Great job, you made it through this long section about electrostatic.
Let us summarize what we have learned.
So in the first lecture, we have considered the Maxwell equations for electrostatic.
So here we are only concerned with the electric fields and all of the time derivatives of these fields
are zero.
So this leaves us with two equations and one of them becomes zero on the right hand side.
This in turn allowed us to establish the electrostatic potential because if you write down the rotation
of a gradient of a scalar function, it's always zero.
So this means it must always be possible in this case to write down the electric field in terms of a
gradient of some scalar function.
And this function is called the electrostatic potential.
Now, when we use the other Maxwell equation, we can derive the song equation, which is essentially
a second order differential equation, because this is here.
There's a class operator that includes the sum of all the partial derivatives with respect to the coordinates
in the second order.
And on the right hand side, we have here the charge density of our.
Furthermore, we have established the electrostatic potential and then also the electric field for a
general charged distribution.
So when you have any charged distribution, you can take these expressions to calculate fire and eat.
So this we have used together with the integral formulation of this Maxwell equation to solve several
several examples.
For example, here we have calculated the electric field of a homogeneously charged sphere.
So since this is rotational symmetric, we could immediately say that our electric field, this vector
here, is always oriented along the radial direction.
And we only had to consider here the absolute value of R, and the result was that inside of this sphere
we get a linear dependence on the radius because the included charge in such an imaginary circle here
increases when we make the radius larger.
However, outside of this circle, the electric field decreases with the power of our to the power of
minus two.
So this is here the famous equation that you probably knew already before this course.
And we could also solve another example by exactly the same method, which is the spherical capacitor.
So here we have two metals that are charged oppositely.
Q and minus.
Q And we could hear calculate the electric field in all three areas and we could calculate the electrostatic
potential.
So for this, it's very important to remember that the electrostatic potential must be continuous here
at these positions.
Furthermore, we could also calculate the capacitance based on the voltage which is given by this expression.
So we have here one of a..
And one of a B, which corresponds to these radii here of Metal One and metal B, uh, metal one.
Metal two.
Then in one of the longer sections, we have discussed the electric dipole moment.
So this is one we have two charges, for example, here, a negative Q and positive Q that are separated
by a vector A..
So then we can define this dipole moment P, which is given by this expression.
And it took us quite a long time to calculate this expression here for the electric field.
And you can see that the orientation of the dipole moment P really plays a role in these two terms.
However, far away from this dipole moment, this field behaves like one of our to the power of three,
because here we get our square, here we get our square, and here we divide by our to the power of
five.
Furthermore, we have considered the case where we take this dipole moment and put it into a homogeneous
and constant electric field.
And what we found is that this dipole moment reorients and it aligns with the electric fields.
So this means on the center point or on the total configuration of these two charges, there's no charge.
There is no force acting here.
However, we have a talk that leads to this reorientation.
And you can see this if you consider here a constant electric field that we get this talk.
However, we do not get the force because the gradient of a constant is zero.
In one of the last sections we have discussed and arrived the energy of electrostatic.
So if you know the electric field, for example, here in these two examples, you can just take this
equation that is based on the continuity equation and just integrate over the electric field squared.
However, when you have not when you do not have the electric field, you can just take the charged
distribution.
And the very last section of this of this part of discourse, we also discussed boundary conditions
and image charges and we found out that the boundary conditions are actually very important because
if you if you have a single charge and vacuum, the electric field configuration looks like this.
It's radial symmetric.
However, once you bring this charge in the vicinity of a metal where you have electrons inside of it,
these electrons will react.
And this is because the boundary of this metal has to fulfill a boundary condition.
It has to fulfill that the electrostatic potential is constant, for example, can be zero.
And in order for this to be true, these field lines here must be perpendicular to this boundary.
And this can only be the case if the electric field profile looks as is shown here.
So the electric fields of this single charge in the vicinity of a metal behaves as if there were another
charge inside of the metal.
And this is called the mirror charge.
And that emerges because the electrons react to this charge outside of the metal.