 Suppose a capacitor is charged to a potential difference, we having a total charge of Q on each plate. 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 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 fairy over here who's doing the charge separation. It doesn't matter. But 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 over 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 four 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 V. 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 Mahesh is, well, to move one coulomb of charge, I have to do V amount of work. That's the meaning of potential difference. But I moved Q amount of charge, so the total work that I did must be Q times V. 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, I want you to think a little bit about, why is this wrong? It's not enough to 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 charging process step by step. So here is 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 charge 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 gonna 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 gonna call almost zero. So we're gonna 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 gonna be some tiny potential difference. I'm gonna use the same formula. Work done equals potential difference times the charge transferred. So some tiny potential difference multiplied by the charge transferred. How much am I transferring the charge? Some tiny amount, I'm just gonna call that DQ. Usually small charges you call DQ. And let's continue this. So now I move the third charge. And as I'm doing that, notice the potential difference has now increased. Ooh, now it's gonna 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 DQ. In the next step, the potential difference is even more higher. Next step, even more, even more. And so in every step, this V keeps on increasing. Finally, finally. Now imagine I'm transferring the last charge. And imagine I'm 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 V. 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, and I'm transferring the last charge, that's when I transfer 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 V? Can you, again, at this point, can you pause the video and come up with an explanation by looking at this summation? It's mathematical now. Why is it not Q times V? All right, hopefully we 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 a fraction of Q? Well, it's an infinitismally small part, meaning we have taken Q and divided into almost an infinite amount of parts. And each tiny infinitism 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 we have to set up this integral. So how do we set up the integral? So whenever you're integrating, you always have a starting point and an 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 somewhere in middle. So let's consider this situation where right now, let's say, the charge on this plate, negative charge minus Q and 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 gonna call that a small v. And now let's assume that we're gonna 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 get 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 gonna 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, let me write that down, capacitance is charge 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 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 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 I want on my plate is capital Q. So from Q to capital Q. And now if we integrate this, we'll get the work done and that'll be our potential energy. So again, good time doing this. The last time I want you to pause the video and see if you can calculate this integral. All right? So since we're integrating Q with respect to DQ, you might, you can recall the integral formula. If you're integrating X power N with respect to X, it's gonna be X power N plus one divided by N plus one. So let me keep that denominator C. So over here you'll get Q power one. So it's gonna be Q power one plus one. That's Q squared divided by one plus one by two. And whenever we're integrating, we're gonna put our limits from zero to Q. So let me make some space. And so we put down our upper limit. So that's Q squared over two C 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 two C. And if I substitute Q is equal to C into E, I can say total charge should be equal to 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 squared V squared, square cancels, you'll get C V squared. And if I substitute for C over here, C is equal to Q over V. And 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 one third or one fourth, because it'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 V. So this is how we calculate the energy stored in a capacitor.