in this video we introduce dielectrics or insulators and we talk about the behavior between
the plates of a capacitor so first let's see what happens qualitatively if we charge up a capacitor
disconnect it from the power source that's the first picture here and then insert a dielectric
substance between the plates that's the second picture here now charges are not really free to
move around in this dielectric that's sort of the definition of an insulator but you can stretch the
electron clouds around the molecules a little bit and i have a downward electric field between these
two plates that means there's an upward force on the electrons around all the molecules in the
insulator and so i can stretch the electron clouds upward and i end up with a picture like this
now inside the dielectric material everything is still neutral because every negative on
these little electric dipoles is right next to a positive on the one above it
but something significant happens at the surface of the insulating material i can see i have
a bunch of electron clouds stretching upward right to that upper surface of the dielectric
and that means there's a small negative charge on that surface likewise at the bottom surface of the
dielectric i've stretched my electron clouds a little bit upward leaving a slight positive
charge on the bottom surface of the dielectric the net result of this polarization is that the
dielectric itself now looks like a parallel plate capacitor with a positive plate on the bottom and
a negative plate on the top so i end up with a small induced electric field pointing upward
so this small induced electric field partially cancels the original electric field
and that's where we begin getting into the mathematics here i'm going to say that my new
electric field is equal to the original divided by some constant and traditionally we use a kappa
for that and that kappa is called the dielectric constant that's just a property of a material and
every different type of material has a different kappa well if the electric field between the
plates has been reduced then i know the potential between the plates has been reduced as well
so what i'm appealing to here is the fact that the electric field between the plates of a capacitor
is given by potential difference over the gap between the plates and i can say the original
potential difference is related to the original electric field by v naught equals e naught times d
what about my new potential it's going to be the new electric field times d but that's e naught
over kappa times d and i recognize the e naught d in the numerator and i see that my potential
is the original potential over kappa so the potential difference across the plates has
also been reduced as a result of the reduced electric field between the plates now let's
examine what happens to the capacitance the original capacitance was the charge on the plates
over the potential difference between them notice i didn't put a subscript on the q
because the q isn't changing and that's key to all these derivations on this particular video
i've disconnected the capacitor from the power source so there's no way i can have
any more charge transfers q is fixed so how about our new capacitance that's q
over our new potential difference but our new potential difference was v naught over kappa
so when i divide by v naught over kappa that's the same as multiplying by kappa
and i end up with a factor of kappa out in front of q over v naught which is the original
capacitance so i see that my capacitance is increased by a factor of the dielectric
constant now that formula is particularly important because it will apply whether or
not we're dealing with a capacitor that that's been disconnected from the power source or if
it's still connected to the power source you can always say that if you put the dielectric
in there the capacitance goes up by a factor of the dielectric constant for example if you
were given the dimensions of a capacitor and the dielectric constant between the plates
you could compute the capacitance based on the geometry of the capacitor and our old formula was
epsilon 0 a over d and now to generalize i'm just going to put a kappa in there for the dielectric
constant of the material between the plates we'll get to one of those examples in another video
for now looking at a quick example of comparing a parallel plate capacitor with no dielectric
to what happens if you put in a dielectric with constant 2.0 in between the plates and
we're primarily concerned with energy in this example so i have an original
capacitance of one microfarad i charge to 5 volts potential difference and i disconnect
it from the power source a compute the energy in the capacitor before the dielectric is inserted
you can use one half cv squared for this and c is going to be 1 micro farad that's 10 to the
negative 6 ferrads and v is 5 volts and when i run the numbers on this i get 1.25 times 10 to
the negative fifth joules in b i'm asked for the new capacitance after the dielectric is inserted
well i know the new capacitance is just going to be kappa times the original capacitance that's
twice the original capacitance which was one microfarad so now it's two microfarads
in part c i'm asked to find the energy in the capacitor after the dielectric has been inserted
so if i write down the formula one half cv squared
i know that my new capacitance is given by kappa times the original that's two microfarads
but v has changed you have to be really careful about that provided i'm not connected to the power
source anymore v is going to drop by a factor of kappa we derived a little formula for that
when the dielectric is inserted so i end up with one half
kappa c naught times v naught over kappa all squared one factor of kappa is going
to cancel and i end up with a one over kappa out in front times one half c naught v naught squared
in other words the potential energy goes down by a factor of kappa just one more formula for the list
okay so all i have to do then is insert kappa equals two and i see that i get one half the
original potential energy i guess i could call that p e not i'll change it in part a
and that comes out to 0.625 times 10 to the negative fifth joules i just wanted to keep
the power of 10 constant there because it makes comparison easier now keep in mind that everything
changes if this thing is connected to a power source because that power source is going to
maintain the potential difference at a constant value the way it's going to do that is by pumping
more charge onto the plates when you add the dielectric and i'll post a link to a video where
i do that calculation if you find the physics content on zack's lab helpful click on the zak's
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