In this lecture, I want to present to you how you can calculate the energy in electricity based on
our established equations for the electric field and for the electrostatic potential.
So to derive the equation for the energy and electrostatic, we need to go back to one of the very first
slides of this cause this was in one of the first sections on the Maxwell's equations where we have
established here this continuity equation for the energy density.
And here we have found out that the energy density is given by such a term where you have the absolute
value of the electric field squared, plus the absolute value of the magnetic field squared.
But since we are in electrostatic, we do not have these magnetic fields, so we can focus only here
on this term.
Now, since this is the energy density, we need to integrate over the whole space of the whole volume
to get the total energy.
So this is the expression.
And if we know what the electric field independence of the position vector art looks like, then we
can quite easily determine the energy of the whole system and we are done.
But also, I want to give you here an alternative method or an alternative expression on how you can
calculate the energy if you do not know the electric field yet and if you only know the charged density
distribution.
And to do this, we use this and that's here where we have introduced the electrostatic potential and
we ride the electric field in terms of this gradient of this scale scalar potential.
So he can see I have only replaced one of these eyes.
So this is square.
So it's E times E, obviously.
And so I have really expressed one of these ease in terms of minus Knobler five.
So the next step, what I do here is I first show you the results, so when you look at this term here,
you have to let Nabala act on both of these terms here on the field and on the scale of potential.
Fine.
So you have to apply the product rule, which gives you two terms.
So the first term would be e scalar product, not acting on Fi's.
This is exactly this term here.
But then you get another term, which is five times or five times the divergence of this electric field
with a minus sign.
So since this term is not here, we have to get rid of it again to compensate it.
So we add this term here.
So we know now that these two equations are equal.
Now, in the next step, I use this relation here, which is one of the maximal equations that the divergence
of the electric field can be expressed in terms of the charged density.
And also so let's hear this right term.
And also I change I express this left term here.
And so if we go to one dimension, maybe you better understand what I did here, some what you see is
we have an integral of a derivative and if you integrate a derivative, you get the initial function
at a certain position.
And so what this does here is we basically.
Get rid of one of the integration's, so this is a three dimensional integration, so these are actually
three integrals here.
And when we get rid of this gradient, we are left with only two integrals.
So this here is a surface integral.
And this is maybe a bit too complicated, but let me tell you that this term here describes the surface
charges of a system.
So, for example, if we have a sphere and we have some charged separation, then this term contributes
an energy.
But if we make our system infinitely large or if we have a system that is finite but does not have any
surface charges, then this term here is zero.
And for simplicity, we want to consider this case.
So we are only left with this right hand term, which is quite easy to derive when you look here.
We have just used the Maxwell equation.
So I hope that you understand this.
So if we have no surface charges, then this is here the energy of the whole system.
And you can see we do not need the electric field anymore, but we only need to charge distribution
and the electrostatic potential.
So now we can go even one step further, because we already know what the electrostatic potential looks
like and that it only depends on the charged distribution.
So we take this expression and put it in here and you see that this is the general expression for our
energy, and it only depends on the charged distribution.
So you have such a product and please notice that you have to you cannot simply square this term here,
but you have to integrate of a different position vectors.
So you have to do two integrations, one after another, the first integration of R and the other one
over the other arm.
And the reason is that we have this term here where we have a difference, quite a distance of two precision
vectors.
Now, if we have parking charges, we can integrate over these charge distributions because these charge
distributions will then be zero almost everywhere and will only be finite at certain points.
And what we get is then that we integrate of our role.
And Debe, so we get just total charge of this point and we have to add up all of these powers and we
cannot consider himself into action because then the position vector distance would be zero and we would
divide by zero.
So that's not possible.
And if we only have two charges, then we get this famous expression that you probably know already
is here.
Now, the energy of two point charges.