Let´s end with a good example on how to use capacitors
and this question asks if we have a 12-volt battery connected to 3 different capacitors,
where these 3 capacitors are in parallel with their 3 capacitances
of 1, 2 and 3 microfarads, how much total energy is stored
on the capacitors at equilibrium and then we change it a little bit by adding a dielectric.
Teflon turns out to be a common dielectric.
We add this dielectric which has a dielectric constant of 2
and put it just in the 1 and 2 microfarad capacitors
and then we ask what is the new total charge stored in all 3 capacitors.
So give this a try, we have equations now for the capacitance.
The capacitance with a dielectric included as well as for the energy stored
and see what you get. We´re gonna try that here
and see what happens if we have these 3 capacitors.
Again, what´s going on here is we have some battery.
So be the positive side of our battery and this is connected to 3 capacitors that are all in parallel.
So we have a capacitor here, a capacitor here, and a capacitor here.
So what's happening here, since we have 3 capacitors in parallel, where we have 1 microfarad,
2 microfarads and 3 microfarads. The total capacitance once again for capacitors in parallel
will simply be the sum of these three. So this is the total capacitance.
In this case, the total capacitance is 6 microfarads. So now, we´re almost done.
We have an equation for the energy stored in a capacitor.
It is 1/2 the capacitance times the voltage squared
and we can always see how this comes from the equations starting with the charge
may also had the equation 1/2 the charge times the voltage.
But we also saw that the charge Q is equal to the capacitance times the voltage.
And so there´s always 2 equations or two ways to write the energy in the capacitor.
One is 1/2 Q times V, the other one is 1/2 Q which is CV times V
which is 1/2 the capacitance times the voltage squared.
So certainly, always be aware that there are be more than one way
to write the energy stored in a capacitor. So now that we know the voltage
and we know the total capacitance which we just found here,
we can say that the energy stored is 6 microfarads times the voltage
which were already given in this problem is 12 volts squared.
And so now we have 1/2 times 6 microfarads times 144 volts squared as our unit.
Now here´s where we have to be a little bit careful.
We can put all of these numbers together and see something like we could take the 1/2
and apply it to the 144, for example, or 1/2 times 6 and C3 times 144
and again we have units of microfarad times volts squared
but if we simplify this and we get an answer,
so hopefully you get something like I got which 432.
We have to be careful with our units here.
We know this unit here farads and volt squared is our energy unit, that´s joules.
But we have this micro term here and don´t forget that micro is simply a way of saying
10 to the minus 6th, one-one millionth and so this would be 432 microjoules.
This is the amount of energy stored in our 3 capacitors.
For the second part of this problem, we´re going to do one small difference
which is that we´re going to add a dielectric to a couple of our capacitors.
So we´re going to do the exact same type of analysis
but we´re going to have a different capacitance now,
not for all but for some of the capacitors that we have. So let´s see what that looks like.
In this case, what we have is a dielectric just added to the 1 and 2 microfarad capacitors
which is going to change our analysis but only slightly
so now instead of 1 microfarad and 2 microfarads with our 3 microfarads
we now have the double value because we have K being equal to 2.
So now we have 2 instead of 1. We have 4 instead of 2
but we do not have any change to our 3 microfarad capacitor
because we did not add any dielectric here.
So now, we have a slightly different total capacitance which is now 9.
So using this new value of 9, we can see that the energy stored in our capacitor
is going to follow the exact same logic except instead of 3 times 144,
we´ll get 4.5 which is half of 9 times 144 and then again the units will be the same.
We still have microfarads times volt squared.
And if you get the same thing I got, you'll see 648 and again we have microjoules as our units.
So this is an example of how we could do a capacitance problem
both with adding capacitors in parallel in this case
as well as adding dielectric into a capacitor and see what happens to a quantity
relating to capacitors like the energy. This also concludes our discussion of capacitors.
And now that we have both resistors and capacitors discussed,
we´re ready to discuss some more complicated circuits
and see how we can put all these together.
Thanks for listening.
Thanks for listening.