 We have three charges kept like this. These are fixed charges. Let's say now imagine I Take this plus two coulomb charge and slowly move it along this path and Bring it to this point so that now it is at a distance of three Meters these are all in meters from the charges the question we want to try and answer in this video is in Moving this charge how much work did that external agent? That is me. How much work did Mahesh do in Moving the charge from here to here and we also want to calculate how much work did the electric field? Do when the charge went from here to here? How do we do that? Well, we can always go back to our basics and say hey work done is force into displacement And I can try and do that over here the problem is that the force keeps continuously changing Right as I come closer and closer to the charge the force might start increasing Which means I have to do some integral and look at that path taken off. No, we're not going to do any integrals over here So to avoid integrals we introduced the concept of potential energy But whenever I used to do these calculations I used to always get confused with the science Is it final potential energy minus initial or is it initial minus final potential energy and I used to always be confused So here's what I like to do now. I always always like to go back to gravity because that helps me It's a very simple case. So let's go back to gravity and think a little bit about it So imagine I have a stone lying on the ground What happens when I again Mahesh I move it up What happens? Well, let's say I did I Did hundred joules of work in pushing it up. I do positive work in pushing it up What happens to the potential energy? Well, if the initial potential energy was zero when I do work I add a potential energy into this rock and so the potential energy of the rock Increases by a hundred joules, right and therefore work done by me Mahesh work done by the external force that Equals the final potential energy Minus the initial potential energy of the system. So let me write that down. So the work that I do That equals the final potential energy minus the initial potential energy and What about the work done by the electric field? Well, I go back over here ask myself What is the work done by gravity over here? Well, when I'm pushing up when I'm putting a force on it upwards gravity is always putting a force on it downwards So gravity is putting a force in the opposite direction, which means gravity is doing negative work So the work done by gravity would be negative hundred Same will be the case over here the work done by the Electric field instead of gravitational field will be the negative of this. So that would be initial potential energy Minus the final potential energy. Does it make sense don't ever ever mug this up because we will forget that Always go back to gravity take a simple case and that will always help us in coming back over here So this means all I have to do now is think about what was the initial potential energy of the system? What is the final potential energy of the system and then we can figure out the work done? So encourage you to pause this video before we move forward and try it yourself All right, so let me let me draw what the initial put initial system looks like This is what initially it looked like 10 10 meters far away from each other Then when I once I did that work the final system looks like this And so the energy of this would be the final potential energy And if I subtract that I'll get the work done by me one thing you may be curious about is hey What about the path taken doesn't that matter? No, it doesn't because electric fields are what we call conservative fields It doesn't matter what path you take all that matters is the what is the initial state and the final state the work done is path Independent. All right, so let's go ahead and figure out these potential energies. Let's make some space over here All right a quick reminder of how we calculate potential energies of a system We've seen this in the previous video the way I like to remember is first think about what is the potential energy of Two charges the potential energy of a system of just two charges would be k You can call it one two k q1 q2 divided by r One two with r12 is a distance between these two And now if you want to calculate potential energy of system of three charges you take two at a time It will be the potential energy of this plus potential energy of this system plus potential energy of this system Same would be the case over here and once you substitute we can subtract and we can get an answer So again, if you haven't tried before maybe now would be a great time to pause and try All right, let's see. Let's start with the final potential energy Over here. So that's going to be k I'm going to keep the k as it is and I'm going to put a bracket because k will be common everywhere So first let's take these two charges. So q1 q2 is going to be minus five times minus five. That's 25 Divide by the distance between them. That's 10 plus Then let's take these two charges. It's going to be minus five times two q1 q2 minus five times two. That's minus 10 Divide by this distance. That's going to be three plus Let's take these two charges. We'll get the same answer as these two charges. So again, it'll be minus 10 over 3 And I'm not simplify Let me just go ahead and write what the initial potential energy was going to be sometimes things cancel out So that's going to be again k I'm going to take these two charges first. We'll get the same answer as what we got over here. So it's 25 by 10 plus When I did these two this time, I'll get minus 10 divide by 10 So minus 10 divide by 10 and I can I get the same over here minus 10 Divide by 10. I have to now subtract this. So when I subtract this cancels out and So finally, what do I get? Okay, you can do this yourself. This is just now subtraction Let me quickly do it. What I'll do is this will be minus 20 by 3 and this is minus 20 by 10 So I'll take that minus 20 out and what I have here is one over three And what I'll have over here is minus one over 10 because we're subtracting Minus one over 10 you can check you can pause and check so we'll get Minus 20 times k times. What is this 10 minus 3 7 7 by 30. So this cancels and that'll end up with minus 14 Over 3 and now I'll substitute for k. You might remember k is 9 times 10 to the power 9 So 3 cancels out Let me write down over here the final answer the change in potential energy is going to be minus 42 So let me write that somewhere over here. Okay minus 42 times 10 to the power 9 Joules that's the change oops that's the change in Potential energy Therefore the work done by me Would be just this number minus 42 Billion Joules 10 power 9 is I'm just saving space over 10 power 9 is billions I'll say minus 42 billion Joules and what is the work done by electric field It's going to be the opposite of that. So I have opposite sign. So it's gonna be plus 42 billion Joules and by the way, this this is only happening to be true because I'm slowly moving it without any acceleration Okay, I think I forgot to mention that in the question. I'm slowly moving it without any acceleration. That's important That's why the total work done would be zero because there is no acceleration. There is no change in kinetic energy Let's quickly try one more. We have three identical charges kept at the vertices of an equilateral triangle and let's say an external agent again me Let's say I come in and I move all the three charges now. I Move all the three charges in the given path and such that they form a new equilateral triangle of having edge length R And the question now is in doing so if if I move it very slowly without any acceleration What is the total work done by the electric field? So can you pause the video and give it a shot? All right, so just like what we saw earlier the work done by the electric field is the negative of the work done by me So it's gonna be negative of final minus initials gonna be initial minus final. So it's gonna be initial potential energy Minus final potential energy And the initial system is gonna look like and the initial potential energy will be the potential energy of this system The final potential energy will be the potential energy of this system. So I can just go ahead and calculate So initial potential energy is going to be just like before I can take two charges at a time K Q 1 Q 2 by R But notice this is since they're all equal. It's gonna be just three times this number. So it's gonna be Three times K Q 1 Q 2 by their distance So it's Q square by 2r plus Q square by 2r plus Q square by 2r that's gonna be three times I'm just trying to save some time and space over here and The final potential energy is gonna be very similar instead of 2r. It's gonna be R. It's gonna be 3k Q square Over R and now the work done by the electric field is going to be This minus this initial minus final you to be very careful about the signs over here So everything there are 3k Q square by R is common What remains now inside? Let me write that let's not do too many things in our head So 3k Q square by R is common when I subtract what remains is a half here minus what remains is the one here So that means I end up with Minus half so I get minus 3k Q square By 2r this is the work done by the electric field in moving in changing that configuration