Electric potential energy of charges | Physics | Khan Academy

– [Narrator] So here’s something
that used to confuse me. If you had two charges, and we’ll keep these straight
by giving them a name. We’ll call this one Q1
and I’ll call this one Q2. If you’ve got these two charges
sitting next to each other, and you let go of them,
they’re gonna fly apart because they repel each other. Like charges repel, so
the Q2’s gonna get pushed to the right, and the Q1’s gonna get pushed to the left. They’re gonna start
gaining kinetic energy. They’re gonna start speeding up. But if these charges are
gaining kinetic energy, where is that energy coming from? I mean, if you believe in
conservation of energy, this energy had to come from somewhere. So where is this energy coming from? What is the source of this kinetic energy? Well, the source is the
electrical potential energy. We would say that
electrical potential energy is turning into kinetic energy. So originally in this system, there was electrical potential energy, and then there was less
electrical potential energy, but more kinetic energy. So as the electrical
potential energy decreases, the kinetic energy increases. But the total energy in this system, this two-charge system,
would remain the same. So this is where that
kinetic energy’s coming from. It’s coming from the
electrical potential energy. And the letter that
physicists typically choose to represent potential energies is a u. So why u for potential energy? I don’t know. Like PE would’ve made sense, too, because that’s the first two letters of the words potential energy. But more often you see it like this. We’ll put a little subscript e so that we know we’re talking about electrical potential energy and not gravitational
potential energy, say. So that’s all fine and good. We’ve got potential energy
turning into kinetic energy. Well, we know the formula
for the kinetic energy of these charges. We can find the kinetic
energy of these charges by taking one half the
mass of one of the charges times the speed of one
of those charges squared. What’s the formula to find the
electrical potential energy between these charges? So if you’ve got two or more charges sitting next to each other, Is there a nice formula to figure out how much electrical
potential energy there is in that system? Well, the good news is, there is. There’s a really nice formula that will let you figure this out. The bad news is, to derive
it requires calculus. So I’m not gonna do the calculus
derivation in this video. There’s already a video on this. We’ll put a link to that
so you can find that. But in this video, I’m just
gonna quote the result, show you how to use it, give you a tour so to
speak of this formula. And the formula looks like this. So to find the electrical potential energy between two charges, we take
K, the electric constant, multiplied by one of the charges, and then multiplied by the other charge, and then we divide by the distance between those two charges. We’ll call that r. So this is the center to center distance. It would be from the center of one charge to the center of the other. That distance would be r,
and we don’t square it. So in a lot of these formulas, for instance Coulomb’s law,
the r is always squared. For electrical fields, the r is squared, but for potential energy,
this r is not squared. Basically, to find this
formula in this derivation, you do an integral. That integral turns the
r squared into just an r on the bottom. So don’t try to square this. It’s just r this time. And that’s it. That’s the formula to find the electrical potential
energy between two charges. And here’s something
that used to confuse me. I used to wonder, is this the
electrical potential energy of that charge, Q1? Or is it the electrical potential
energy of this charge, Q2? Well, the best way to think about this is that this is the
electrical potential energy of the system of charges. So you need two of these charges to have potential energy at all. If you only had one, there
would be no potential energy, so think of this potential
energy as the potential energy that exists in this charge system. So since this is an
electrical potential energy and all energy has units of
joules if you’re using SI units, this will also have units of joules. Something else that’s important to know is that this electrical
potential energy is a scalar. That is to say, it is not a vector. There’s no direction of this energy. It’s just a number with
a unit that tells you how much potential
energy is in that system. In other words, this is good news. When things are vectors, you have to break them into pieces. And potentially you’ve got
component problems here, you got to figure out how much
of that vector points right and how much points up. But that’s not the case with
electrical potential energy. There’s no direction of this energy, so there will never be any
components of this energy. It is simply just the
electrical potential energy. So how do you use this formula? What do problems look like? Let’s try a sample problem
to give you some feel for how you might use this
equation in a given problem. Okay, so for our sample problem, let’s say we know the
values of the charges. And let’s say they start from rest, separated by a distance
of three centimeters. And after you release them from rest, you let them fly to a
distance 12 centimeters apart. And we need to know one more thing. We need to know the mass of each charge. So let’s just say that
each charge is one kilogram just to make the numbers come out nice. So the question we want to know is, how fast are these
charges going to be moving once they’ve made it 12
centimeters away from each other? So the blue one here, Q1, is
gonna be speeding to the left. Q2’s gonna be speeding to the right. How fast are they gonna be moving? And to figure this out, we’re gonna use conservation of energy. For our energy system,
we’ll include both charges, and we’ll say that if
we’ve included everything in our system, then the total initial
energy of our system is gonna equal the total
final energy of our system. What kind of energy did
our system have initially? Well, the system started
from rest initially, so there was no kinetic
energy to start with. There would’ve only been
electric potential energy to start with. So just call that u initial. And then that’s gonna have
to equal the final energy once they’re 12 centimeters apart. So the farther apart,
they’re gonna have less electrical potential energy
but they’re still gonna have some potential energy. So we’ll call that u final. And now they’re gonna be moving. So since these charges are moving, they’re gonna have kinetic energy. So plus the kinetic energy of our system. So we’ll use our formula for
electrical potential energy and we’ll get that the initial
electrical potential energy is gonna be nine times 10 to the ninth since that’s the electric constant K multiplied by the charge of Q1. That’s gonna be four microcoulombs. A micro is 10 to the negative sixth. So you gotta turn that
into regular coulombs. And then multiplied by Q2,
which is two microcoulombs. So that’d be two times
10 to the negative sixth divided by the distance. Well, this was the initial
electrical potential energy so this would be the initial
distance between them. That center to center distance
was three centimeters, but I can’t plug in three. This is in centimeters. If I want my units to be in joules, so that I get speeds in meters per second, I’ve got to convert this to meters, and three centimeters in
meters is 0.03 meters. You divide by a hundred, because there’s 100
centimeters in one meter. And I don’t square this. The r in the bottom of
here is not squared, so you don’t square that r. So that’s gonna be equal to it’s gonna be equal to another term that looks just like this. So I’m gonna copy and paste that. The only difference is
that now this is the final electrical potential energy. Well, the K value is the same. The value of each charge is the same. The only thing that’s different is that after they’ve flown apart, they’re no longer three centimeters apart, they’re 12 centimeters apart. So we’ll plug in 0.12 meters, since 12 centimeters is .12 meters. And then we have to
add the kinetic energy. So I’m just gonna call this k for now. The total kinetic energy of the system after they’ve reached 12 centimeters. Well, if you calculate these terms, if you multiply all this
out on the left-hand side, you get 2.4 joules of initial
electrical potential energy. And that’s gonna equal, if you calculate all of this in this term, multiply the charges, divide by .12 and multiply by nine
times 10 to the ninth, you get 0.6 joules of
electrical potential energy after they’re 12 centimeters apart plus the amount of kinetic
energy in the system, so we can replace this
kinetic energy of our system with the formula for kinetic energy, which is gonna be one half m-v squared. But here’s the problem. Both of these charges are moving. So if we want to do this correctly, we’re gonna have to take into account that both of these charges
are gonna have kinetic energy, not just one of them. If I only put one half times
one kilogram times v squared, I’d get the wrong answer because I would’ve neglected
the fact that the other charge also had kinetic energy. So we could do one of two things. Since these masses are the same, they’re gonna have the same speed, and that means we can write this mass here as two kilograms times
the common speed squared or you could just write two
terms, one for each charge. This is a little safer. I’m just gonna do that. Conceptually, it’s a little
easier to think about. Okay, so I solve this. 2.4 minus .6 is gonna be 1.8 joules, and that’s gonna equal one
half times one kilogram times the speed of that
second particle squared plus one half times one
kilogram times the speed of the first particle squared. And here’s where we have
to make that argument. Since these have the same mass, they’re gonna be moving
with the same speed. One half v squared plus one half v squared which is really just v squared, because a half of v squared
plus a half of v squared is a whole of v squared. Now if you’re clever, you
might be like, “Wait a minute. “This charge, even though
it had the same mass, “it had more charge than this charge did. “Isn’t this charge gonna be moving faster “since it had more charge?” No, it’s not. The force that these charges
are gonna exert on each other are always the same, even if
they have different charges. That’s counter-intuitive, but it’s true. Newton’s third law tells
us that has to be true. So if they exert the
same force on each other over the same amount of distance, then they will do the same
amount of work on each other. And if they have the same mass, that means they’re gonna
end with the same speed as each other. So they’ll have the same speed,
a common speed we’ll call v. So now to solve for v, I just take a square root of each side
and I get that the speed of each charge is gonna
be the square root of 1.8. Technically I’d have to divide that joules by kilograms first, because
even though this was a 1, to make the units come out right I’d have to have joule per kilogram. And if I take the square root,
I get 1.3 meters per second. That’s how fast these
charges are gonna be moving after they’ve moved to the point where they’re 12 centimeters
away from each other. Conceptually, potential
energy was turning into kinetic energy. So the final potential energy was less than the initial potential energy, and all that energy went
into the kinetic energies of these charges. So we solved this problem. Let’s switch it up. Let’s say instead of starting
these charges from rest three centimeters apart, let’s say we start them from
rest 12 centimeters apart but we make this Q2 negative. So now instead of being
positive 2 microcoulombs, we’re gonna make this
negative 2 microcoulombs. And now that this charge is negative, it’s attracted to the positive charge, and likewise this positive charge is attracted to the negative charge. So let’s say we released these from rest 12 centimeters apart, and we allowed them to
fly forward to each other until they’re three centimeters apart. And we ask the same question, how fast are they gonna be going
when they get to this point where they’re three centimeters apart? Okay, so what would change
in the math up here? Since they’re still released from rest, we still start with no kinetic energy, so that doesn’t change. But this time, they didn’t
start three centimeters apart. So instead of starting with
three and ending with 12, they’re gonna start 12 centimeters apart and end three centimeters apart. All right, so what else changes up here? The only other thing that
changed was the sign of Q2. And you might think, I
shouldn’t plug in the signs of the charges in here, because that gets me mixed up. But that was for electric
field and electric force. If these aren’t vectors,
you can plug in positives and negative signs. And you should. The easiest thing to do is just plug in those
positives and negatives. And this equation will just tell you whether you end up with a
positive potential energy or a negative potential energy. We don’t like including
this in the electric field and electric force formulas because those are vectors, and if they’re vectors,
we’re gonna have to decide what direction they point and
this negative can screw us up. But it’s not gonna screw
us up in this case. This negative is just gonna tell us whether we have positive potential energy or negative potential energy. There’s no worry about
breaking up a vector, because these are scalars. So long story short, we
plug in the positive signs if it’s a positive charge. We plug in the negative sign
if it’s a negative charge. This formula’s smart
enough to figure it out, since it’s a scalar, we
don’t have to worry about breaking up any components. In other words, instead of two up here, we’re gonna have negative
two microcoulombs. And instead of positive
two in this formula, we’re gonna have negative
two microcoulombs. So if we multiply out the left-hand side, it might not be surprising. All we’re gonna get is negative 0.6 joules of initial potential energy. And this might worry you. You might be like, “Wait a minute, “we’re starting with
negative potential energy?” You might say, “That makes no sense. “How are we gonna get kinetic
energy out of a system “that starts with less than
zero potential energy?” So it seems kind of weird. How can I start with less than
zero or zero potential energy and still get kinetic energy out? Well, it’s just because this term, your final potential energy term, is gonna be even more negative. If I calculate this term, I end
up with negative 2.4 joules. And then we add to that the
kinetic energy of the system. So in other words, our system is still gaining kinetic energy because it’s still
losing potential energy. Just because you’ve got
negative potential energy doesn’t mean you can’t
have less potential energy than you started with. It’s kind of like finances. Trust me, if you start
with less than zero money, if you start in debt, that doesn’t mean you can’t spend money. You can still get a credit
card and become more in debt. You can still get stuff,
even if you have no money or less than zero money. It just means you’re gonna
go more and more in debt. And that’s what this
electric potential is doing. It’s becoming more and more in debt so that it can finance an
increase in kinetic energy. Not the best financial
decision, but this is physics, so they don’t care. All right, so we solve
this for the kinetic energy of the system. We add 2.4 joules to both sides and we get positive 1.8
joules on the left hand side equals We’ll have two terms because
they’re both gonna be moving. We’ll have the one half times one kilogram times the speed of one
of the charges squared plus one half times one
kilogram times the speed of the other charge squared, which again just gives us v squared. And if we solve this for v,
we’re gonna get the same value we got last time, 1.3 meters per second. So recapping the formula for
the electrical potential energy between two charges is gonna be k Q1 Q2 over r. And since the energy is a scalar, you can plug in those negative signs to tell you if the potential
energy is positive or negative. Since this is energy, you
could use it in conservation of energy. And it’s possible for systems to have negative electric potential energy, and those systems can still convert energy into kinetic energy. They would just have to make sure that their electric
potential energy becomes even more negative.

20 comments on “Electric potential energy of charges | Physics | Khan Academy”

  1. Stupidity Master says:

    In Current and Electricity what is the higher potential… Guys this thing is messing me soo much..
    I don't know why but i can simply get this thing clear.. My Exams are just around the corner.
    I have Questions of which I need some simple answers.
    1.What basically is the electricity??
    2.Do electrons flow from Positive terminal to the negative terminal or from the negative terminal to the positive terminal.
    3. What does actually flow: Protons or Electrons?
    Please reply to me…
    Thank you in advance.

  2. Nick says:

    Thank You!

  3. Gary Lai says:

    Nobel prize in teaching

  4. Samar surya says:

    Hey khan, to be honest,the level of clarity,depth of understanding,way of explanation are simply awesome man..your videos are mind blowing..My humble request to you "PLEASE DONT STOP ANYWHERE"..keep uploading more and more videos in physics

  5. Anish Mulchandani says:

    Not the best financial decision, but this is physics. So, they don't care.

  6. Nisreen Alqaisi says:

    What if the masses weren't the same? would you need more information to solve for their velocities?

  7. Mohanad Abdulrahim says:

    What does a negative potential energy mean?

  8. Umar Patel says:

    "This is physics so they don't care" lololololol

  9. Nicholle Willis says:

    I really loved this video because the speaker is so funny. ^^

  10. haziq hassnain says:


  11. Raspberry says:

    I'm still a bit confused as to why the speeds are the same 🙁

  12. سراج الليل says:

    thanks for the explanation, but I am still confused about something:
    in a series circuit made of a battery/cell providing an electromotive force of 9 V, and a bulb, the electrons give the bulb all the energy they are carrying which is 9 V, so how do they continue moving after leaving the bulb if they lost all their energy?

  13. Thomery says:

    This is so helpful!! Thank you so much 😊

  14. Bohan Shan says:

    I really liked how he explained what each of the values meant. Keep up the good work!

  15. Zhe Kang says:

    thanks for saving my grades

  16. Lin哈尼 says:

    Thank you

  17. VON says:

    So basically using signs in this equation only gives better understanding and does not affect the answer, Making it of no use eh

  18. Reeti Chauhan says:

    How is no one talking about his ability to draw perfect arrows?? ( well, obviously apart from how good he teaches)

  19. vraj Patel says:

    Great video but I’m confused
    1. other resources use r^2 instead of r. Why is that

  20. Luiza Saunders says:

    Apologies to Sal for this comment, even though it is intended, if anything, as constructive criticism – but the way David explains things is so much clearer. He does not feel the need to fill in every bit of time with words (which I'm sorry to say oftentimes comes off as rambling with Sal), but speaks eloquently, letting us take it in. Also I can tell the graphics are cleaner as well, I'm sure Sal can implement this. But the slower and better organised mode of speaking, the clear explanations, for me they make David preferable for me. Please make more videos

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