# 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.

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.

Thank You!

Nobel prize in teaching

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

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

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

What does a negative potential energy mean?

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

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

amazing…

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

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?

This is so helpful!! Thank you so much 😊

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

thanks for saving my grades

Thank you

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

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

Great video but I’m confused

1. other resources use r^2 instead of r. Why is that

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