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Electric potential energy

Introduction to electric potential energy. Created by Sal Khan.

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  • blobby green style avatar for user Ananyo Khan
    around sal says that to get the charge moving downwards, we have to exert a force of 10N. But if we exert that force in the downward direction, seeing that the metal plate is ALSO exerting a force by the same amount, won't the charge just stay stationary over there (like suspended in the electric field)???
    PLEASE HELP ME ON THIS ONE :(
    VERY IMPORTANT
    (62 votes)
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    • male robot donald style avatar for user Bhaskar
      for one moment a force little greater than 10N is applied, so that the body gains a velocity. after that the force can be equal to 10N.(In that case net force will be zero and the charged body will continue to go with the velocity it has gained at first.
      (14 votes)
  • blobby green style avatar for user Memuhilan Senthil
    please explain the functions of a battery.how does it function?the electrons move from - side to the bulb and to the +side? PLEASE explain it elaborately
    (4 votes)
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    • blobby green style avatar for user googol88
      My understanding of the situation is that a battery has two metal plates inside along with some acid. As the metal sits in the acid, it strips protons off of one plate (zinc, for example). The plate thus builds up a negative charge, and the acid a positive charge. This positive charge in the acid begins removing electrons from the other plate (carbon, for example)causing it to have a positive charge. Thus, the two plates of metal inside the battery have different charges.

      When the battery is inserted into a device and the circuit completed, the electrons can flow through the wire, evening the charge and powering the device. This process repeats until the metal in the battery has corroded from the acid (and proton/electron stripping) to a certain point, and your battery is dead.
      (31 votes)
  • old spice man green style avatar for user alvinfu
    I got a question about electric potential energy, though may not be related to Sal's video. When a positive charge is brought near a positive point charge. The work done will be changed to the electric potential energy and stored in the charge. However, when a negative charge is brought away from the positive charge, the negative charge gains electric potential energy.When r keeps increasing, the electric potential energy stored in the negative charge will be extremely large?? Hope you guys can understand my question....
    (10 votes)
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    • old spice man green style avatar for user alvinfu
      I understand now! for opposites charges, work is done to pull them away from each other. it changes into PE and stored inside the charge. But their separation is getting larger as well, the attractive force becomes smaller, then the work needed is smaller too. So the pulling force decreases. It means that the electric PE increases at a decreasing rate!
      (14 votes)
  • blobby green style avatar for user shafi1248
    At Sal says that a field of 5N/C is quite strong. According to Google a Newton is about 1/5 of a pound. So the field is something like 1 pound/coloumb.

    Why is this so strong? I was under the impression that a coloumb was a fairly large amount of charge. 1 pound for a high amount of charge does not seem so strong.
    (7 votes)
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  • blobby green style avatar for user Eric Marshall
    why does the temperature of a solid conductor increase when the conductor is carrying current?
    (5 votes)
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  • blobby green style avatar for user siddharth kashyap
    im always confused at this fact .
    if an object weighing mg newtons needs to be pulled up,it requires an opposite force EQUAL to mg newton.(should'nt the force be more than the downward force)(because if force to pull up would be equal it would be suspended in air {if applied from rest}).
    (3 votes)
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  • duskpin sapling style avatar for user aruja
    At Sal said that if we let go of the particle all of 30J would be converted to kinetic energy.
    I get that if left, the particle would move away from the positively charged plate, but its still in the electric field so it should have some potential energy remaining, right?
    (2 votes)
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    • spunky sam blue style avatar for user Sal Khan
      From the author:Yes, the 30J is really the increase in potential from it's starting point to the ending point 3 meters closer to the plate. If we let go at the ending point, the particle would start accelerating in the direction of the field (upward), and, in this case, when it passes its starting, all of that increase in potential energy will be converted to kinetic energy.

      The particle, however, still has the field acting on it and will still be accelerated upwards (so you could say that it still has a positive potential relative to a point further away from the plate).
      (2 votes)
  • blobby green style avatar for user yoohoiohaha
    The potential is constant throughout a given region of space . Is the electrical field zero or non zero
    (3 votes)
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  • piceratops ultimate style avatar for user Omkar Rajwade
    at ,sal says that the work done is equal to force of gravity into h,but shouldn't the force be a little bit more if we want to lift it upward against the force of gravity?
    (6 votes)
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    • piceratops seed style avatar for user Ian
      Yes. In later in the video he goes on to say that exact same thought. His equation shows 10n/m but the faster you wanted to go the more you would apply. (10n - SLIGHT SPEEED).
      (0 votes)
  • old spice man blue style avatar for user Shayan Nik Akhtar
    i have 2 questions
    1.what is the intuition behind a constant like G(gravitational constant)and K(coulombs constant),where do we get the values of them?
    2.when sal said that we will apply a force of 10N DOWNWARDS wouldn't that force be balanced by the upward force since the field is also applying a force of 10N on the charge in the opposite direction that is UPWARDS?
    (2 votes)
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Video transcript

Let's review a little bit of what we had learned many, many videos ago about gravitational potential energy and then see if we can draw the analogy, which is actually very strong, to electrical potential energy. So what do we know about gravitational potential energy? If we said this was the surface of the Earth-- we don't have to be on Earth, but it makes visualization easy. We could be anywhere that has gravity, and the potential energy would be due to the gravitational field of that particular mass, but let's say this is the surface of the Earth. We learned that if we have some mass m up here and that the gravitational field at this area-- or at least the gravitational acceleration-- is g, or 9.8 meters per second squared, and it is h-- we could say, I guess, meters, but we could use any units. Let's say it is h meters above the ground, that the gravitational potential energy of this object at that point is equal to the mass times the acceleration of gravity times the height, or you could view it as the force of gravity, the magnitude of the force of gravity. You know, it's a vector, but we can say the magnitude of the vector times height. And so what is potential energy? Well, we know that if something has potential energy and if nothing is stopping it and we just let go, that energy, at least with gravitational potential energy, the object will start accelerating downwards, and a lot of that potential energy, and eventually all of it, will be converted to kinetic energy. So potential energy is energy that is being stored by an object's situation or kind of this notional energy that an object has by virtue of where it is. So in order for something to have this notional energy, some energy must have been put into it. And as we learned with gravitational potential energy, you could view gravitational potential energy as the work necessary to move an object to that position. Now, if we're talking about work to move something into that position, or whatever, we always have to think about, well, move it from where? Well, when we talk about gravitational potential energy, we're talking about moving it from the surface of the Earth, right? And so how much work is required to move that same mass-- let's say it was here at first-- to move it from a height of zero to a height of h? Well, the whole time, the Earth, or the force of gravity, is going to be F sub g, right? So essentially, if I'm pulling it or pushing it upwards, I'm going to have to have-- and let's say at a constant velocity-- I'm going to have to have an equal and opposite force to its weight to pull it up. Otherwise, it would accelerate downwards. I'd have to do a little bit more just to get it moving, to accelerate it however much, but then once I get it just accelerating, essentially I would have to apply an upward force, which is equivalent to the downward force of gravity, and I would do it for a distance of h, right? What is work? Work is just force times distance. Force times distance, and it has to be force in the direction of the distance. So what's the work necessary to get this mass up here? Well, the work is equal to the force of gravity times height, so it's equal to the gravitational potential energy. Now this is an interesting thing. Notice we picked the reference point as the surface of the Earth, but we could have picked any arbitrary reference point. We could have said, well, from 10 meters below the surface of the Earth, which could have been down here, or we could have actually said, you know, from a platform that's 5 meters above the Earth. So it actually turns out, when you think of it that way, that potential energy of any form, but especially gravitational potential energy-- and we'll see electrical potential energy-- it's always in reference to some other point, so it's really a change in potential energy that matters. And I know when we studied potential energy, it seemed like there was kind of an absolute potential energy, but that's because we always assume that the potential energy of something is zero the surface of the Earth and that we want to know the potential energy relative to the surface of the Earth, so it would be kind of, you know, how much work does it take to take something from the surface of the Earth to that height? But really, we should be saying, well, the potential energy of gravity-- like this statement shouldn't be, you know, this is just the absolute potential energy of gravity. We should say this is the potential energy of gravity relative to the surface of the Earth is equal to the work necessary to move something, to move that same mass, from the surface of the Earth to its current position. We could have defined some other term that is not really used, but we could have said potential energy of gravity relative to minus 5 meters below the surface of the Earth, and that would be the work necessary to move something from minus 5 meters to its current height. And, of course, that might matter. What if we cut up a hole and we want to see what is the kinetic energy here? Well, then that potential energy would matter. Anyway, so I just wanted to do this review of potential energy because now it'll make the jump to electrical potential energy all that easier, because you'll actually see it's pretty much the same thing. It's just the source of the field and the source of the potential is something different. So electrical potential energy, just actually we know that gravitational fields are not constant, we can assume they're constant maybe near the surface of the Earth and all that, but we also know that electrical fields aren't constant, and actually they have very similar formulas. But just for the simplicity of explaining it, let's assume a constant electric field. And if you don't believe me that one can be constructed, you should watch my videos that involve a reasonable bit of calculus that show that a uniform electric field can be generated by an infinite uniformly charged plane. Let's say this is the side view of an infinite uniformly charged plane and let's say that this is positively charged. Of course, you can never get a proper side view of an infinite plane, because you can never kind of cut it, because it's infinite in every direction, but let's say that this one is and this is the side view. So first of all, let's think about its electric field. It's electric field is going to point upward, and how do we know it points upward? Because the electric field is essentially what is-- and this is just a convention. What would a positive charge do in the field? Well, if this plate is positive, a positive charge, we're going to want to get away from it. So we know the electric field points upward and we know that it's constant, that if these were field vectors, that they're going to be the same size, no matter how far away we get from the source of the field. And I'm just going to pick a number for the strength of the field. We actually proved in those fancy videos that I made on the uniform electric field of an infinite, uniformly charged plane that we actually proved how you could calculate it. But let's just say that this electric field is equal to 5 newtons per coulomb. That's actually quite strong, but it makes the math easy. So my question to you is how much work does it take to take a positive point charge-- let me pick a different color. Let's say this is the starting position. It's a positive 2 coulombs. Once again, that's a massive point charge, but we want easy numbers. How much work does it take it to move that 2-coulomb charge 3 meters within this field? How much work? So we're going to start here and we're going to move it down towards the plate 3 meters, and it's ending position is going to be right here, right? That's when it's done. How much work does that take? Well, what is the force of the field right here? What is the force exerted on this 2-coulomb charge? Well, electric field is just force per charge, right? So if you want to know the force of the field at that point-- let me draw that in a different color. The force of the field acting on it, so let's say the field force, or the force of the field, actually, is going to be equal to 5 newtons per coulomb times 2 coulombs, which is equal to 10 newtons. We know it's going to be upward, because this is a positive charge, and this is a positively charged infinite plate, so we know this is an upward force of 10 newtons. So in order to get this charge, to pull it down or to push it down here, we essentially have to exert a force of 10 newtons downwards, right? Exert a force of 10 newtons in the direction of the movement. And, of course, just like we did with gravity, we have to maybe do a little bit more than that just to accelerate it a little bit just so you have some net downward force, but once you do, you just have to completely balance the upward force. So just for our purposes, you have a 10-newton force downward and you apply that force for a distance of 3 meters, the work that you put to take this 2-coulomb charge from here to here, the work is going to be equal to 10 newtons-- that's the force-- times 3 meters. So the work is going to equal 30 newton-meters, which is equal to 30 joules. A joule is just a newton-meter. And so we can now say since it took us 30 joules of energy to move this charge from here to here, that within this uniform electric field, the potential energy of the charge here is relative to the charge here. You always have to pick a point relative to where the potential is, so the electrical potential energy here relative to here and this is electrical potential energy, and you could say P2 relative to P1-- I'm using my made-up notation, but that gives you a sense of what it is-- is equal to 30 joules. And how could that help us? Well, if we also knew the mass-- let's say that this charge had some mass. We would know that if we let go of this object, by the time it got here, that 30 joules would be-- essentially assuming that none of it got transmitted to heat or resistance or whatever-- we know that all of it would be kinetic energy at this point. So actually, we could work it out. Let's say that this does have a mass of 1 kilogram and we were to just let go of it, right? We used some force to bring it down here, and then we let go. So we know that the electric field is going to accelerate it upwards, right? It's going to exert an upward force of 5 newtons per coulomb, and the thing's going to keep [COUGHS]-- excuse me-- keep accelerating until it gets to this point, right? What's its velocity going to be at that point? Well, all of this electrical potential energy is going to be converted to kinetic energy. So essentially, we have 30 joules is going to be equal to 1/2 mv squared, right? We know the mass, I said, is 1, so we get 60 is equal to v squared, so the velocity is the square root of 60, so it's 7 point something, something, something meters per second. So if I just pull that charge down, and it has a mass of 1 kilogram, and I let go, it's just going to accelerate and be going pretty fast once it gets to this point. Anyway, I'm 12 minutes into this video, so I will continue in the next, but hopefully, that gives you a sense of what electrical potential energy is, and really, it's no different than gravitational potential energy. It's just the source of the field is different. See you soon.