So maybe the last exercise was a bit difficult when you tried to solve it yourself, because it was
the very first time that you are dealing with these problems.
But I hope that once you have seen the solution, it all made more sense to you and that you can now
deal much better with these kinds of problems.
So this is why here is another exercise.
This time we will calculate the electric field and the electrostatic potential of a spherical capacitor.
So, yeah, these are the first two tasks, and then we will also calculate the voltage between the
two metals of the capacitor and we will calculate the capacitance.
But like in the previous example, the only real difficult task is task number one and then maybe task
number two.
The other two tasks are quite easy.
So this is what our capacitor looks like.
We have metal, number one, and green and metal number two in blue.
And we have here they are both spheres, basically, or surfaces of spheres.
And they have the same middle point, same zero position.
And to charge on both of these metals of the capacitor is equal in magnitude, but opposite insigne.
So the main difference to the previous example is that there is no charge density inside here, so you
really only have this charge density on this metal on the surface of the sphere.
And now we do not only have one sphere, but we have two spheres and they have opposite charges.
So please go ahead and post a video and really try to solve this problem yourself for the region inside
of this green metal, then for the region between the two metals and then for the region outside of
the blue metal.
OK, so now I will present to you the solution.
So the first task was to calculate the electric field of this capacitor for all three regions.
So we begin with the region inside of the first metal and green.
So this is for our smaller than a and here the solution is really very simple because there is no charge
and closed.
If we so so let's say we we consider here sphere of reference inside of this green metal.
And so inside of the sphere of reference, there's no charge included.
So this means he was zero and even row of R zero everywhere.
And so this means that this integral here on the right hand side will be zero.
And the only way this is possible is that E is equal to zero.
So there's no electric field inside this metal.
No one.
Now, for the second region, we consider the region between these two metals.
So this is for A smaller are smaller B.
So this time, the right hand side integrates up to cue to capital cute because you can see there's
a charge on this on the screen metal in total, we have to charge.
Q So we have that this integral is equal to Cube.
Now we can of course, consider again this radial symmetry.
So this means we can pull out this E and we just have to calculate here to surface integral over the
S, which is just the area of the surface of the sphere that we are considering and where the radius
is somewhere between A and B..
So this means the surface is four PI Times Square, as we had previous examples.
So this is the equation that we get and now we just solve for E.
So this means each of our is equal to this prefecture.
Times Cube divided by square times, this radio position vector.
And maybe you remember, this is exactly the same field as we had in the previous example outside of
the sphere, so it does not matter at all if this sphere here is filled or not.
It's just this whole electric field just depends on the enclosed charge, but not really on the distribution
of the charge as long as we have some radial symmetry.
Now for the third region, we go outside of the sphere, so we have here our larger than B and here
we have the same argument as in the first region.
We have these two charges, plus Q and minus Q and if you consider a sphere that is even larger, the
enclosed charge is zero.
Those means here we get to zero and then this electric field must also be zero.
And so this is our solution to the first task we have these electric fields and these three regions,
and this is what the magnitude of the electric field looks like in these regions.
The second task was to calculate the electrostatic potential, and so since we have these solutions
here for the electric field, we just have to integrate over the electric field to get the electrostatic
potential.
But also we have to be careful because we have to consider the integration constant to make the electrostatic
potential continuous.
So what we get for the inside and the outside region is, of course, that the electrostatic potential
is a constant number because if we integrate of a zero, we had zero plus an integration constant.
And so as we get to your two constants, that would probably not be the same.
So we have to determine these later and for the middle region, for R between A and B. We have to integrate
this term here.
So we get minus one of our and then we get to another minus signs.
So this will be the solution and then we get another integration constant.
And so now what we have to consider or what we have to determine are these three integration constants,
because here we have constant here we have some constant and here we have one of our dependence and
some constant.
And on these on these points, A and B, these electrostatic potentials have to be continuous.
So this is you, of course, the electrostatic potential and not the electric field.
So what we have to do first is we just have to set one of these constants to some value.
And I think it always makes sense to me to go to our to infinity and to make sure that the electrostatic
potential is zero.
In this case, it's not really something you have to do, but it's often done.
And I think it makes sense.
And to you, it's very easy because it's constant.
We have constant electrostatic potential for large hours.
So we just take this constant and set it to zero.
So now we know what our electric and our electrostatic potential looks like at this position, B, so
it's zero.
And so this means when we take this solution here and set our equal to B, then we must also get zero.
So this means that this constant must be equal to this prefecture here times Q divided by B and we need
to have A minus sign.
So we can write it down like this and then we have this solution, we can determine what this electrostatic
potential looks like at the position are equal to a test would be, of course, you one over a minus
one of a B and D prefecture.
So this constant here for the first part must be exactly this constant.
And so this is our whole solution.
We have now the electrostatic potential for all three parts.
And since we have chosen the ticket, the electricity potential constant here for this region to be
zero, it is now very, very easy to solve the third task.
It is that we have to calculate the voltage between the two metals and the voltage is just a difference
in the electrostatic potential.
And so in case you have chosen different constants here, this is absolutely fine, it will it will
still be the correct result, but it could be different because all of these curves here could be shifted
by some constant.
But now for the third task, you should get exactly the same value as I get.
So here is my solution.
So we had inside this electrostatic potential, which is just as constant.
So there is no our dependence.
And outside we had also know our dependence here.
We had that the electrostatic potential is zero.
And so the difference in electrostatic potentials is just this value.
So our voltage difference is kieu divided by four pipes and zero and then one over a minus one of a
B.
And now the solution for the fourth task is also quite straightforward.
So here we had to calculate the capacitance of our capacitor and this capacitance is just the ratio
of the stored charge divided by the voltage you.
So, yeah, we of course have to charge.
Q And we know our voltage you so we can just set it in here and now this Qs cancel out and we can make
the rest look a bit more beautifully.
So this is the solution that we get.
It's for Pi Epsilon zero times B divided by B minus A.