an expression in which E stands for r, that
shows me that I am using this method here
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actually am successful.
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You can easily imagine that you can go through
all points of the room
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achieve that you have balls of different diameters
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consider and for each ball
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-and if you can now let any radius of the
sphere-
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then you can reach any point in space.
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We can actually reach any vector r in this
way too.
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Well, that's wonderful that we can already
write this integral down like this.
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Now we simply continue to write on the other
side of the Coulomb-Gaussian integral.
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1 through epsilon_0 - that's the term we brought
over here to the other side,
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originally there was an epsilon_0 in front
of it.
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So 1 through epsilon_0 volume integral rho_V
d_V and a spherically symmetric one
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I will of course use this volume integral
in
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run spherical coordinates, so run in spherical
coordinates and then I have
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just an integral from 0 to r rho_V r´ r´
^ 2 dr´, an integral over phi, that's easy
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from 0 to 2 d_phi and an integral over theta,
that goes from 0 to pi sin_theta d_theta.
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Also r´ ^ 2 d_r d_phi sin theta d theta.
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That is precisely the volume element in spherical
coordinates.
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We can easily carry out the two rear integrals.
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The integral 0 to 2 pi d_phi, that obviously
just gives a term 2 pi.
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The integral 0 to pi sin_theta d_thea gives
2, as you can easily calculate.
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So that here we simply give the two back integrals
a factor of 4 pi.
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So that here stands 4 pi through epsilon 0.
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And the integral rho_v r´ ^ 2 dr´ simply
stops.
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If we now compare the terms and solve the whole
thing for E of r, then we have to
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yes 4 pi r ^ 2 still bring it to the other
side.
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That means, we then simply have an expression,
E of r equals 1 by epsilon_0 r ^ 2 times
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Integral 0 to r rho_v from r´ r´ ^ 2 d_r´.
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At this point - without knowing how exactly
spherical symmetric distribution
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looks- we would have to leave it that way for
now.
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That is, as far as we can calculate that if
we do not use spherical symmetry
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don't know anything about rho_v yet.
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But of course there are simple cases for which
you can calculate this further.
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And the simplest of these cases is simply that
we are using a homogeneous charged sphere
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Radius large R and then just no charges outside,
ie that rho_V of r´
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rho_V would be constant for all r´ less or
equal to R and 0 for r´ greater than R.
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In that case, of course, this integral becomes
relatively simple, I have to
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Keep the case distinction, but both cases
are of course simple.
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I can, if I go from 0 to r and am still completely
inside the ball, I would
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just pull the constant rho_v in front of it
and then have an integral r´ ^ 2 d_r.
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That is 1/3 r 'to the power of 3.
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And within the limits of 0 to r that of course
results in 1/3 r to the power of 3.
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And the terms in front of it are simply taken
over.
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Here you could shorten it - you can bring it
to the form - rho_V times r by 3 epsilon_0.
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Or, of course, I can also bring the total
charge of the ball into play
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- that's the volume of the sphere times the
charge density - and the volume of the sphere
is 4 pi / 3 R to the power of 3.
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If I put all this in, I would get a term charge
of the total sphere times r by 4 pi
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epsilon_0 R to the power of 3.
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In the event that my small r is bigger than
capital R, I have to be careful - I can
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Do not pull rho_V directly in front of it without
further ado - but that has to do with the
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Integration path different values.
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So I would have to split the integral into
an integral from 0 to large R.
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On the section I can actually pull the value
rho_V forward as a constant.
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And then there would be a second integral from
large R to small r.
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However, this then runs completely in the charge-free
area, where rho_V is equal to 0
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and so this second part does not contribute.
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The first part then gives formally the same
expression as in the first case
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00:19:11.040 --> 00:19:15.040
has been. Only that instead of lowercase r
the upper limit is uppercase R.
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Here, too, I can bring the charge back into
play and then get it
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- bring the total charge Q into play - and
get the expression Q by 4 pi epsilon_r ^ 2.
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A term that is already known from point charge
.
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That is also a characteristic that when I
am outside such a spherically symmetrical
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Charge distribution am that outside the field
then appears exactly like the field
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a point charge of the appropriate size at
the origin.
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Well, that's a very, very simple example.
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Calculated here to show you what the basic
procedure is.
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So we can - and that is, so to speak, a solution
method - from the
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Coulomb-Gauss integral in some cases calculate
the E-field directly.
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This is a good and also intuitive solution,
but only if this is the case
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left integral easily becomes an expression
for E of r for all r im
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Solution volume can be transformed.
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Whether the right integral can then really
be solved analytically is first
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sometimes secondary. That is of course very
nice if that is the case and we always will
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00:20:46.000 --> 00:20:49.000
strive to get an analytical solution.
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Simply because you can of course read physical
relationships from an analytical solution
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00:20:56.000 --> 00:21:06.060
can. But in case of doubt you can of course
always solve the right integral numerically.
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The numerical integration is not a big problem.
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It is stable at least if there are not any
critical singularities in the
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Have integration area. Those are all things
you can do in numerical math
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will then also discuss.
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00:21:22.060 --> 00:21:27.060
But of course an analytical solution is always
preferable - I have already said -
215
00:21:27.060 --> 00:21:30.060
because we can read off analytical connections
there.
216
00:21:30.060 --> 00:21:39.060
And in case of doubt, the numerical solution
will only give you numerical values or
217
00:21:39.060 --> 00:21:45.060
provides colorful pictures, but not directly
physical connections.
218
00:21:45.060 --> 00:21:55.060
This left integral can usually only be evaluated
if
219
00:21:55.060 --> 00:22:00.020
there are strong symmetries, as we have just
seen with the examples.
220
00:22:00.020 --> 00:22:08.020
Further examples, where this works, would be
the charged ring, which is infinitely long
homogeneous
221
00:22:08.020 --> 00:22:13.020
loaded cylinders or similar items.
222
00:22:13.020 --> 00:22:18.020
You will certainly look at something like
that again as an exercise.
223
00:22:18.020 --> 00:22:27.020
But we need, because we are already limited
to certain problems, we
224
00:22:27.020 --> 00:22:31.020
So on the one hand need more general methods.
225
00:22:31.020 --> 00:22:38.020
On the other hand, maybe also methods for other
special cases.
226
00:22:38.020 --> 00:22:47.020
More general methods help us further, if we
just don't get into one of the special cases
227
00:22:47.020 --> 00:22:50.020
fit in with our problem.
228
00:22:50.020 --> 00:22:56.020
But it is often the case that one is just about
to consider a particular situation
229
00:22:56.020 --> 00:23:03.080
one - in a faster way - comes to a simple
solution, which then sometimes also
230
00:23:03.080 --> 00:23:08.080
can serve as an approximate solution for a
more general case.
231
00:23:08.080 --> 00:23:15.080
Yes, that is, the task will really be that
we find such more general solutions
232
00:23:15.080 --> 00:23:23.080
find, but also that we are just looking at
methods for other special cases.
233
00:23:23.080 --> 00:23:32.080
And we continue in exactly the same direction
in which we next take the so-called
234
00:23:32.080 --> 00:23:36.080
Coulomb integral - now not the Goulomb-Gauss
integral, but
235
00:23:36.080 --> 00:23:41.080
Look at the Coulomb integral and thus also
the scalar potential.
236
00:23:41.080 --> 00:23:47.080
We know - yes, in the end already from school
and you already have it in the
237
00:23:47.080 --> 00:23:54.080
first semester learned how the electric field
of a point charge looks like.
238
00:23:54.080 --> 00:24:04.040
These are always such expressions q times r
minus r_j through r minus r_j amount to the
power of 3
239
00:24:04.040 --> 00:24:13.040
-if a cargo is at location r_j.
240
00:24:13.040 --> 00:24:19.040
We have just seen a special case for the origin.
241
00:24:19.040 --> 00:24:26.040
That would be the generalization for charges
that are located at locations r_j.
242
00:24:26.040 --> 00:24:33.040
And well, if I know that for a charge, then
I can from the superposition principle
243
00:24:33.040 --> 00:24:38.040
-from the superposition- actually do the same
for n-charges.
244
00:24:38.040 --> 00:24:40.040
Why can I superpose?
245
00:24:40.040 --> 00:24:45.040
The Maxwell equations as we introduced them
are linear equations and
246
00:24:45.040 --> 00:24:50.040
because they are linear equations, the sums
of solutions are also the
247
00:24:50.040 --> 00:24:54.040
Maxwell's equations again a solution to Maxwell's
equation.
248
00:24:54.040 --> 00:25:00.000
I can always do that as long as I look at
the fields.
249
00:25:00.000 --> 00:25:02.000
I can always superpose solutions to Maxwell's
equations .
250
00:25:02.000 --> 00:25:14.000
So if I now have n point charges q_j at position
r_j, then I know this field
251
00:25:14.000 --> 00:25:17.000
of these n individual charges already.
252
00:25:17.000 --> 00:25:22.000
The whole thing applies if there are no further
boundary conditions in the finite
253
00:25:22.000 --> 00:25:28.000
- We will come back to the boundary conditions
later - but that too here
254
00:25:28.000 --> 00:25:34.000
mentioned once. That is, we still assume that
we essentially have the free space.
255
00:25:34.000 --> 00:25:41.000
Yes, we can then actually write down E as
1 through 4 pi epsilon_0 and
256
00:25:41.000 --> 00:25:50.000
just the sum j equals one to n, q_j r minus
r_j through r minus r_j amount to the power
of 3.
257
00:25:50.000 --> 00:25:56.000
If we can do that for n charge carriers ,
we can of course also do the transition
258
00:25:56.000 --> 00:26:01.060
to a continuous load carrier distribution.
259
00:26:01.060 --> 00:26:13.060
The sum then turns into an integral and takes
the place of the discrete charges
260
00:26:13.060 --> 00:26:17.060
just a differential charge.
261
00:26:17.060 --> 00:26:22.060
And I can divide the differential charge through
the charge carrier density
262
00:26:22.060 --> 00:26:29.060
Express volume element. So roh_V at the place
r´ d³ r´.
263
00:26:29.060 --> 00:26:33.060
And otherwise it's just translated one to one.
264
00:26:33.060 --> 00:26:38.060
As I said: the sum becomes an integral and
the discrete charge carriers become
265
00:26:38.060 --> 00:26:46.060
ultimately roh_V d³ r´. The whole thing is
written down here again .
266
00:26:46.060 --> 00:26:53.060
Yes and now this term is actually interesting,
which is repeated here again and again
267
00:26:53.060 --> 00:27:01.020
appears, r minus r 'by the amount of r minus
r' to the power of 3 a little more precisely
268
00:27:01.020 --> 00:27:08.020
consider. It is easy to calculate, really
easy, that this r minus r 'through
269
00:27:08.020 --> 00:27:17.020
r minus r 'amount to the power of 3 is nothing
other than the negative of the gradient with
respect to r
270
00:27:17.020 --> 00:27:22.020
from 1 through r minus r '.
271
00:27:22.020 --> 00:27:31.020
You can of course use that here just for fun
, that is, I can do this here
272
00:27:31.020 --> 00:27:36.020
replace accordingly in the integral.
273
00:27:36.020 --> 00:27:56.020
And then there is an expression here, this
gradient, which is related to the variable
r
274
00:27:56.020 --> 00:28:04.080
and the 1 through r minus r´ -da is of course
also included with the r´.
275
00:28:04.080 --> 00:28:10.080
That means, if I put that under the integral,
then I can use the gradient with respect to
the
276
00:28:10.080 --> 00:28:15.080
Drag variables r and also the minus sign in
front of the integral.
277
00:28:15.080 --> 00:28:23.080
1 through r minus r´ remains of course under
the integral.
278
00:28:23.080 --> 00:28:31.080
If I consider that E is equal to minus the
gradient phi , then that compensates for each
other
279
00:28:31.080 --> 00:28:40.080
just with this minus sign here so that I can
see that the phi of r is nothing else,
280
00:28:40.080 --> 00:28:48.080
as 1 through 4 pi epsilon_0 integral rho_V
of r´ d³_r´ through r minus r´.
281
00:28:48.080 --> 00:28:54.080
We call this quantity phi the scalar potential.
282
00:28:54.080 --> 00:29:00.040
We already got to know it once during the
axiomatic introduction , when it came to being
completely
283
00:29:00.040 --> 00:29:03.040
different, it was introduced quite differently.
284
00:29:03.040 --> 00:29:08.040
We will meet again afterwards - now in the
next step - again
285
00:29:08.040 --> 00:29:16.040
introduce more formal. But let's keep in mind
that we actually have this scalar potential
as well
286
00:29:16.040 --> 00:29:22.040
can deduce from observation, so to speak.
287
00:29:22.040 --> 00:29:31.040
The observation would be that we know the field
of point charges, from that the field
288
00:29:31.040 --> 00:29:38.040
formally write down a charge distribution and
then do a little math
289
00:29:38.040 --> 00:29:45.040
this potential - this potential function,
this scalar potential - as a function of
290
00:29:45.040 --> 00:29:49.040
write down the charge carrier density distribution.
291
00:29:49.040 --> 00:30:02.000
Side note: this integral, that is, this Coulomb
integral, we can of course do that
292
00:30:02.000 --> 00:30:07.000
also use it to now use the
293
00:30:07.000 --> 00:30:14.000
Attempt to calculate the scalar potential
directly through integration.
294
00:30:14.000 --> 00:30:20.000
This will also work analytically in a few selected
cases.
295
00:30:20.000 --> 00:30:26.000
In other cases you will only be able to do
it numerically this way.
296
00:30:26.000 --> 00:30:31.000
We'll definitely hold it down as an important
formula .
297
00:30:31.000 --> 00:30:40.000
Next, so you don't necessarily have to watch
the video about the axiomatic introduction
yet
298
00:30:40.000 --> 00:30:47.000
have to take a look at the scalar potential
again , perhaps very briefly
299
00:30:47.000 --> 00:30:53.000
formally introduced. You may remember , we
had that back then about that
300
00:30:53.000 --> 00:30:56.000
Lemma made by Poincaré.
301
00:30:56.000 --> 00:31:04.060
That -I'm leafing through this completely-
knows three formulations or three formulations
302
00:31:04.060 --> 00:31:12.060
includes. First of all, the first statement
that when I'm on one easy
303
00:31:12.060 --> 00:31:18.060
contiguous area defined vortex-free vector
field that I have this
304
00:31:18.060 --> 00:31:23.060
can then write as a gradient of a potential
field.
305
00:31:23.060 --> 00:31:32.060
This means that from rotation a equals 0 it
follows that a equals the gradient of a potential
field f.
306
00:31:32.060 --> 00:31:39.060
If I have a source-free vector field on a
convex area, then I can
307
00:31:39.060 --> 00:31:41.060
represent this by rotating a vector potential.
308
00:31:41.060 --> 00:31:49.060
In other words: from divergence beta equals
0, beta equals rotation alpha.
309
00:31:49.060 --> 00:31:57.060
And if I have a scalar field density, then
this can be seen as a divergence of a
310
00:31:57.060 --> 00:31:59.060
Vector field are represented.
311
00:31:59.060 --> 00:32:07.020
In other words, if I have a gamma - that is
this field density - then it is
312
00:32:07.020 --> 00:32:13.020
yes a volume integrand or can be understood
as a volume integrand, then is
313
00:32:13.020 --> 00:32:18.020
this gamma is also the divergence of a vector
field beta.
314
00:32:18.020 --> 00:32:23.020
The vector arrow is missing here! Please pay
attention to it.
315
00:32:23.020 --> 00:32:26.020
I will try to correct it in the final documents.
316
00:32:26.020 --> 00:32:36.020
Yes, and of course we can jump in here and
say: well for the rotation
317
00:32:36.020 --> 00:32:44.020
from E we have exactly such a statement of
the form the rotation vanishes
318
00:32:44.020 --> 00:32:47.020
-so rotation E equals 0 in the statics.
319
00:32:47.020 --> 00:32:54.020
That is, we can write E as minus gradient
phi , analogous to the expression we have here
.
320
00:32:54.020 --> 00:33:02.080
The minus that is written here - is a pure
convention - I could just as well use it
321
00:33:02.080 --> 00:33:08.080
pulling in the scalar potential - you do n't
do it - and also here this potential, this
one
322
00:33:08.080 --> 00:33:14.080
Potential function or potential field what
appears here is what has just been introduced
323
00:33:14.080 --> 00:33:17.080
electrical scalar potential.
324
00:33:17.080 --> 00:33:21.080
These are exactly these two ways how to get
to the scalar potential.
325
00:33:21.080 --> 00:33:31.080
Once more out of the empirical and on the other
hand purely formally via rotation E.
326
00:33:31.080 --> 00:33:36.080
equals 0 and Poincarè to E equals minus gradient
phi.
327
00:33:36.080 --> 00:33:48.080
Yes, well, that is, we always have rotation
E equal to 0 for static problems.
328
00:33:48.080 --> 00:33:56.080
Hence, as just said, E equals minus gradient
phi.
329
00:33:56.080 --> 00:34:01.040
Now let's look at the vacuum.
330
00:34:01.040 --> 00:34:08.040
But we could also say at this point, as long
as we have simple materials,
331
00:34:08.040 --> 00:34:17.040
namely those for which the dielectric shift
is simply a scalar
332
00:34:17.040 --> 00:34:20.040
results from the electric field.
333
00:34:20.040 --> 00:34:26.040
So where D is equal to epsilon times E and
epsilon is now equal to epsilon 0 times
334
00:34:26.040 --> 00:34:28.040
epsilon_r and epsilon_r is a scalar.
335
00:34:28.040 --> 00:34:39.040
We can simply write down the difference E as
1 through epsilon times rho_V.
336
00:34:39.040 --> 00:34:44.040
We can use that
337
00:34:44.040 --> 00:34:53.040
that is, for E in divergence E we can now
write phi for E minus gradient.
338
00:34:53.040 --> 00:35:00.000
So here we have the difference of minus gradient
phi equal to 1 by epsilon times rho_V.
339
00:35:00.000 --> 00:35:07.000
Divergence gradient, that's the Laplace operator.
340
00:35:07.000 --> 00:35:10.000
We bring the minus to the other side,
341
00:35:11.000 --> 00:35:14.000
so here we have a differential equation
342
00:35:14.000 --> 00:35:17.000
get second order for the scalar potential.
343
00:35:17.000 --> 00:35:23.000
Laplace phi is equal to minus 1 through epsilon
rho_V.
344
00:35:23.000 --> 00:35:27.000
This important equation is called the Poisson
equation.
345
00:35:27.000 --> 00:35:34.000
There is another simple special case, namely
if I am now in an area
346
00:35:34.000 --> 00:35:40.000
where I have no charge carrier density, i.e.
rho_V is equal to 0, then we have that
347
00:35:40.000 --> 00:35:47.000
Special case Laplace phi equals 0 and that
is the so -called Laplace equation.
348
00:35:47.000 --> 00:35:58.000
The advantage compared to rotation E equal
to 0 and difference E equal to 1 through epsilon
rho_V
349
00:35:58.000 --> 00:36:03.060
-the advantage of Poisson's equation or the
Laplace equation is-
350
00:36:03.060 --> 00:36:10.060
that only one equation is needed here to solve
the problem.
351
00:36:10.060 --> 00:36:20.060
As soon as I have found a phi , E is equal
to minus the gradient phi for the associated
E-field.
352
00:36:20.060 --> 00:36:26.060
And because that is then a gradient field,
rotation E equal to zero is automatically
fulfilled.
353
00:36:26.060 --> 00:36:32.060
That's an advantage, I do n't have to worry
about the second equation anymore.
354
00:36:32.060 --> 00:36:39.060
As is so often the case, advantages do not
come entirely without disadvantages and disadvantages
355
00:36:39.060 --> 00:36:45.060
is just here that I am at this point of two
first-order differential equations
356
00:36:45.060 --> 00:36:52.060
now have a second order differential equation.
357
00:36:52.060 --> 00:37:00.020
We'll solve that later or get to know methods
of how to solve this.
358
00:37:00.020 --> 00:37:09.020
But first one more remark and we wo n't do
that in this block either
359
00:37:09.020 --> 00:37:13.020
solve, that comes later. A note on what you
can immediately see here.
360
00:37:13.020 --> 00:37:19.020
Let's imagine we go from phi to a phi plus
another constant and
361
00:37:19.020 --> 00:37:28.020
then look at the E-field that leads to this
changed scalar potential phi plus constant
362
00:37:28.020 --> 00:37:33.020
belongs. Then this constant will not play
a role in the gradient formation,
363
00:37:33.020 --> 00:37:40.020
ie, the gradient of phi plus the constant is
equal to the gradient of phi and therefore
E is also of phi
364
00:37:40.020 --> 00:37:43.020
Constant equal to E of phi.
365
00:37:43.020 --> 00:37:50.020
That means, for the fields it makes no difference
whether I add a constant to the scalar potential
366
00:37:50.020 --> 00:37:58.020
add. In other words, this scalar potential
phi is only up to one
367
00:37:58.020 --> 00:38:01.080
additive constant determined.
368
00:38:01.080 --> 00:38:11.080
Yes, let's look next at forces and also fields
and so-called
369
00:38:11.080 --> 00:38:14.080
Equipotential surfaces. Let's start with the
force.
370
00:38:14.080 --> 00:38:22.080
In case of doubt, you already know the force
on a test charge q in field E
371
00:38:22.080 --> 00:38:29.080
the school, F of r that is the Coulomb force,
that is q times E.
372
00:38:29.080 --> 00:38:40.080
If I had such a test charge q and the one
in the electrical field from the beginning
373
00:38:40.080 --> 00:38:48.080
Let go of the rest, then it would move naturally
- would start to move -
374
00:38:48.080 --> 00:38:51.080
because there is a force acting on them.
375
00:38:51.080 --> 00:39:02.040
That is to say, this charge then passes through
a certain trajectory and the trajectory of
such a trajectory
376
00:39:02.040 --> 00:39:08.040
fictitious test charge, which describes a so-called
field line.
377
00:39:08.040 --> 00:39:12.040
The charge - the test charge - is a positive
test charge here,
378
00:39:12.040 --> 00:39:19.040
that is, it moves from plus to minus and so
we also have one
379
00:39:19.040 --> 00:39:24.040
Field line orientation from plus to minus.
380
00:39:24.040 --> 00:39:36.040
We designate places with the same scalar potential
as -or form an- equipotential surface.
381
00:39:36.040 --> 00:39:47.040
If we consider a small displacement ds on an
equipotential surface,
382
00:39:47.040 --> 00:39:53.040
that is, we have an equipotential surface and
are now entering the equipotential surface
383
00:39:53.040 --> 00:40:02.000
a little bit from one place to another, then
the potential changes
384
00:40:02.000 --> 00:40:05.000
not because we are on an equipotential surface.
385
00:40:05.000 --> 00:40:16.000
That means, d_phi -which is nothing else than
phi of r plus ds minus phi of r- must be equal
to 0,
386
00:40:16.000 --> 00:40:27.000
because equipotential surface. And well, that's
no different - by definition actually - than
gradient phi
387
00:40:27.000 --> 00:40:31.000
scalar multiplied by these same bits ds.
388
00:40:31.000 --> 00:40:33.000
That is precisely the definition of the derivative.
389
00:40:33.000 --> 00:40:49.000
So if d_phi equals 0 and d_phi equals gradient
phi times ds, then those are
390
00:40:49.000 --> 00:40:54.000
Trivial cases : I didn't move at all - I can
neglect a special case,
391
00:40:54.000 --> 00:41:00.060
does not matter. Or I don't have a gradient
phi at all .
392
00:41:00.060 --> 00:41:04.060
In other words, the phi is the same everywhere
- I do n't need to look at it either.
393
00:41:04.060 --> 00:41:10.060
And then the only possibility that this becomes
0 is that the gradient phi and ds
394
00:41:10.060 --> 00:41:20.060
stand perpendicular to each other. On the other
hand , gradient phi is equal to minus because
of E.
395
00:41:20.060 --> 00:41:22.060
Gradient phi parallel to E.
396
00:41:22.060 --> 00:41:33.060
In other words: the E-field lines are therefore
perpendicular to the equipotential surfaces.
397
00:41:33.060 --> 00:41:39.060
This is an important statement that we will
use quite often .
398
00:41:39.060 --> 00:41:45.060
The equipotential surfaces, which should also
be noted here, are usually perfect conductors.
399
00:41:45.060 --> 00:41:51.060
That is, what we already have here as a statement,
that the E-field on the surface of a
400
00:41:51.060 --> 00:41:54.060
perfect conductor is vertical.
401
00:41:54.060 --> 00:42:05.020
We can also briefly think about the independence
of the path.
402
00:42:05.020 --> 00:42:08.020
That too is something that will not be new
to them, but we are writing
403
00:42:08.020 --> 00:42:09.020
just put it here again.
404
00:42:09.020 --> 00:42:16.020
Of course we can now use any - any two -
405
00:42:16.020 --> 00:42:22.020
Consider paths C_1 and C_2 between r_0 and
r.
406
00:42:22.020 --> 00:42:29.020
That means, here is the point r_0, here is
the point r and we just have two paths C_1
407
00:42:29.020 --> 00:42:37.020
and C_2 which go from this point r_0 to the
point r.
408
00:42:37.020 --> 00:42:44.020
Because rotation E equals 0, for every closed
path C, the path integral C
409
00:42:44.020 --> 00:42:49.020
times E_ds - path integral over C E_ds -disappears-
is equal to 0.
410
00:42:49.020 --> 00:42:54.020
And such a closed path would be if I walk
over path C_1
411
00:42:54.020 --> 00:42:59.020
there and back via path C_2.
412
00:42:59.020 --> 00:43:03.080
C_2 back means yes, I'm going the way minus
C_2.
413
00:43:03.080 --> 00:43:08.080
So path C_1 plus minus C_2 is a closed path.
414
00:43:08.080 --> 00:43:20.080
This means that the integral over the entire
path C_1 and then minus C_2 is equal to 0.
415
00:43:20.080 --> 00:43:27.080
And that is only possible if the two integrals
are the same via these two ways.
416
00:43:27.080 --> 00:43:37.080
That means, we then immediately see: integral
from r_0 to r via C_1 E ds is equal to integral
from r_0 to
417
00:43:37.080 --> 00:43:44.080
r over C_2 E ds and that of course applies
to any paths.
418
00:43:44.080 --> 00:43:54.080
In other words, no matter how I lead my way
from r_0 to r, the integral E ds will always
be
419
00:43:54.080 --> 00:43:56.080
have the same value.
420
00:43:56.080 --> 00:44:01.040
Something like this is also called a conservative
field, a term you get from physics
421
00:44:01.040 --> 00:44:02.040
surely already know.
422
00:44:02.040 --> 00:44:09.040
We can take advantage of this if we now have
a situation where
423
00:44:09.040 --> 00:44:13.040
Equipotential surfaces play a role.
424
00:44:13.040 --> 00:44:22.040
So let's assume that this curve would be an
equipotential surface here , with one
425
00:44:22.040 --> 00:44:29.040
Potential phi of r_0 and we have here accordingly
an equipotential surface with
426
00:44:29.040 --> 00:44:32.040
a potential phi of r.
427
00:44:32.040 --> 00:44:41.040
And we want to calculate the integral E ds
from point r_0 to point r.
428
00:44:41.040 --> 00:44:46.040
Then of course we can now take any path through
the volume that
429
00:44:46.040 --> 00:44:49.040
but would not be particularly skillful.
430
00:44:49.040 --> 00:44:58.040
It is more skillful to divide the path into
posts that go along the
431
00:44:58.040 --> 00:45:07.000
Lead potential surface and contributions that
lead along a field line.
432
00:45:07.000 --> 00:45:13.000
So here this red curve, that should be a field
line, the black line, that is
433
00:45:13.000 --> 00:45:21.000
times an equipotential surface - here above
- and the green arrows that indicate
434
00:45:21.000 --> 00:45:25.000
just now to how I choose the integration path.
435
00:45:25.000 --> 00:45:32.000
First of all from r_0 to r_1 along an equipotential
surface, then I bend in the right one
436
00:45:32.000 --> 00:45:35.000
Angle to it because, now it goes on on a field
line.
437
00:45:35.000 --> 00:45:39.000
Come from r_1 to r_2.
438
00:45:39.000 --> 00:45:43.000
Here I turn again at a right angle, because
that's a field line on one
439
00:45:43.000 --> 00:45:51.000
Equipotential surface and then go to the equipotential
surface r_2 to the point r.
440
00:45:51.000 --> 00:45:58.000
The advantage of having it is very obvious.
441
00:45:58.000 --> 00:46:05.060
Along the paths r_0 to r_1 and r_2 to r -that
is, the paths that lie on the equipotential
surface-
442
00:46:05.060 --> 00:46:09.060
they don't contribute anything to the integral.
443
00:46:09.060 --> 00:46:18.060
Because there is always ds perpendicular to
E - because everywhere the E field is perpendicular
to the equipotential surface.
444
00:46:18.060 --> 00:46:25.060
And even if I then walk along the field line
, the situation is simple.
445
00:46:25.060 --> 00:46:32.060
Along the field line, the path is always aligned
with the field strength,
446
00:46:32.060 --> 00:46:43.060
ie, E scalar multiplied by ds is always equal
to the simple product E ds
447
00:46:43.060 --> 00:46:52.060
-without vectors- and that is then just minus
the amount of the gradient phi ds without
vector.
448
00:46:52.060 --> 00:47:01.020
And so I can then just write the total integral,
i.e. the integral from r_0 to r.
449
00:47:01.020 --> 00:47:11.020
I can just see that there are still mistakes
creeping in here: They are not closed
450
00:47:11.020 --> 00:47:14.020
Path integrals. Not here either.
451
00:47:14.020 --> 00:47:18.020
Up here, these are not closed path integrals
either.
452
00:47:18.020 --> 00:47:25.020
So the integral r_0 to r E ds is then equal
to minus the integral
453
00:47:25.020 --> 00:47:29.020
from r_1 to r_2 gradient phi ds.
454
00:47:29.020 --> 00:47:36.020
And the integral over gradient phi, that is
then simply phi taken from the
455
00:47:36.020 --> 00:47:40.020
corresponding boundaries of the integral.
456
00:47:40.020 --> 00:47:48.020
I know the potential at these points, which
is phi of r_0 and phi of r.
457
00:47:48.020 --> 00:47:57.020
This means that the integral E ds simply becomes
a potential difference here .
458
00:47:57.020 --> 00:48:06.080
And because such potential differences naturally
occur frequently in electrostatics
459
00:48:06.080 --> 00:48:14.080
it is advisable to introduce another size
accordingly .
460
00:48:14.080 --> 00:48:19.080
The voltage U_21 is then defined as precisely
this potential difference,
461
00:48:19.080 --> 00:48:25.080
ie, U_21 is equal to phi_2 minus phi_1 and
that is equal to minus
462
00:48:25.080 --> 00:48:30.080
the integral from r_1 to r_2 E ds.
463
00:48:30.080 --> 00:48:34.080
It is also correct here, this is not a ring
integral, but a path integral.
464
00:48:34.080 --> 00:48:41.080
And I'm not telling you anything new if I tell
you that the unit has this
465
00:48:41.080 --> 00:48:43.080
Voltage equals volts.
466
00:48:43.080 --> 00:48:52.080
We can also introduce work in a simple way
with what we are now
467
00:48:52.080 --> 00:48:56.080
already know. We consider a test charge q.
468
00:48:56.080 --> 00:49:05.040
The electrostatic force F_e, equal to q times
E, naturally acts on them in the E-field .
469
00:49:05.040 --> 00:49:09.040
We've already used that: it's just the Coulomb
force.
470
00:49:09.040 --> 00:49:19.040
And in order to hold the cargo in place, I
of course have to have an opposing one
471
00:49:19.040 --> 00:49:21.040
use mechanical force.
472
00:49:21.040 --> 00:49:32.040
This mechanical force is then F mechanically
equal to minus F_e and the
473
00:49:32.040 --> 00:49:38.040
mechanical power on your part I need that
again ... Just a little moment, here
474
00:49:38.040 --> 00:49:45.040
The sun is shining right now. I have to blend
in a little.
475
00:49:59.040 --> 00:50:00.000
So sorry.
476
00:50:00.000 --> 00:50:04.000
As nice as it is when you have the sun, it
sometimes bothers you,
477
00:50:04.000 --> 00:50:07.000
when it shines straight in the face.
478
00:50:07.000 --> 00:50:15.000
So, we actually need this mechanical force
to infer the work.
479
00:50:15.000 --> 00:50:22.000
Because the work is force times distance, i.e.
integral over the mechanical force ds.
480
00:50:22.000 --> 00:50:30.000
We can translate that to the Coulomb force,
and that's what we need here
481
00:50:30.000 --> 00:50:35.000
negative sign. We know the Coulomb force,
that is q times E, that q can precede that
482
00:50:35.000 --> 00:50:36.000
Be drawn integrally.
483
00:50:36.000 --> 00:50:41.000
And the integral E ds - we already know that
- is equal to the potential difference and
484
00:50:41.000 --> 00:50:43.000
thus equal to the tension.
485
00:50:43.000 --> 00:50:51.000
So that we can express the mechanical work
directly via this test charge q
486
00:50:51.000 --> 00:50:53.000
and the potential difference.
487
00:50:53.000 --> 00:50:59.000
And that is the test charge q multiplied by
the corresponding voltage.
488
00:50:59.000 --> 00:51:05.060
When we have the work, we can of course also
make statements
489
00:51:05.060 --> 00:51:09.060
to make electrostatic energy.
490
00:51:09.060 --> 00:51:17.060
Because we can define the energy to be confined
to a finite space
491
00:51:17.060 --> 00:51:24.060
Charge configuration of work corresponds to
charges from infinity to this
492
00:51:24.060 --> 00:51:27.060
Configuration to be added.
493
00:51:27.060 --> 00:51:36.060
The finite space is important at this point
because we can then make statements about
it
494
00:51:36.060 --> 00:51:40.060
Infinite potential.
495
00:51:40.060 --> 00:51:47.060
That disappears in infinity, as can be seen
from the expression of the potential here
496
00:51:47.060 --> 00:51:50.060
looks nice when the charge distribution is
actually on a
497
00:51:50.060 --> 00:51:54.060
spatially finite area is limited.
498
00:51:54.060 --> 00:52:00.020
Before we analyze this in more detail for
continuous charge distribution , do
499
00:52:00.020 --> 00:52:07.020
let's start with a little simpler and think
about it first for N
500
00:52:07.020 --> 00:52:16.020
Point charges anchored at locations r_i.
501
00:52:16.020 --> 00:52:24.020
First of all, we consider: what is the potential
for the first i minus 1
502
00:52:24.020 --> 00:52:29.020
Point charges from these N point charges?
503
00:52:29.020 --> 00:52:41.020
They sit with their charges q_j at locations
r_j and the potential at location i is then
504
00:52:41.020 --> 00:52:48.020
simply the sum of the Coulomb potentials -1
by 4 pi epsilon_0 sum j equals 1 to i
505
00:52:48.020 --> 00:52:53.020
minus 1 of q_j by r minus r_j amount.
506
00:52:53.020 --> 00:52:58.020
According to the definition , we are now interested
in what happens if we now have the ith charge
507
00:52:58.020 --> 00:53:01.080
bring r_i to their place?
508
00:53:01.080 --> 00:53:06.080
And for that work has to be done.
509
00:53:06.080 --> 00:53:13.080
We come from the infinite, move it to r_i
and the work for it - for the i-th
510
00:53:13.080 --> 00:53:19.080
Charge - is then q_i times the potential at
location r_i.
511
00:53:19.080 --> 00:53:25.080
The overall work - when we introduce ourselves,
we keep doing it until we get there
512
00:53:25.080 --> 00:53:32.080
Have N charges together - then the total work
is simply the sum of i
513
00:53:32.080 --> 00:53:38.080
equals 2 to N about these individual works.
514
00:53:38.080 --> 00:53:48.080
I equals 1 - there should be a 1 here - i equals
1 we have omitted here, because,
515
00:53:48.080 --> 00:53:53.080
when we bring the first load to its place,
then there is still no other load.
516
00:53:53.080 --> 00:54:01.040
That is, the room is completely field-free
and therefore the work for this would be first
517
00:54:01.040 --> 00:54:03.040
Charge actually also 0.
518
00:54:03.040 --> 00:54:06.040
That's why we don't need to take this term
with us .
519
00:54:06.040 --> 00:54:19.040
And yes, if we add that up, there is also
a small mistake here, here must be
520
00:54:19.040 --> 00:54:25.040
just like here below are W_e- and by adding
up we come to that
521
00:54:25.040 --> 00:54:30.040
Total work and that then also corresponds to
the total electrostatic energies that we W_e
522
00:54:30.040 --> 00:54:37.040
call. So that would be this expression that
stands here with this double sum:
523
00:54:37.040 --> 00:54:43.040
simply i equals 2 to N times this expression
summed up here .
524
00:54:43.040 --> 00:54:49.040
Now this double sum is a bit unwieldy when
we come to that later
525
00:54:49.040 --> 00:54:52.040
want to go over continuous charge distribution.
526
00:54:52.040 --> 00:54:54.040
That's why we're rewriting them.
527
00:54:54.040 --> 00:54:59.040
To do this, let's make it clear which terms
are actually used here.
528
00:54:59.040 --> 00:55:05.000
So for i equal to 2, the j-sum would just run
over j equal to 1.
529
00:55:05.000 --> 00:55:11.000
For i equal to 3, we would have j from 1 to
2.
530
00:55:11.000 --> 00:55:15.000
For i equal to 4 we would have j = 1, 2, 3
etc.
531
00:55:15.000 --> 00:55:19.000
So you see: if you imagined it all as a matrix,
we would have it
532
00:55:19.000 --> 00:55:25.000
the upper triangle occupies but the main diagonal
would be missing.
533
00:55:25.000 --> 00:55:33.000
So the terms j equal to i they don't exist
here because the j-summation is already at
i minus
534
00:55:33.000 --> 00:55:45.000
1 stops. That is, if we write this now as a
double sum, really about all of them
535
00:55:45.000 --> 00:55:49.000
Terme - there is another typo here, too, I
apologize for it.
536
00:55:49.000 --> 00:55:59.000
So here i should be 1 to N and j should be
1 to N- then
537
00:55:59.000 --> 00:56:03.060
we counted all terms twice.
538
00:56:03.060 --> 00:56:08.060
I compensate for this by adding 1 to 8 pi
instead of 1 by 4 pi.
539
00:56:08.060 --> 00:56:13.060
And besides, we would have counted the main
diagonal in this matrix -
540
00:56:13.060 --> 00:56:15.060
so the terms i equal j.
541
00:56:15.060 --> 00:56:20.060
And I exclude that by simply multiplying again
under the sum with
542
00:56:20.060 --> 00:56:29.060
1 minus delta_ij, that is 0 for i and 1 otherwise.
543
00:56:29.060 --> 00:56:40.060
After we have written this down for the discrete
case of N charges, we can
544
00:56:40.060 --> 00:56:47.060
also move to a continuous distribution.
545
00:56:47.060 --> 00:56:54.060
It should be mentioned once again that for
N point charges we have the terms i equal to
j
546
00:56:54.060 --> 00:57:00.020
have not taken, there is this 1 minus delta_ij
under the sum.
547
00:57:00.020 --> 00:57:05.020
But with a continuous charge distribution
, I can't make it that easy.
548
00:57:05.020 --> 00:57:09.020
With the integral, I can't easily add any
small pieces
549
00:57:09.020 --> 00:57:16.020
let out. That is, if we now simply transfer
the formula to a continuous one
550
00:57:16.020 --> 00:57:21.020
Charge distribution, then of course the double
sum becomes the double integral and
551
00:57:21.020 --> 00:57:25.020
instead of the discrete charges, there is now
the charge distribution
552
00:57:25.020 --> 00:57:28.020
- once at place r and once at place r '.
553
00:57:28.020 --> 00:57:32.020
And then it is integrated accordingly.
554
00:57:32.020 --> 00:57:38.020
It has to be mentioned that this is not 100
percent, so not perfect
555
00:57:38.020 --> 00:57:39.020
equivalent expression is.
556
00:57:39.020 --> 00:57:46.020
But it is obvious that in the continuous case
it is just as written down as
557
00:57:46.020 --> 00:57:48.020
we wrote it down here.
558
00:57:48.020 --> 00:57:58.020
Now we can see here, of course, that with
1 by 4 we pi epsilon_0 integral rho v of r
'
559
00:57:58.020 --> 00:58:06.080
by r minus r´ d³ r´ nothing other than phi
of r are here.
560
00:58:06.080 --> 00:58:09.080
This is exactly what we wrote down as a term
for phi of r .
561
00:58:09.080 --> 00:58:12.080
That is, we can just write the whole thing
as half
562
00:58:12.080 --> 00:58:17.080
Integral rho_V of r phi of r d³ r.
563
00:58:17.080 --> 00:58:27.080
As I said, let's go back to the difference
when we do that now
564
00:58:27.080 --> 00:58:34.080
would discuss: What are the terms for i and
j for the point charges?
565
00:58:34.080 --> 00:58:39.080
That actually leads to the concept of self-energy
being a point landing.
566
00:58:39.080 --> 00:58:45.080
If we were to continue doing that - what we
can, but not do - if we do that
567
00:58:45.080 --> 00:58:51.080
would continue, then sooner or later we would
actually come back to
568
00:58:51.080 --> 00:58:54.080
certain contradictions come in.
569
00:58:54.080 --> 00:58:58.080
That is, this topic of self-energy of a point
charge is not yet real
570
00:58:58.080 --> 00:59:05.040
Clarified without contradiction. Honestly,
that's not really surprising either, because
571
00:59:05.040 --> 00:59:11.040
the point charge itself is only a perfectly
idealized concept.
572
00:59:11.040 --> 00:59:13.040
There is no point charge.
573
00:59:13.040 --> 00:59:20.040
Each charge is also spatially distributed
in some way and therefore I have to
574
00:59:20.040 --> 00:59:26.040
not really surprised if we keep looking at
point charges
575
00:59:26.040 --> 00:59:32.040
as a model, to come to something else or to
motivate something that we then
576
00:59:32.040 --> 00:59:35.040
also run into such contradictions.
577
00:59:35.040 --> 00:59:37.040
So we won't look at that any further.
578
00:59:37.040 --> 00:59:44.040
Now of course we can look at this expression
here that we just found
579
00:59:44.040 --> 00:59:48.040
and think about whether we can rewrite it
for something else .
580
00:59:48.040 --> 00:59:59.040
Yes, we can, because we know the Poisson equation
-Laplace phi equal to minus 1
581
00:59:59.040 --> 01:00:01.000
by epsilon_0 times rho_V.
582
01:00:01.000 --> 01:00:07.000
Or here just resolved to rho_V: rho_V equals
minus epsilon_0 Laplace phi.
583
01:00:07.000 --> 01:00:18.000
This is a term that we can use here for rho_V
without further ado.
584
01:00:18.000 --> 01:00:24.000
Then a Laplace phi times phi appears here.
585
01:00:24.000 --> 01:00:32.000
And we can replace Laplace phi mal phi in
some other way if we identify ourselves
586
01:00:32.000 --> 01:00:35.000
look at. What is divergence phi gradient phi?
587
01:00:35.000 --> 01:00:47.000
We can simply calculate that, which is then
Gradient phi Gradient phi minus phi times
Laplace phi.
588
01:00:57.000 --> 01:01:09.060
And with this expression we can of course use
the phi mal Laplace phi, which is created
here,
589
01:01:09.060 --> 01:01:13.060
if I use Poisson's equation, also replace it.
590
01:01:13.060 --> 01:01:26.060
That means, we'll just keep writing - now
I just have to see if that
591
01:01:26.060 --> 01:01:33.060
Signs are correct, but is correct, exactly-
the minus here comes from here that
592
01:01:33.060 --> 01:01:41.060
half is half and then it is simply inserted
and there must be a plus here and
593
01:01:41.060 --> 01:01:44.060
then the minus is correct here at this point.
594
01:01:44.060 --> 01:01:48.060
Minus times minus is plus, but there must be
a plus here.
595
01:01:48.060 --> 01:01:53.060
I then correct this in the slides that I upload
to OPAL again.
596
01:01:53.060 --> 01:01:59.060
Well, we can just use that.
597
01:01:59.060 --> 01:02:05.020
Once the Poisson equation, once exploiting
this vector identity and then we have
598
01:02:05.020 --> 01:02:10.020
here in front an integral over the divergence
of something - a volume integral over the
599
01:02:10.020 --> 01:02:16.020
Divergence of anything - we can transform
that into a surface integral over phi
600
01:02:16.020 --> 01:02:23.020
Gradient phi then just taken over the surface
and then we have to look at the
601
01:02:23.020 --> 01:02:25.020
Look at dependencies.
602
01:02:25.020 --> 01:02:30.020
phi has a position dependence 1 through r,
gradient phi then has 1 through r square.
603
01:02:30.020 --> 01:02:36.020
And gradient phi would then accordingly have
1 by r to the power of three.
604
01:02:36.020 --> 01:02:44.020
And if we then integrate that in infinity
over the surface of the volume, then
605
01:02:44.020 --> 01:02:47.020
it only grows with r square.
606
01:02:47.020 --> 01:02:56.020
So that we see that this integral actually
disappears at infinity here in front.
607
01:02:56.020 --> 01:03:00.080
So that only this second term remains.
608
01:03:00.080 --> 01:03:06.080
Then the two minus signs cancel each other
out and epsilon_0 simply stays there
609
01:03:06.080 --> 01:03:17.080
Integral gradient phi quadrat dr. Gradient
phi to the square -You know that E equals
minus
610
01:03:17.080 --> 01:03:25.080
Gradient phi is - and because it says grad
phi we can simply use the amount of E to
611
01:03:25.080 --> 01:03:30.080
quadrat so that we can now do another nice
one
612
01:03:30.080 --> 01:03:37.080
Have found expression to actually convey the
energy of the electrostatic field
613
01:03:37.080 --> 01:03:40.080
to indicate the field.
614
01:03:40.080 --> 01:03:47.080
If I have a volume integral here to calculate
the energy, then it stands to reason that
615
01:03:47.080 --> 01:03:53.080
that I also introduce an energy density and
that is natural
616
01:03:53.080 --> 01:03:56.080
simply defined by the integrand.
617
01:03:56.080 --> 01:04:03.040
That means, that would be epsilon_0 half times
the amount of E to the square.
618
01:04:03.040 --> 01:04:17.040
And with this expression we would be at the
end of today's unity.
619
01:04:17.040 --> 01:04:25.040
And it only remains for me to thank you very
much for your attention and then see
620
01:04:25.040 --> 01:04:29.040
we will hopefully meet in the next event.