Capacitors and Dielectrics

hello today I'm responding to some questions that were raised about my video in the a level revision series on capacitors the link to the specific point is shown here I had said that the capacitor which is consisting of parallel plates the capacitance or the value of the capacitance is proportional to the distance in fact it's proportional to 1 over D the separation between the plates it's also proportional to the area of as it were overlap so C is proportional to the area of overlap of the plate and finally I have said that C is proportional to the dielectric which is a piece of insulating material that is sandwiched between the plates and so C is proportional to epsilon and from that we had drawn the overall conclusion that the capacitance equals epsilon which is the dielectric times area the area of overlap divided by D the distance between the plates and the questions I have been asked is how does the dielectric play a role how is the energy stored in the capacitor is the dielectric itself affected and what happens if you remove the plate from the dielectric after charging and replace it with a new set of plates can you recover the charge well we shall see that the last question doesn't really arise but I'll try to explain as we go through first let's just explain what this term epsilon actually is epsilon is usually expressed as K times epsilon naught where K is the relative permittivity and epsilon naught is the permittivity of free space we've met this before in the formula which tells you what the forces between two charged particles the force is equal to Q 1 Q 2 to where q1 q2 of the charges on the two particles divided by four PI epsilon naught R squared where R is the separation of the two charges and epsilon naught is this epsilon naught the permittivity of free space and Epsilon in other words the dielectric constant of the insulator between the capacitors is equal simply to a constant for that Capac for that dielectric times the permittivity of free space sorry about my dog barking the value of epsilon naught is eight point eight five times ten to the minus twelve farads per meter and you'll see you need that term to be fair ads per meter because the capacitance is in fur heads area is in meter squared distance is in meter so area divided by distance is meter so farad's per meter is what epsilon has to be so let's remind ourselves what happens when a capacitor is charged here is a battery and we connect it to a parallel plate capacitor what happens is that the electrons flow onto this plate sorry that's not mine as it is - no plus the electrons are on that plate they will repel the electrons from this plate and leave behind positive charges and so the capacitor is charged because the electrons flow onto this plate they repel the electrons from that plate and leave positive charges no current flows across the gap very important to stress that now let's say that the charge on this plate per unit area is Sigma minus and the charge per unit area on this plate is Sigma plus and they will of course be equal and opposite now we remove the battery and in those circumstances the charged capacitor remains charged the charge can't go anywhere we've taken the capacity battery away there's nowhere for the charge to flow it must just sit there and because there's a positive charge on this plate and a negative charge on this plate there will be an electric field in that direction between the two plates now let's insert the dielectric so here's the capacitor again this plate is positively charged this plate is negatively charged and now I'm going to insert the dielectric and that will be an insulator and the point about an insulator is that there is not a large amount of there are not enough large amount of free electrons to flow that doesn't mean to say that there aren't any it just means that there aren't very many so what actually happens to the dielectric well the electrons that are free in the dielectric will be repelled from this positive charge and so there will be an induced negative charge right at the surface of the dielectric remember the dielectric is an insulator so there will not be a flow of charge inside but right at the edges the electrons will be repelled by this positive charge and likewise sorry the the positive charges will be repelled the electrons will be attracted by this positive charge and likewise at this edge the electrons will be repelled leaving an induced positive charge so here you've got the surface charge on the plate which we're going to call the free charge and here you've got the negative free charge on this plate but here you've got a negative induced charge on the dielectric on the surface of the dielectric and here is a negative a sorry a positive induced charge on the surface of the dye electric at this end that means that you've got an electric field which we might call the free electric field between the positive and the negative plates but you also now got an induced field between the positive and negative edges of the dielectric and that's let's call that e induced so the net field in this direction in the downwards direction is going to equal the free field between the two plates minus the induced field between the two edges of the dielectric now let's take it make an assumption which turns out to be a reasonable assumption that the charge per unit area in the induced part of the dielectric is equal to a constant times the free charge in other words the amount of charge that is induced on this surface is a direct proportionality to the amount of charge on the plate itself well if that's the case you can also say that the e induced equals B III in other words the induced field between the edges of the dielectric is a direct proportion of the free field that is flowing between the plates since the strength of the field is proportional to the amount of charge you'll be able to see that in my video on fields in the a level revision series and that means that since E induced equals EF minus e induced and E induced is B times a free that means that een ed he Qin to a 3 into 1 minus B and one minus B is K the dielectric constant all the motive permittivity that's the K that we established all the way back up here so that means that epsilon equals K times epsilon naught the permittivity of the dielectric or the dielectric constant is equal to the relative permittivity times the permittivity of free space for glass K is 5 for water K is 80 so since the capacitance equals epsilon which is K epsilon zero times the area of overlap divided by D what that means is that if you put a glass dielectric between the two plates then the capacitance will be five times greater than if you had not had the glass between the plates now the impact of all that depends on precisely what you do here's the first experiment that we're going to do we take our battery we're going to now put an ammeter in the circuit and here is the capacitor and when we switch on the battery then we're going to see the capacitor charge up negative this end positive at that end and there will be a certain capacitance which will equal epsilon a over D and epsilon for the moment is one because this is just free space we haven't put the dielectric in yet now we insert the glass dielectric between it's sandwiched between the two plates and of course there will be the induced charges at the edges but what we say is that the dielectric constant is now five times higher because epsilon equal k times Epsilon free space so if K is 5 then C will be 5 times higher so simply by putting the dielectric in the capacitance will increase by 5 times and C is Q over V now V is the voltage on the battery and that's not changing because the battery is still connected but C has increased fivefold so immediately you can see that Q will increase fivefold in other words a lot more charge will flow onto these plates and you will see it current flow in the ammeter because more charge as soon as you put that dielectric in more charge will flow that means the current will flow and there will be more charge on the plates what does that mean in terms of energy stored well in my previous video on capacitors I said that the energy is 1/2 QV I'll demonstrate why that's the case in a moment but since the charge has increased fivefold but the voltage remains the same that means that the energy will increase fivefold so before you put the dielectric in the capacity with the capacitor would have a certain stored energy which would be half QV when you put the dielectric in the charge increases fivefold and so the stored energy potential energy increases fivefold now let's take a second case this time we're going to charge up the capacitor in the usual way by putting a battery across the two plates and we shall end up with the 2 plates charged positive and negative but now we disconnect the battery so the capacitor is just standing there isolated the charge cannot go anywhere it is fixed and it is trapped now we introduce the dielectric use a glass dielectric again and once again we shall get induced charges right at the edges of the dielectric once again C equals epsilon a over D and effectively the dielectric has increased fivefold so the capacitance has increased fivefold and C his Q charge over voltage but now the charge cannot alight up here the charge cannot increase fivefold because here it could because there's more charge to come from the battery but here there is no source of any extra charge that short charge is now fixed and trapped you can't get any more charge so if the capacitance increases by five fold the only way that can happen is if the voltage decreases fivefold because the charge can't change so that means that the voltage across the plates goes down by five and since the electric field is equal to the voltage times the distance the electric field is effectively volts per meter that means that this field across the plates has fallen to a value of one-fifth of what it was before what about the energy I'm guilty of using Eve for both electric field and energy up here I should have said that this is energy is half QV and here energy is 1/2 times Q times V now in this example what does that mean well Q is unchanged there's no source of any extra charge and it can't go anywhere so Q is a constant V we've just said decreases by five so the energy goes down by five fold but here's the question and I think this is probably what the questioner was getting at where has that energy gone and the answer is that that lost energy has it work has gone into an electrostatic force between the capacitor and the dielectric think of it you see as you start to put that dielectric in there becomes an induced charge on the surfaces and now you've got a situation where the positive charge on the plate is going to attract the negative charge that is on the surface of the dielectric and down here the negative charge on the plate is going to attract the positive charge on the dielectric so actually that capacitor will pull the dielectric in and it is using energy to do that and that's where this lost energy is gone and if you want to get that dielectric out from between the capacitor whilst it is still charged you will actually have to physically do work to get it out you will have to give it energy to get it out from between the two plates and then you will restore the capacitor to its original energy because it will go back to having its original voltage and thus its original e field now it might just be worth explaining what bad energy is that's in the capacitor and we'll think of it not in terms of how you put energy in but how you might get energy out for electricity the power is equal to the voltage times the current if you have an ordinary circuit battery resistance and a current flowing and you can say and the battery is V you can say that of course vehicles IR that simply Ohm's law but the power is V times I the voltage times the current energy therefore which is power times time is VI which is the power times T in seconds because power is energy divided by time now I T is simply charge because current is charge flowing in a particular and q is equal to VC that's how we measure capacitance capacitance you remember is charge per volt so that means that we can take this formula here energy and we can say that energy is vit well V we can say is Q over C and I T is Q and that's Q squared over C but there is a mistake here because of course what we're talking about is discharging a capacitor so here's a fully charged capacitor and we discharge it through say a resistance which might be a kind of a fresh bulb or something like that and the charge will flow and it may take T seconds and the current will flow but of course it wouldn't be the same current throughout because as the charge flows there will be less of it and the charge will actually reduce from the maximum charge to zero so in fact the average charge that flows throughout that whole process is half the total charge on the capacitor so in fact the energy is not Q squared over C but 1/2 Q squared over C because as I've said the the charge doesn't remain the same as it flows through here the charge is actually going down to zero so on average you've only got half the charge flowing around the wire and of course because Q equals VC you can also say that energy equals 1/2 Q times V and since that is the energy that is dissipated through the resistance as you discharge the capacitor then it must also be the potential energy that sits on the capacitor or is stored within the capacitor before you discharge it so when you charge up hasset er its total potential energy is half QV and if you then discharged it that energy will be discharged through all of you discharge it through so I hope that deals with the questions that I've been asked let's just review the questions and a summary of the answers question one was how does the dielectric play a role well the answer is the surface of the dielectric has an induced charge that induced charge creates a field in the opposite direction and thus reduces the overall field and what we find is that the capacitance therefore varies how is the energy stored in the capacitor well I try to explain that here it is stored as potential energy such that it can be released if you connect the capacitor to some sort of circuit that energy will flow as a current through a resistor and discharge in that way is the dielectric affected yes it is the dielectric has an induced charge on it and it's that induced charge that creates a field within the dielectric that opposes the field between the plates and that's what changes the capacitance the final question what happens if you remove the plates from the dielectric after charging well I suppose removing the plates from the dielectric is the same as removing the dielectric from the plates and the answer is if you take the dielectric out then you are just left with a fully charged capacitor once again remember the charge can't go anywhere and so the capacitance goes back to what it was before you put the dielectric in in the first place I hope that helps

Channel: DrPhysicsA

Views: 54,433

Rating: 4.9333334 out of 5

Keywords: capacitor, dielectric, permitivity, energy, charge

Id: 4niec5qZtgY

Length: 21min 48sec (1308 seconds)

Published: Sun Mar 18 2012