So now that has come to an exercise for you, we want to calculate the electric field and the electrostatic
potential for a charged sphere so the tasks that you have to solve are calculate the electric field
of a homogeneously charged sphere.
And as a hint, you have to distinguish the inside region of the sphere and the outside region of the
sphere.
And you should use the rotational symmetry of the problem to make it a lot simpler for you to solve
the integral.
And you have to consider that homogeneously charged means that this row here it is charged density is
constant.
So it means you can pull it out of the integral.
And then the second task is quite easy when you have determined the electric field from the electric
field to calculate the electrostatic potential.
So if you want to have another Hynde's, please keep on watching the video, otherwise you can now stop
the video and go ahead and solve the problem yourself.
So the bigger hint is that, of course, we have already solved this problem for the region outside
of the sphere.
So this was as a solution to Coulomb stop where we have integrated here over the charged density and
got to that charge.
Q And on the left hand side, we could pull out to the electric field because we have this dependence
artist, a cemetery where the electric field is always pointing along the radio direction.
So this vector here is parallel to the vector from the integral.
And so this term here would just be the surface of the sphere that we are considering for our integration.
And so this gave us the electric field, which is proportional to you, divide divided by our square.
So the first part we have actually already solved, but still it's a nice exercise to do it yourself
once again.
And now what is about the region inside of the sphere?
So this you really have to do yourself.
So please post a video and just continue when you have tried for a few minutes.
So since this is the first time you are doing this, please don't expect to be able to solve it just
right away, because even though it may be easy, once you have seen the solution, sometimes it can
be quite difficult to get the right idea and to come up with a solution.
So please don't give up.
Just really spend some minutes trying to solve this.
And when you've done it or when you need more hints, please continue the video.
And now I will present you the solution.
So everything that we need is the Maxwell equation and its integral formulation.
So this one here, then we need the symmetry of the problem, which allows us to write down the electric
field as a scalar function multiplied by this vector are.
So it means the electric field is always pointing along the radio direction.
And so this means that this is always parallel to this D.R.C. here.
Now, in previous light, I have already shown you what the solution looks like outside of the sphere,
so this means this small R is larger than this capital R, which is the radius of the sphere.
And so our right hand side of this solution is always the volume of this sphere.
So it is four divided by three times PI times capital R to the power of three.
And in this charge density and of course this one, we could also write down as the inclosed charge,
but it's not really necessary.
Now, the left hand side you can see here, I've already pulled out this E of our so we are just left
with the surface integral of the way of the sphere that we are considering here.
So this would be this black Dasht sphere with the radius small R and not capital R, so the surface
of a sphere is four pi times the radius square.
So this is what we have here.
So as you have seen on the previous slides, this is already a solution.
We just have to divide by Epsilon zero by four pi r square.
And so this is the solution that we come up with so we can express it like this or we can reintroduce
the inclose charge and then we get the result that we had previously.
Now, if you have the solution, it is not so difficult to find a solution for the inside region.
The main difference here is that the right hand side of this equation looks different because here the
right hand side of this equation was always the volume of our charge, so of our charged sphere.
So this is because this road density is only finite for this sphere.
And in this wide region here, it would be zero.
So it would not contribute here to this integral.
However, now if we look on the inside, this black Dasht sphere here would be inside of this blue sphere.
So this means now the radius of the black sphere really plays a role.
So this means here for this integral, we do not have to consider the radius of the black out of the
blue sphere capital R, but we have to consider the radius of the black sphere smaller.
So this means here the left hand side is exactly the same, but the right hand side is now four pi four
divided by three times pi times smaller to the power of three times real.
And this means that this is now our solution, so why we had to hear a one of our square dependent's
we have now here and our dependence, and this is, of course, because here instead of this capital,
are we have now a small R and so these power of three and this power of two, they council and we are
left with power of one.
Now we can look at this electric field and you can see that on the inside we get our dependence.
So it's a linear function.
It increases, which is, of course, because when we go far away from the center of this sphere, we
have more a charge generating these electric fields.
So this means here we have this linear increase in the electric field.
And when we are outside of the sphere, the electric field decreases with the power of one over R to
the power of two.
And at the position are we can also calculate the value.
This would be divided by three epsilon zero times capital R.
And as you can see, this is this value we get from both solutions, of course, because the electric
field has to be steady.
So here we would get capital R straightforwardly and here we would get R to the power of three divided
by R to the power of two, which gives us the same value.
Now, the second task was calculate the electrostatic potential, and so here we just have to take our
solutions from inside and outside of the sphere for the electric field.
And now we have to calculate the electrostatic potential by this equation here.
So basically, we just have to integrate over the electric field.
So here we get as a solution that we get here an exponential of this hour that is increased by one.
So instead of R to the power of minus two, we get R to the power of minus one or we can write it down
as one of R and then we would also get a minus sign.
But he is also a minus sign.
So these cancel out.
So this is what we get for the electrostatic potential.
And of course, you can remember that we could add here any constant to this electrostatic potential
because we can shift our zero position of the electromagnetic potential as we wish.
But here I have chosen it such that it is zero when we take the position vector to infinity.
So if you go with R to infinity, this whole thing here would go to zero, which is something that,
yeah, looks good or makes sense physically, but it's actually not that important.
It's important that you have this term and then you would have here another constant possibly.
Now for the inside region, we also have to integrate here.
So what we get is we have to increase this exponent here so we get all square and then we have to divide
by two.
And then also we get you just minus sign from this minus sign.
So this is what we get for the electrostatic potential.
But also we get here another constant.
And now this time the constant is really important because at the position are is equal to capital.
Are these two electrostatic potentials need to match each other.
So it must be a continuous function.
So if we have assumed here that our constant is zero, we can now determine the constant here so that
these to match at the position are equal capital R so you can see that this solution and we can also
write it like this.
And if you now put in here to Capital R, you will see that you will get here two times Capital R Square.
So you get real divided by three Epsilon, zero Times Capital R Square and you, you get the same thing
because you used to cancel out and we are only left with a capital R Square.
And so this is what the electrostatic potential looks like.
So it keeps on decreasing yappers basically because of the shape of our electric field on the previous
slide.
And then we have here a continuous function and to position R, which is really important.
So I'm just saying this one more time.
You could add here a different constant and then you would have to add here the same constant additional
to this constant.
So when you plot your electrostatic potential, it should look exactly like this, but this whole function
could be shifted by any value.