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Electric field

We can think of the forces between charges as something that comes from a property of space. That property is called the electric field. Created by Sal Khan.

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  • aqualine ultimate style avatar for user Justin Z
    At , how do two protons or positive charges repel each other? They do repel each other, but how?
    (32 votes)
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  • male robot hal style avatar for user SARTHAK
    Is it possible to have vector addition of gravitational and electric field to get a total field ?
    (14 votes)
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  • piceratops seedling style avatar for user Karushen
    In a way, does Coulomb's Law mean that every particle, no matter how far away they are from each other, has some effect/force on the other?? Just that its too small to really make a difference...
    (20 votes)
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  • blobby green style avatar for user Prag Sangha
    Why positive electric fields of lines are radially outwards?
    (10 votes)
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    • purple pi purple style avatar for user Asmita Pathak
      The test charge (a unit charge that is used to test the effect of the source charge) is assumed to be positive. So, the source charge, which is the charge from which the electric fields are generated, pushes the test charge outwards. And so, the electric field lines are radially outwards.
      Totally sure now. Hope that helped. :)
      (17 votes)
  • spunky sam blue style avatar for user Khubaib Abdullah
    I did not get what exactly an electric field is? Can someone give a simple definition?
    (12 votes)
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  • blobby green style avatar for user kshitijneekhra
    The Lightning Which We See In The Sky Is Said To Be A Result Of Static Electricity And Static Electricity Does Not Move Or Flow Then How Does Electricity From Clouds Reach The Ground
    (6 votes)
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    • leafers seedling style avatar for user Kyle Vastine
      Lightning is a fairly complex process. You can model a cloud as a half circle with a round uniform top and a flat bottom. There will be a charge buildup in the cloud as a result of the evaporation and condensation cycle. The top of the cloud will have a net positive charge and the bottom obtains a net negative charge. The net negative charge on the bottom of the cloud will cause Earth's surface beneath the cloud to accumulate a net positive charge (the negative charge at the bottom of the cloud exerts a force towards the earth that repels electrons on Earth's surface causing a local positive charge to accumulate).

      When there is a difference in charge potential separated by a distance (the negative underside of the cloud located say 1 mile from Earth's surface with a local positive charge) a voltage difference occurs. As charge accumulates in the cloud the total difference in electrical potential (voltage) between the cloud and Earth also increases. This causes the electric field between the underside of the cloud and the surface of Earth to become stronger.

      At a certain point the electric field will grow strong enough to cause the ionization of air to occur. As a molecule ionizes it loses electron(s) from its valence shell. These electrons will experience a force from the electric field created by the cloud and Earth that causes them to accelerate towards the positive charge on Earth surface.

      The lightning 'strikes' quickly because the potential builds until it reaches the point at which the air around it can ionize but once it reaches this point the large difference in potential energies created by the cloud and Earth accelerates these electrons seemingly instantaneously. The electrons 'flow' in a sense because the air ionizes, lowering the effective resistance, which results in the distinctive lightning bolt (the air being ionized) that you see.

      I could be wrong on some of the details but I'm fairly sure this is an accurate description of what causes lightning and what it really is.
      (5 votes)
  • spunky sam blue style avatar for user Chimaa Tamim
    If the electric field lines generated by a positively charged particle are radially outwards and those of an electron are radially inwards,then this is visualised as though the positively charged particle's electric field force is pushing the negative charge further away from itself while being simultaneously pulled towards the negative charge whose electric field lines point inwardly towards itself.Right?What I'm mainly confused about is that if opposite charges attract each other,then the respective electric fields' force vectors of the positive and negative charges should ppint inwardly towards each charge;say A and B are two oppositely charged particles,hence both particles are pulling upon each other,shouldn't we represent the force vector of the force of A on B as pointing inwardly towards A and the force vector of B on A as pointing inwardly towards B.If that's true,then why did Sal at schematize the electric field lines issued from A (which is +ve) point outwardly on B(which is -ve) as though it is repelling it.
    Thanks in advance
    (4 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      No problemo.

      Most important thing in your work is to understand what the arrows indicate.

      The direction of the arrow indicates the directon of the force acting on 'a positive charge'.

      Thats why they are 'outwards' for a proton and 'inwards' for an electron.

      and thats why a proton sitting in an electrons electric field would feel a force attracting it towards the centre of the electron. because the arrows point in and the proton is positive...

      and for an electron sitting in a protons electric field, if the arrows from the proton point outwards, then the force is inwards... this is because the arrow is the direction of force that would act on a positive charge.... electron is negative, so the force is in opposite direction...ie inwards.


      ok??
      (5 votes)
  • marcimus pink style avatar for user Mehak
    When u put distance instead of radius.. we actually measure the distance between the centres of the charges or the distance between their respective outer surfaces ?
    (3 votes)
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  • primosaur ultimate style avatar for user Shreya
    To explain the charge interaction between a charged and neutral object we use the concept of polarisation. So if I talk about a positively charged balloon and a wall, then the electrons are attracted towards the balloon and get crowded on the wall near the balloon. Why don't they enter the balloon?
    (3 votes)
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  • leaf green style avatar for user Daniel Jung
    I have a question in this video. About 6min 45s, you talked about the direction of the force. And I agree that the q particle will be move farther from the Q particle. But my question is, would the Q particle is also going to be move farther from the q particle?? and if it does, which particle is going to be exerted more force??
    (3 votes)
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    • leaf green style avatar for user Ahmed Jadoon
      Yes Q particle is also going to move away as q does.
      I have tried to know how much force both charges exert on each other.
      Lets say Q particle has 2 Coulomb charge and q has 1 Coulomb charge.You can calculate the electric field created by charges Q and q as E(Q)=F/q= k.Q/d2 and E(q)=F/Q= k.q/d2 respectively.In this way you get E(Q)=1.8*10^10 N/C
      E(q)=9*10^9 N/C

      We can calculate force exerted on charge q by charge Q as F=E(Q)*q= 1.8 x 10^10 N
      And force exerted on charge Q by charge q as
      F=E(q) * Q = 9*10^9 x 2= 1.8 x 10^10 N

      So we got to know both charges experiences same Force due to their somehow different electric fields.
      Let me know if it helps.My first ever answer on Khan Academy since I joined yesterday.
      (5 votes)

Video transcript

Let's imagine that instead of having two charges, we just have one charge by itself, sitting in a vacuum, sitting in space. So that's this charge here, and let's say its charge is Q. That's some number, whatever it is. That's it's charge. And I want to know, if I were to place another charge close to this Q, within its sphere of influence, what's going to happen to that other charge? What's going to be the net impact on it? And we know if this has some charge, if we put another charge here, if this is 1 coulomb and we put another charge here that's 1 coulomb, that they're both positive, they're going to repel each other, so there will be some force that pushes the next charge away. If it's a negative charge and I put it here, it'll be even a stronger force that pulls it in because it'll be closer. So in general, there's this notion of what we can call an electric field around this charge. And what's an electric field? We can debate whether it really exists, but what it allows us to do is imagine that somehow this charge is affecting the space around it in some way that whenever I put-- it's creating a field that whenever I put another charge in that field, I can predict how the field will affect that charge. So let's put it in a little more quantitative term so I stop confusing you. So Coulomb's Law told us that the force between two charges is going to be equal to Coulomb's constant times-- and in this case, the first charge is big Q. And let's say that the second notional charge that I eventually put in this field is small q, and then you divide by the distance between them. Sometimes it's called r because you can kind of view the distance as the radial distance between the two charges. So sometimes it says r squared, but it's the distance between them. So what we want to do if we want to calculate the field, we want to figure out how much force is there placed per charge at any point around this Q, so, say, at a given distance out here. At this distance, we want to know, for a given Q, what is the force going to be? So what we can do is we could take this equation up here and divide both sides by this small 1, and say, OK, the force-- and I will arbitrarily switch colors. The force per charge at this point-- let's call that d1-- is equal to Coulomb's constant times the charge of the particle that's creating the field divided by-- well, in this case, it's d1-- d1 squared, right? Or we could say, in general-- and this is the definition of the electric field, right? Well, this is the electric field at the point d1, and if we wanted a more general definition of the electric field, we'll just make this a general variable, so instead of having a particular distance, we'll define the field for all distances away from the point Q. So the electric field could be defined as Coulomb's constant times the charge creating the field divided by the distance squared, the distance we are away from the charge. So essentially, we've defined-- if you give me a force and a point around this charge anywhere, I can now tell you the exact force. For example, if I told you that I have a minus 1 coulomb charge and the distance is equal to-- oh, I don't know. 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. 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? 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? 2.5 times 10 to the third or 2,500 newtons per coulomb. 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 2,500 newtons per coulomb. So the coulombs cancel out, and you'll have 2,500 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? 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 can't possibly draw them at every single point. So at that point, that's the vector. That's the electric field vector. But then if we go a little bit further out, the vector is going to be-- it falls off. This one should be shorter, then this one should be even shorter, right? You could pick any point and you could actually calculate the electric field vector, and the further you go out, the shorter and shorter the electric field vectors get. And so, in general, there's all sorts of things you can draw the electric fields for. Let's say that this is a positive charge and that this is a negative charge. Let me switch colors so I don't have to erase things. If I have to draw the path of a positive test charge, it would go out radially from this charge, right? But then as it goes out, it'll start being attracted to this one the closer it gets to the negative, and then it'll curve in to the negative charge and these arrows go like this. And if I went from here, the positive one will be repelled really strong, really strong, it'll accelerate fast and it's rate of acceleration will slow down, but then as it gets closer to the negative one, it'll speed up again, and then that would be its path. Similarly, if there was a positive test charge here, its path would be like that, right? If it was here, its path would be like that. If it was here, it's path would be like that. If it was there, maybe its path is like that, and at some point, its path might never get to that-- this out here might just go straight out that way. That one would just go straight out, and here, the field lines would just come in, right? A positive test charge would just be naturally attracted to that negative charge. So that's, in general, what electric field lines show, and we could use our little area method and see that over here, if we picked a given area, the electric field is much weaker than if we picked that same area right here. We're getting more field lines in than we do right there. So that hopefully gives you a little sense for what an electric field is. It's really just a way of visualizing what the impact would be on a test charge if you bring it close to another charge. And hopefully, you know a little bit about Coulomb's constant. And let's just do a very simple-- I'm getting this out of the AP Physics book, but they say-- let's do a little simple problem: Calculate the static electric force between a 6 times 10 to the negative sixth coulomb charge. So 6 times-- oh, no, that's not on an electric field. Oh, here it says: What is the force acting on an electron placed in an external electric field where the electric field is-- they're saying it is 100 newtons per coulomb at that point, wherever the electron is. So the force on that, the force in general, is just going to be the charge times the electric field, and they say it's an electron, so what's the charge of an electron? Well, we know it's negative, and then in the first video, we learned that its charge is 1.6 times 10 to the negative nineteenth coulombs times 100 newtons per coulomb. The coulombs cancel out. And this is 10 squared, right? This is 10 to the positive 2, so it'll be 10 to the minus 19 times 10 to the positive 2. The force will be minus 1.6 times 10 to the minus 17 newtons. So the problems are pretty simple. I think the more important thing with electric fields is to really understand intuitively what's going on, and kind of how it's stronger near the point charges, and how it gets weaker as it goes away, and what the field lines depict, and how they can be used to at least approximate the strength of the field. I will see you in the next video.