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The previous lecture we have discussed, the Maxwell equations and I have shown that several of the



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very early laws of electrodynamics are straightforwardly included in the Maxwell's equations.



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In this lecture, I will show you that even the Cullum's law can be derived from the Maxwell equations



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quite easily.



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So we have discussed in the first section of this course that Cooloola Law was one of the laws that



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was found experimentally quite early.



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So already in the 17th, 60s and 70s and 80s, Robinson and Coulomb established that force between two



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charged object depends on the distance and follows a one of our square law.



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While Robinson found that the exponent is minus two point zero six, it's much more reasonable.



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That is an integer number.



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And also we have here as prefect, which is one of our four PI times this constant Epsilon zero.



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And now I want to show you where this all comes from.



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So we take one of our integral Maxwell equations.



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So this is where we relate to electric field, to the included charge density in a volume V..



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So in general, what this could mean is you have some charge distribution row, which depends on the



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back to ah, and this can have some very in homogeneous shape, for example, that can have holes here



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and then there can be another charge and it can have some funny shape.



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So what you would do then is you would take volume.



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So consider that this is here actually three dimensional and not two dimensional, and you take then



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the surface of this volume and integrate over this volume and you can thereby determine the electric



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field.



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So here we want to consider a more special example, a specific case, and what we want to say is that



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our charged distribution is not random and not that general as in this case, but it is rotational symmetric.



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So it is like a sphere.



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So it can change in intensity starting from zero going along this radius here by the test to be radio



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symmetric.



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So what this means is we get this typical shape for the electric field so the electric fields will be



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parallel to the radial vector e r, so it will be pointing along the radial directions.



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So what we can do then is we can write down our electric field as vector here in terms of a scalar of



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our.



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So this is also a scalar and not a vector anymore.



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And we multiply it just by this radial vector.



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And also, as I mentioned, please remember that these are not two dimensional shapes, as you might



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think, following from the sketch here, but it is actually a three dimensional sphere.



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So what we have to do then is we have to understand what this this means here.



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So this is the infinitesimal element of the surface.



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So it has a magnitude and it has an orientation which is characterized by the normal vector.



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So the vector the as is equal to the normal vector times this scalar disease.



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And now we know that we have to radiometric field.



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We also know that the electric field is always parallel to end and therefore it's always parallel to



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the esq.



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So this means this term here to scale our product Yakitori Times vector, the S is just the scalar version.



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It's just a scalar of E times as scalar AFTRS.



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So what we can do then is we can pull out of our because we want to consider here some, some volume,



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some fixed volume of some fixed size.



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So it does not any more depend on R and we therefore can pull out of R so we can write Epsilon Zero



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Pallotta of our.



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So there's this one here and then we have just this closed integral over the surface of DV and integrate



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over the S which is this volume element or sorry, not the volume element but the surface element is



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really important.



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So we integrate the surface infinitesimal element over the whole surface.



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So what we get is just the area of the surface of the sphere.



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And I hope that you can remember this.



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The area of the sphere is four pi times the radius square.



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So we get Epsilon zero E times for PI.



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Our square is equal to the inclosed charge in this sphere.



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So as long as this sphere that we consider here is larger than the radius of this charge, then this



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equation here holds and then we can just solve it for each of our and we get one over four percent and



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zero times Q over our square.



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Or we could also write it down in this pictorial representation of the vector of he is equal to this



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magnitude here, multiplied by it is radial vector.



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So as we can see, we found that it is one of our square independence and we have even established the



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prefecture.



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And now, of course, if we want to calculate the force that acts on another charge, we have to multiply



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this other charge times the electric field.



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So then we get here it is Q1 times, Q2, one of our square dependent's that we had on the previous



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slide.