The previous lecture we have discussed, the Maxwell equations and I have shown that several of the
very early laws of electrodynamics are straightforwardly included in the Maxwell's equations.
In this lecture, I will show you that even the Cullum's law can be derived from the Maxwell equations
quite easily.
So we have discussed in the first section of this course that Cooloola Law was one of the laws that
was found experimentally quite early.
So already in the 17th, 60s and 70s and 80s, Robinson and Coulomb established that force between two
charged object depends on the distance and follows a one of our square law.
While Robinson found that the exponent is minus two point zero six, it's much more reasonable.
That is an integer number.
And also we have here as prefect, which is one of our four PI times this constant Epsilon zero.
And now I want to show you where this all comes from.
So we take one of our integral Maxwell equations.
So this is where we relate to electric field, to the included charge density in a volume V..
So in general, what this could mean is you have some charge distribution row, which depends on the
back to ah, and this can have some very in homogeneous shape, for example, that can have holes here
and then there can be another charge and it can have some funny shape.
So what you would do then is you would take volume.
So consider that this is here actually three dimensional and not two dimensional, and you take then
the surface of this volume and integrate over this volume and you can thereby determine the electric
field.
So here we want to consider a more special example, a specific case, and what we want to say is that
our charged distribution is not random and not that general as in this case, but it is rotational symmetric.
So it is like a sphere.
So it can change in intensity starting from zero going along this radius here by the test to be radio
symmetric.
So what this means is we get this typical shape for the electric field so the electric fields will be
parallel to the radial vector e r, so it will be pointing along the radial directions.
So what we can do then is we can write down our electric field as vector here in terms of a scalar of
our.
So this is also a scalar and not a vector anymore.
And we multiply it just by this radial vector.
And also, as I mentioned, please remember that these are not two dimensional shapes, as you might
think, following from the sketch here, but it is actually a three dimensional sphere.
So what we have to do then is we have to understand what this this means here.
So this is the infinitesimal element of the surface.
So it has a magnitude and it has an orientation which is characterized by the normal vector.
So the vector the as is equal to the normal vector times this scalar disease.
And now we know that we have to radiometric field.
We also know that the electric field is always parallel to end and therefore it's always parallel to
the esq.
So this means this term here to scale our product Yakitori Times vector, the S is just the scalar version.
It's just a scalar of E times as scalar AFTRS.
So what we can do then is we can pull out of our because we want to consider here some, some volume,
some fixed volume of some fixed size.
So it does not any more depend on R and we therefore can pull out of R so we can write Epsilon Zero
Pallotta of our.
So there's this one here and then we have just this closed integral over the surface of DV and integrate
over the S which is this volume element or sorry, not the volume element but the surface element is
really important.
So we integrate the surface infinitesimal element over the whole surface.
So what we get is just the area of the surface of the sphere.
And I hope that you can remember this.
The area of the sphere is four pi times the radius square.
So we get Epsilon zero E times for PI.
Our square is equal to the inclosed charge in this sphere.
So as long as this sphere that we consider here is larger than the radius of this charge, then this
equation here holds and then we can just solve it for each of our and we get one over four percent and
zero times Q over our square.
Or we could also write it down in this pictorial representation of the vector of he is equal to this
magnitude here, multiplied by it is radial vector.
So as we can see, we found that it is one of our square independence and we have even established the
prefecture.
And now, of course, if we want to calculate the force that acts on another charge, we have to multiply
this other charge times the electric field.
So then we get here it is Q1 times, Q2, one of our square dependent's that we had on the previous
slide.