Energy stored in capacitor derivation (why it's not QV)

suppose a capacitor is charged to a potential difference we having a total charge of q on each plates the goal of this video is to figure out what is the total potential energy stored in this capacitor so let's start by asking ourselves what do we mean by this potential energy and how do we calculate that well before the capacitor was charged if i look at the uncharged situation there are no charges there are no potential difference we could say the potential energy is zero so in the process of charging someone must have separated those charges maybe positive charges from here went to here creating a positive negative charge separation and doing that requires work right because an electric field would be generated let me show you so imagine we separate positives and negatives like this so after after separating the charges adding additional positive charges like this let me write that adding additional charges like this requires work because there's repulsion just think about it if i try to add another positive charge over here it gets repelled by this plate of positive charge so you need to overcome that repulsion and therefore some work must have been done by the external agent and that external agent can be anyone that could be a battery that could be a ferry over here who's doing the charge separation it doesn't matter but somebody did that somebody did that work and that work done in transferring a total of q amount of charges q amount of charges from this plate to that that work done itself represents the potential energy stored so we can say that the potential energy stored in the capacitor is the work done by the external agent and let's say i am the external agent hours here so work done by mahesh that is the potential energy so i have to calculate what is the amount of work i do so how do we calculate that well if you go back to our basics of work done we could say it's force times the distance but by now you probably know we don't want to use that instead whenever we're talking about work done in electricity we can use the concept of potential difference so remember potential difference itself is a measure of work done so for example if the potential difference over here was 10 volt it basically means that now if i want to transfer an additional one coulomb of charge from here to here i have to do 10 joules of work okay so here's the question we know that the potential difference between the two plates is we how much work needs to be done in moving a coulomb from one plate to another i also know that the total charge transferred from here to here is q the question now is what is the total work i did in transferring that charge can you pause the video and think a little bit about it all right so what i would do what my common sense is telling me is that the work done by maheshes well to move one coulomb of charge i have to do v amount of work that's the meaning of potential difference but i move to q amount of charge so the total work that i did must be q times we and that itself must be the potential energy stored in the capacitor but that's wrong okay that's wrong and this is the first part i want to you know i want you to think a little bit about why is this wrong it's not enough to you know just accept that this is wrong but really to deeply understand why this is wrong so can you again pause the video and think about why is this wrong all right i hope you did some deep thinking now to answer this question let's look at this you know charging process step by step so here's our uncharged capacitor and let's imagine i am going to do the work and i'm going to take a very simple approach i'm going to start taking positive charges from here and start putting it onto this plate every time i do that i create a chart separation so let me bring in some charges so you can see positive but beneath that i've also put negative charges so this is neutral you can see there's a negative charge over here so i'm just showing some positives and negative charges right now the plate is neutral and what we'll do is we'll slowly in each step move the charges and look at how much work am i doing so let's move the first charge when i'm moving the first charge from here to here i ask myself hey in doing so how much work do i do well this is a very tiny amount of charge i don't know maybe you can say delta q it's not the entire charge q maybe a small delta q amount of charge what is the potential difference right now before i moved it the potential difference right now is zero because there is no charge and therefore to move that first charge it's almost free that kind of makes sense because i don't see any repulsion coming from this plate i know as i move it i still get some attraction and i still get some i don't have to do some work but if we neglect that i can say pretty much the work done in the first when i'm moving the first charge is almost zero so let me write that i'm going to write how much work i'm doing over here so the work that mahesh is doing in the first step the work done i'm just going to call almost zero so we're going to do it step by step okay now let's move the second charge now when i move the second charge do i have to do some work yes i do because now i feel repulsion from this charge in fact as a result of this charge separation there is a tiny potential difference generated it's not as big as this it's really really small but a tiny potential difference has generated so i have to do some work against that potential and so now i have to do some work so the work done in the second step is going to be some tiny potential difference i'm going to use the same formula work done equals potential difference times the charge transferred so some tiny potential difference multiplied by the charge transfer how much am i transferring the charge some tiny amount i'm just going to call that dq usually small charges you call dq and let's continue this so now i move the third charge and i'm as i'm doing that notice the potential difference has now increased now it's going to become even harder for me so in every step as i transfer more and more charges it becomes harder and harder next step it becomes even harder because the potential difference has even increased more i feel more repulsion it becomes harder and harder and harder so what i see so what i see is that in every step the potential difference starts increasing so in the next step the potential difference is a little higher that's why i'm writing a little bigger v to show that the potential difference is now a little higher times tq in the next step the potential difference is even more higher next step even more even more and so in every step this week keeps on increasing finally finally now imagine i'm transferring the last charge i imagine i'm almost almost done and i'm transferring the last charge okay all one billion charges have been transferred now what's the potential difference well when i'm transferring the last charge my capacitor is almost completely charged now the potential difference is we and so my last charge that i'm transferring i have to muscle through a lot of repulsion and i'm doing the maximum amount of work the work that i do now goes through the potential difference capital v so now and only now in the last when i'm transferring the last charge that's when i transferred the charge to the big potential difference and so clearly if i add up all these charges all the dqs i should get the total charge transfer q i should get that but can you now see why the total work done is not just q into e can you again at this point can you pause the video and come up with an explanation by looking at this summation it's it's mathematical now why is it not q times v all right hopefully you tried the main reason is because not all the charges went through the potential difference capital v only that last dq went through it and that's why because not the rest of the charges went through smaller potential difference the work done for the rest of the charges is smaller the total work done should be less than qv does that make sense okay so now how do we calculate this summation well here's a clue this dq if you ask me we know that dq is a small amount of charge right so it's a small fraction of q how small is the fraction of q well it's an infinitesimally small part meaning we have taken q and divided into almost an infinite amount of parts and each tiny infinitesimal is what we call as dq that means this is an infinite summation and whenever you're doing something like this we call that an integral so though so we have to set up this integral so how do we how do we set up the integral so whenever you're integrating uh you'll always have a starting starting point and ending point in our case we have a starting charge of zero and we have an ending charge of capital q so whenever you're integrating you start with some somewhere in middle so let's consider this situation where right now let's say the charge on this plate negative charge minus q the charge on this plate is plus q it's not as much as capital q it's less than that some value some random value and as a result of this there is some potential difference which is less than this number so i'm going to call that as small v and now let's assume that we're going to transfer an additional charge and that's how we usually do an integral so let's now calculate how much work tiny amount of work that we need to do in transferring an additional charge from here so let me an additional charge of dq from here to here how much work do i need to do well that work done that small amount of work that i have to do would equal same formula it's going to be the potential difference v times the charge dq and i could ask now hey what is that potential difference do i know what that potential difference is well think about it if i know the capacitance and let's say the capacitance of this capacitor is c then we know the capacitance formula capacitance is how charge divided by voltage so at any point voltage so let me write that down capacitance is charged by voltage so at any point voltage should equals charge divided by capacitance so i know that my voltage at this point should be the charge right now divided by capacitance so this is the tiny amount of work i did in moving a transferring a small amount of charge dq now to calculate the total work done i integrate this expression so the total work done by mahesh that's going to be an integral of this expression q by c of dq so basically i'm writing the same thing but in a proper integral notation and from where to where what is the initial uh initial charge i'm taking the initial charge is zero i'm starting from zero right from q equals zero and to what is the final charge i want on my plate the final charge on my plate is capital q so from q to capital q and now if you integrate this we'll get the work done and that will be our potential energy so again good time then is the last time i want you to pause the video and see if you can calculate this integral [Music] all right so since we are integrating q with respect to dq you might you can you can recall the integral formula if you're integrating x power n with respect to x it's going to be x power n plus 1 divided by n plus 1. so let me keep that denominator c so over here you'll get q power 1 so it's going to be q power 1 plus 1 that's q squared divided by 1 plus 1 by 2. and whenever we're integrating we're going to put our limits from 0 to q so let me make some space and so we put down our upper limit so that's q squared over 2c minus the lower limit that is zero and therefore this is our work done and hence and now let me write that somewhere over here the total energy stored must be in a capacitor must be q squared divided by 2 c and if i substitute q is equal to c into e i can say total charge should be total capacitance into voltage if i substitute that i can also write this equals let's see i'll end up with half if i put cv c square v square square cancels you'll get c v squared and if i substitute for c over here c is equal to q over v then i'll get and you can do that yourself you'll get it as half q v so half q times v and you can check that all i'm doing is substituting these values over here and changing the variables and what's interesting is that whatever we had predicted wrongly the actual answer is half of that and although that's not very logical you can't logically i mean it's hard for me to logically explain why it's half not 1 3 or 1 4 because there's an integral coming but at least it makes sense that it's less than what we predicted because of what we explained earlier not all the charges are moving through that entire potential difference capital we so this is how we calculate the energy stored in a capacitor