Potential difference and electric field relationship help

Electric field (video) | Khan Academy

potential difference and electric field relationship help

It might make more sense to think of a rearrangement of this relationship: At any point in an electric field, the amount of electric potential energy is the electric. Description: To understand the relation between the potential difference and the electric field, take two points A and B in the electric field with a very small. The electric field is by definition the force per unit charge, so that multiplying the field times This association is the reminder of many often-used relationships.

The distance is equal to let's say-- let's make it easy. Let's say 2 meters. So first of all, we can say, in general, what is the electric field 2 meters away from? So what is the electric field out here? This is 2, right? And it's going to be 2 meters away. It's radial so it's actually along this whole circle. What is the electric field there?

Well, the electric field at that point is going to be equal to what? And it's also a vector quantity, right? Because we're dividing a vector quantity by a scalar quantity charge. So the electric field at that point is going to be k times whatever charge it is divided by 2 meters, so divided by 2 meters squared, so that's 4, right, distance squared. And so if I know the electric field at any given point and then I say, well, what happens if I put a negative 1 coulomb charge there, all I have to do is say, well, the force is going to be equal to the charge that I place there times the electric field at that point, right?

So in this case, we said the electric field at this point is equal to-- and the units for electric field are newtons per coulomb, and that makes sense, right? Because it's force divided by charge, so newtons per coulomb.

potential difference and electric field relationship help

So if we know that the electric charge-- well, let me put some real numbers here. Let's say that this is-- I don't know. It's going to be a really large number, but let's say this-- let me pick a smaller number. Let's say this is 1 times 10 to the minus 6 coulombs, right?

potential difference and electric field relationship help

If that's 1 times 10 to the minus 6 coulombs, what is the electric field at that point? Let me switch colors again. What's the electric field at that point? Well, the electric field at that point is going to be equal to Coulomb's constant, which is 9 times 10 to the ninth-- times the charge generating the field-- times 1 times 10 to the minus 6 coulombs. And then we are 2 meters away, so 2 squared. So that equals 9 times 10 to the third divided by 4.

So I don't know, what is that? So we know that this is generating a field that when we're 2 meters away, at a radius of 2 meters, so roughly that circle around it, this is generating a field that if I were to put-- let's say I were to place a 1 coulomb charge here, the force exerted on that 1 coulomb charge is going to be equal to 1 coulomb times the electric fields, times 2, newtons per coulomb.

So the coulombs cancel out, and you'll have 2, newtons, which is a lot, and that's because 1 coulomb is a very, very large charge. And then a question you should ask yourself: If this is 1 times 10 to the negative 6 coulombs and this is 1 coulomb, in which direction will the force be?

Electric Potential and Electric Field

Well, they're both positive, so the force is going to be outwards, right? So let's take this notion and see if we can somehow draw an electric field around a particle, just to get an intuition of what happens when we later put a charge anywhere near the particle. So there's a couple of ways to visualize an electric field. One way to visualize it is if I have a-- let's say I have a point charge here Q.

What would be the path of a positive charge if I placed it someplace on this Q? Well, if I put a positive charge here and this Q is positive, that positive charge is just going to accelerate outward, right? It's just going to go straight out, but it's going to accelerate at an ever-slowing rate, right? Because here, when you're really close, the outward force is very strong, and then as you get further and further away, the electrostatic force from this charge becomes weaker and weaker, or you could say the field becomes weaker and weaker.

But that's the path of a-- it'll just be radially outward-- of a positive test charge. And then if I put it here, well, it would be radially outward that way. It wouldn't curve the way I drew it. It would be a straight line. I should actually use the line tool. If I did it here, it would be like that, but then I can't draw the arrows. If I was here, it would out like that. I think you get the picture. At any point, a positive test charge would just go straight out away from our charge Q.

And to some degree, one measure of-- and these are called electric field lines. And one measure of how strong the field is, is if you actually took a unit area and you saw how dense the field lines are. So here, they're relatively sparse, while if I did that same area up here-- I know it's not that obvious.

I'm getting more field lines in. But actually, that's not a good way to view it because I'm covering so much area. Let me undo both of them. You can imagine if I had a lot more lines, if I did this area, for example, in that area, I'm capturing two of these field lines.

Well, if I did that exact same area out here, I'm only capturing one of the field lines, although you could have a bunch more in between here. And that makes sense, right? Because as you get closer and closer to the source of the electric field, the charge gets stronger.

Another way that you could have done this, and this would have actually more clearly shown the magnitude of the field at any point, is you could have-- you could say, OK, if that's my charge Q, you could say, well, really close, the field is strong. So at this point, the vector, the newtons per coulomb, is that strong, that strong, that strong, that strong. We're just taking sample points.

You're just finding the magnitude when you find this formula. The way you find the direction is just by knowing that the field created by a positive is always radially away from that positive.

But even though this formula just gives you the magnitude, that's still really useful. So we're gonna use this. This gives you the magnitude of the electric field from a point charge at any point away from that point charge. Let's solve some examples here. Let's use this thing. Let's say you had a positive two nanoCoulomb charge, and you wanted to determine the size and direction of the electric field at a point three meters below that charge.

We wanna know, what's the size and direction of the electric field right there. To get the size, we could use the new formula we've got, which says that the electric field created by a charge, Q, is gonna be equal to k times that Q over r squared.

We'll use that down here. We get k, which is always nine times 10 to the ninth, and then we multiply by the charge creating the field, which in this case, it's this positive two nanoCoulomb charge.

Nano stands for 10 to the negative ninth, and then we divide by the distance from our charge to the point where we wanna find the field. You can't forget to square this. People forget to square this all the time. It doesn't come out right, so you gotta remember to square the r, and if we solve this, 10 to the ninth times 10 to the negative ninth goes away. And then nine divided by three squared is just nine divided by nine, so all of that's gonna go away. And all we really get for the electric field is that it's gonna be two Newtons per Coulomb at this point here.

So that is the magnitude. This gives us the magnitude of the electric field at this point in space, that's how you get the size of it. How do we get the direction?

Magnitude of electric field created by a charge

We just ask, what was creating this field? It was a positive charge.

potential difference and electric field relationship help

Positive charges always create fields that point radially away from them, and at this point, radially away from this positive is gonna point straight down. So we've got an electric field from this two nanoCoulomb charge that points straight down, and has a value of two Newtons per Coulomb. What did this number mean? It means if we put another charge at that point in space, some little charge q, then there would be two Newtons for every Coulomb of charge that you put there, since we know that electric field is the amount of force per charge.

Notice that even though this electric field came out to be positive, it was pointing down because all we're getting out of this calculation is the magnitude of the electric field.

potential difference and electric field relationship help

Let's do one more. Let's try this one out. Let's say you had a negative four microCoulomb charge, and you wanted to determine the size and direction of the electric field at a point six meters away, to the left of that charge. We use the same formula. We'll say that the electric field created by this negative charge is gonna equal k, which is always nine times 10 to the ninth. And then multiplied by the charge creating that field, which in this case is negative four microCoulombs, but I am not gonna plug in this negative sign because I know all this formula's giving me is the magnitude of the electric field.

I'm not gonna get tricked into thinking that this negative sign would tell me the direction of the electric field. I mean, it does tell you the direction. It tells you that it points radially inward, but it's safest, gotta take my word on this, safest to just leave that negative sign out and know that this is just the magnitude of the electric field.

So we have four microCoulombs. Micro stands for 10 to the negative sixth. And then we divide by the distance from the center of that charge to the point we wanna determine the electric field at, which is right here, and we square it. That's six meters, and we cannot forget to square. So this electric field's gonna be Newtons per Coulomb at that point in space.

That's the magnitude of the electric field. Or, in other words, that's the size of the electric field at that point. We're gonna decide this by thinking carefully about it.

In other words, we don't include this negative, not because direction isn't important. We don't include this negative because direction is so important, we're gonna make sure we get this right.

What I say is that I've got a negative charge. I know that negative charges create fields that point radially into them. And that means over here at this point to the left, the electric field is pointing radially toward that negative charge, it would point to the right. So we'd have Newtons per Coulomb of electric field, and it would point to the right. Note, if I would've just naively plugged this negative sign in over to here, I would've come out with a negative value for my electric field, and I might've thought, well, negative, that means leftward, right?

So that means it points to the left. And I would've got the wrong direction.

potential difference and electric field relationship help

So that's why we don't do that. All this negative sign is representing, if you were to plug it in, is that it's pointing radially inward, but radially in could mean right if you're over here to the left. It could mean left if you're over here to the right. It could mean up if you're underneath the charge, and it could mean down if you're above the charge.