Now that we´ve discussed both parallel plate capacitors and what they are
as well the energy and voltages on a capacitor,
let´s do one last thing and talk about that dielectric
that we said we could add to a capacitor and that dielectric constant value.
We said that the capacitance for a parallel plate capacitor,
where we had a vacuum between the two plates,
was just epsilon times the area of the capacitor plate
divided by the distance between the plates.
But we also saw on our equation that we could add something
in the capacitor to try to increase the capacitance of the plate.
We do this by increasing and changing the permittivity between the two capacitor plates.
In order to do this, we insert some objects, some sort of substance between the two plates
that has a different effective dielectric constant.
So by adding this material which we call a dielectric,
we change our equation for the capacitance of a capacitor with the dielectric included
with this factor of K and so if we have some material that's put between these two plates
that has a dielectric constant of K, we've increased the capacitance of the capacitor
by that value by changing the permittivity from epsilon nought to K times epsilon nought.
It´s important to know that this K this dielectric constant always has to be greater than one
and also that the capacitance with the dielectric simply be K times the capacitance without the dielectric.
Let´s look at what happens if we added dielectric in a few different circumstances.
For example, if we have a voltage being applied to the two plates of our parallel plate capacitor
and we know that the capacitance of the capacitor with the dielectric
is K times the capacitance of the capacitor without the dielectric.
Then we can look at a few of the variables that we´ve already discussed.
We know that for a given voltage, and that´s why we have here
since we have a battery hooked up to our capacitor, Q equals C times V
meaning that if we change the capacitance to be increased by a factor of K
at a given voltage, the charge stored on the capacitor will increase.
Similarly, since we know that the energy stored in the capacitor is 1/2,
the capacitance times the voltage squared,
at a given voltage meaning V isn't changing,
we increase the capacitance by a adding a capacitor or adding a dielectric rather.
Then the energy that´s stored will increase.
This is different from the opposite scenario.
What if we disconnected our battery so we use the battery initially,
we´ve charged the capacitor and then we disconnect the battery
and then the charge is stuck on or stored in the capacitor?
In this case, let´s see what happens when we increase the capacitance again by adding a dielectric.
We will see something very different happen.
So by rearranging the equation for the voltage since we have a different constant value.
Again we have a constant charge because the charge can´t go anywhere, stuck on the capacitor.
It has nowhere to flow and no wires to go through.
This means that the charge is constant while the voltage will change.
So since voltage is equal to charge divided by capacitance,
if we increase the capacitance, what we´re going to do is decrease the voltage
because the denominator of this expression will increase by a factor of K.
Similarly, if we have a given amount of energy on a capacitor,
then again the charge on the capacitor cannot change because the charge doesn't have anywhere to go.
And so if we increase the capacitance, we´re going to get the opposite effect
by inserting a dielectric into our capacitor.
If there´s no voltage attached, we can actually decrease the amount of energy stored in that capacitor.
With the capacitor, we have one last thing to worry about
which is that a capacitor is a time-dependent object.
What we mean by that is that when we have a capacitor and we hook it up to a battery,
it´s going to need to charge up. There´s going to have be this time process
where the charge is built up on either plate of the capacitor.
And so before we've charge the capacitor, if you can see on the graph here on the top right,
what we have is an x-axis or a horizontal x-axis that's telling us about the time
and the vertical axis which is telling us how much charge is stored on the capacitor.
So for our capacitor, what happens is we start at zero charge, so you can see on the graph here.
We´ve started at a charge on the plate as zero and then as time goes on,
the charge that's on the capacitors because the battery is pushing the charge onto the capacitors
will increase until it reaches our equilibrium value.
It´s going to sort of slowly taper off and try to approach.
The charge value for our given capacitor which we discussed, was the capacitance times the voltage.
On the other hand, if we are discharging a capacitor
so say we started at a full charge and then are now using a capacitor to may be light a light bulb.
We'll start with the full charge and then approach to zero charge as the capacitor discharges.
So on the plates that you see on the bottom here,
what´s happening is the charge on the plates is being pushed away and it's equilibrating.
The negative is going to be the positive and vice versa.
As the current flows, we can light up some applications.
So for example, a light bulb again but then slowly overtime
that will die off and we´ll end up with zero charge on our capacitor.