So, the energy is proportional to the square of

the total charge and inversely proportional to the radius,

and we can interpret the equation as saying that the average of one over

Rij for all pairs of points in the sphere is 60 percent,

three over 5A, which we will derive shortly.

So, take a look at this, the Qr,

which is the total amount of charge of existing sphere is equal

to the volume density of the charge times the volume of the sphere

which is four over three times Pi r to the power three,

and if I do infinitesimal change of this dQ,

then we know it becomes four Pi r squared dr times the charge density.

So, if we put this into our nominator

then dU becomes four Pi Rho square r to the power of

four dr over three Epsilon naught and if you

do integration from zero to the radius of the final sphere A,

then we come up with this equation where you have three over five times Q squared over

four Pi Epsilon naught A. I'm going to ask my teaching assistant,

Melody, what meaning it has,

three over five times Q squared over four Pi Epsilon naught A,

what does that mean?

Melody, how can we visualize this number in other form?

So, I think before we learned

that the energy is proportional to the amount of charge incorporated within a system.

However, and that would be your original q squared over four Pi Epsilon naught.

However, in this case,

it's around 60 percent of what the energy would be of both of those two charges.

Exactly. That's very good.

So, if I add a little bit to what Melody already explained,

the electrostatic energy between

two point charges would charge quantity Q and a distance separated by A,

you can understand the energy will be Q squared over four Pi Epsilon naught A.

So, this could be a reference to compare with.

So, this sphere, which has a total amount of

charge equal to one of the point charge here will have the

same or almost the same order of magnitude of

energy by putting that in a uniformly charged sphere.

It is about 60 percent of the energy accumulated between

two point charge with the same charge quantity. Thank you.

So, the next example that we're going to tackle is

the energy of a capacitor and forces on charged conductors.

So, essentially, the shape and size are the same

as in the case of uniformly charged sphere, but in this case,

we are bringing in a conductor instead of nothing,

and see how the charge will behave,

and how much energy will be stored.

So, how much work is done in charging the capacitor is another way to ask this.

How much work is done in charging the capacitor with Q, for example.

We imagine that the capacitor has been charged by transferring

charge from one plate to other in small increments, dQ.

So, now, we're going to think of uncharged conductor and bringing

charge from infinity and injecting charge small,

bit by bit to make it fully charged.

So, we are thinking of a small increment dQ and we know that

the internal energy or the electrostatic energy of

capacitor is equal to the potential times the charge,

potential times the charge.

So, what we can do is doing infinitesimal change of

the energy dU and because the voltage in the conductor is constant.

So, Melody, voltage should be constant inside a conductor.

Why is that the case?

Because there is no movement of charges and there is

no gradient in potential because the charges will equilibriate to a stationary place,

and therefore the voltage is constant when there's no gradient.

Very good. So, if I complement or add a little bit to Melody's comment,

in a conductor, charge are free to move. They're free to move.

So, if there is any electric field inside,

they will not stop moving around so that violates the electrostatic world.

So, in order for them to be static,

we need to make sure there's no electric field inside a conductor,

and electric field is derivative of electrostatic potential.

So, if there's no electric field anywhere inside the conductor,

there is no gradient of electrostatic potential leading to

constant voltage throughout the space and only place they can reside is on the surface.

So, the voltage will be kept constant and we will

only change the amount of charge that we inject into the system,

and that's why it's VdQ.

Now, we already proved that capacitance times voltage is equal to the charge.

Capacitance times voltage is equal to Q. CV is equal to Q.

Therefore, we can replace V by Q over C,

that's equivalent, and then if we rearrange the equation,

it will be QdQ over C. So, from here,

we're going to do integration from zero to Q because we are

injecting charge bit by bit from zero to the final amount, Q.

If we do this integration,

it will result in one over two Q squared over C,

and again CV equals Q.

So, if I replace Q by CV,

then it will be CV square on the numerator and divided by C,

so it becomes one over two CV square.

We also know that capacitance of a conducting sphere is

about is four Pi Epsilon naught A.

If I replace it with this one here,

then the conducting sphere will have electrostatic energy of one over

two Q squared over four Pi Epsilon naught A. I'm going to ask again,

Melody, what this means as I asked her

about what the energy of uniformly charged sphere means.

So, Melody, what does this mean?

I think it's similar before again.

You can think of the energy as being proportional to the number of charges.

So, if you had two charges of equal energy like before,

then it's about half of what the energy between these two charges would be.

Exactly. So, with the same reference now we are in the position to compare

the electrostatic energy with the same amount of charge injected

into conductor versus forming a uniformly charged sphere.

So, in a nutshell,

spherical shell versus uniformly charged sphere.

We can compare apple to apple.

The energy of a charged conducting sphere where a thin spherical shell of total charge

Q is injected is just five over six times of the energy of a uniformly charged sphere.

So, it means, it takes less energy to charge

a conducting sphere than to form a uniformly charged sphere.

So, let's take a moment to think about it.

If I have the same amount of charge,

similar amount of charge and I want to

put it within the boundary of a sphere with a radius of a,

which way is preferable, Melody?

The first one.

Yes. It takes less energy than this one,

so this will be the preference over this one. Thank you.

Now, we're going to change the gear,

the topic, and play with another example,

which is a parallel plate capacitor,

which is used in your smartphones or even for a grid, electrical grid.

You have parallel plate capacitors to instantly store a lot of charge.

Also in the railway, you can find it.

So, it is pretty universal and we're going to think about the force

between two plates in

the parallel plate capacitors using a different way than the way we used before.

So, the way we used before is thinking about

the charge amount and the force between them, but here,

we're going to think about the energy stored in

the system and perturb the system in a way to displace it a little bit,

and see how much energy is caused,

and using that energy caused,

we're going to calculate the force that is acting between those two plates,

and we call that virtual work.

So, here, as described here,

if we imagine that the spacing of the plates is increased by the small amount,

Delta z, that's imaginary,

we didn't displace it,

but if we think we did,

then the mechanical work done from the outside in moving the place would be

Delta W and that is equal to force times the displacement, Delta z.

All right. So, let's take a look at

the case here where you have a parallel plate capacitor.

The energy, total energy,

U is equal to one over two Q squared over C. We just proved that.

Now, if I take Delta infinitesimal change of U,

then because Q is constant,

the charge we put on the plate are staying there,

all the charge are staying there.

So, Q is constant. So, the change only is coming from the capacitance change,

and if you think about a capacitance,

and do the Delta operation here,

what results is you see F Delta z is equal to Q squared over two times Delta one over C,

and one over C is minus Delta C over C squared.