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Net electric field from multiple charges in 1D

In this video David solves an example problem to find the net electric field created by multiple charges at a point in between them. Created by David SantoPietro.

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  • blobby green style avatar for user Bharadwaj Innamuri
    does it mean that the magnitude of net electric field created halfway between two charges of same magnitude and charge be always zero?irrespective of distance between them?Then how do they repel?
    (19 votes)
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    • leafers sapling style avatar for user Om Bhojraj
      Think of it this way. Two people are pushing each other with the same force. If you go and stand in the middle of them you wont have to apply any sort of force to stand. But the guys on the side, who are actually doing all the pushing will have stop themselves from moving backward with their legs. That's why the charges actually repel if no external force is holding the two charges together.
      (5 votes)
  • blobby green style avatar for user sukruthageorge1999
    Isn't electric field a vector quantity? Then how is it possible to add them up using algebraic sum? shouldn't vector sum be used?
    (4 votes)
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    • blobby green style avatar for user levgenid
      You are right to assume that it is a vector quantity. The reason we are adding them or subtracting them algebraically is that we have already figured out the direction of the vectors before we write them down. Thus we can decide if the vectors have a positive sign (pointing to the right or up if we have decided that these are the positive directions for example) or a negative sign (pointing to the left or down).
      (13 votes)
  • male robot hal style avatar for user yousefr
    Why is the negative charge pointing to the right? Shouldn't it be pointing left, since negative and positive charges attract?
    (2 votes)
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    • leafers tree style avatar for user Jared Keele
      You are right that they attract but when talking about Electric Fields, think about them as roads. When dealing with a negative Q, electric fields always point toward the charge and positive Q field always point away from the charge.
      (7 votes)
  • piceratops seedling style avatar for user Ayushi Jain
    What does one mean by the point that the two same charges are placed in same line and another charge is brought exactly between the charges such that the system is in an equilibrium?
    (2 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      My understanding of the question would be this...

      equilibrium = these three charges have no acceleration. They dont experience a net force. The forces within the system is balanced.

      More detail...

      Take two positive charges. A distance r between them. They will feel a mutual force of repulsion.

      Now place a negative charge half way between them.
      Now the negative charge feels two (equal) forces; one from each of the two positive forces.
      Each of the positive forces continue to feel the positive (Repulsive) force from the other positive charge. PLUS the attractive force from the negative charge.

      [ Hint: Think about what happens to the forces when r is very large and also when very small]


      So; i think the question is asking this... what must the distance r be (the distance between the positive charges) so that all these forces cancel out? ie the system is in equilibrium

      Alternatively, the question may be asking you to find r/2. (the distance from centre to outer charges) but the maths is about the same..

      Hope that makes sense

      and do let us know when you get your answer

      [For a mathematical / visual solution You might try drawing graphs of forces/ distance and total (resultant forces) for each particle]
      :-)
      (6 votes)
  • blobby green style avatar for user anissa814
    What would happen if both charges were negative? They would both point inward, so do they cancel?
    (2 votes)
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    • spunky sam blue style avatar for user Chimaa Tamim
      The same thing would happen as in the case for positively charged particles.
      But this time each negative charge's electric field will point inwardly towards itself.Since both charges are put at opposite ends,then each charge's electric field will be opposite with respect to the other,thus counteracting each other's effect on the particle placed halfway between them.
      (3 votes)
  • duskpin tree style avatar for user Prakhar Gupta
    Is there any other unit or name for newtons per coulombs?
    (1 vote)
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  • duskpin ultimate style avatar for user Arki🖤
    is the direction of the electric field created by a negative charge radially outward in the vicinity of another negative charge?
    (1 vote)
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  • male robot donald style avatar for user GURU
    At , how when the direction is right it is +ve , when left it is -ve ?
    (1 vote)
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    • marcimus pink style avatar for user Marcus Hards
      Great question,
      The + sign indicates the field ( at some point ) is pointing to the right.
      The - sign indicates the field ( at some point ) is pointing to the left.

      This comes in incredibly handy when using mathematics to work out electric field problems. Instead of writing 'the electrical field at this point is pointing to the right'
      or 'this electrical field is pointing to the left at this particular point' we can express both statements with one of two symbols.
      (3 votes)
  • blobby green style avatar for user nkululekobonginkosi005
    Two equal positive charges are held in place at a fixed distance. If you put a third positive charge midway between these two charges, its electrical potential energy of the system (relative to infinity) is zero because the electrical forces on the third charge due to the two fixed charges just balance each other.IS THIS TRUE OR FALSE
    (2 votes)
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    • blobby green style avatar for user graciepaige17
      It is only true when when the charges of the first 2 charges are equal and opposite [i.e. one being negative and other positive, but with the same magnitude value {eg. +2 and -2 or +8 and -8}].. like in the above given case.
      But, if you are to take charges that are not equal and opposite, it is false
      (1 vote)
  • male robot donald style avatar for user salmaniajz1999
    why here you write 10 to power -9 with both 8
    (1 vote)
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Video transcript

- [Instructor] So it turns out solving electric field problems gets significantly harder when there's multiple charges. I mean, theoretically it shouldn't, but people have a lot more problems when there's multiple charges involved. So say the question is this; let's say we wanted to know what's the magnitude and direction of the net electric field, i.e. the total electric field, created halfway between these two charges down here. So you've got a positive eight nanocoulomb charge and a negative eight nanocoulomb charge, and they're separated by six meters from the center to center distance. But what we want to know is what's the total electric field that they both create right there? So each charge is going to create an electric field at this point, and if you add up like vectors, those electric fields, what total electric field would you get? Now at first you might think, well you should just get zero, right? It's very tempting to say that the electric field is just gonna be zero there because you've got a positive eight nanocoulomb charge and a negative eight nanocoulomb charge and those should just cancel, right? But you have to be really careful, turns out that's not true here, this is not gonna be true. And to see why, first you should just draw what is the direction of each field at that point? So this positive eight nanocoulomb charge is gonna create a field at this point that goes radially away from the positive charge, and so it's gonna go to the right. And I'm not even looking, so when I'm trying to find the electric field from this positive charge over here, I'm not even paying attention to this negative charge, I pretend like this negative charge doesn't even exist. Then I just ask what field would this positive charge create? It's still gonna create that field whether this negative charge is over here or not. And now I can do the same thing, I can ask what field would this negative charge create? And I'm gonna pretend like this positive charge isn't even here. So negative charges create a field that go radially in. So over here radially in would point to the right. So these don't cancel. The negative charge created a field radially in, that was to the right, the positive charge created a field radially out of the positive charge, and that was to the right. So not only are these not gonna cancel, these are gonna add up to twice the fields cuz you're gonna add up these vectors, you just add them up if they're in the same direction, and you'll get two times the contribution from one of them. So it's not always the case, in other words it's not always the case that a negative charge and a positive charge have to cancel their electric fields. Those electric fields might point the same direction, so you gotta be careful. So how do we find this net electric field then, what do we do? Well we're gonna say that, all right, this electric field, the first thing I can say is this net electric field is just gonna point in the x direction. So this is just really in the x direction, all I really care about is the electric field in this horizontal direction, and it's gonna be equal to the sum of the electric fields each charge creates there. So we'll do the blue charge first, that's gonna be k times the blue charge divided by r squared. Then we'll do the yellow charge, it's gonna be plus k, the charge of that yellow charge, divided by r squared. So we'll plug in some values here, this k is always nine times 10 to the ninth, and the q of this blue charge was positive eight nanocoulombs, nano is 10 to the negative ninth, I like using nano because then that negative nine cancels with that positive nine. And what distance do I put in here? A lot of people wanna put in six, but that's not what I want. Think about it, I want the net electric field halfway between the two charges, so the r that I care about in this electric field formula is the distance from the charge to the point where I want to determine the electric field, and in that case this is three meters. So for this case, from the charge to the point I'm concerned about finding the field is three meters, not six meters. If we were finding the force these charges exert on each other, then I'd have to use six meters, but that's not what I'm finding, I'm finding the field each charge creates at this halfway point. So I'm gonna plug in three meters down here, and I can't forget to square it. And now I have to be careful, just cuz my charge is positive doesn't necessarily mean that the contribution to the electric field is positive. You have to check, you can't rely on the sign of this charge to tell you whether the contribution's positive or negative. I've gotta look at what direction it points, the direction this positive charge creates a field is to the right. Since that's typically the direction we call positive, then I'm okay with calling this entire term here positive. Then we're gonna have another term. I'm gonna leave off the plus or minus cuz, I mean, it might be plus, it might be minus, we'll leave that off for a second, we'll have to decide when we know what direction it goes. So we do nine times 10 to the ninth, and then the charge is negative eight nanocoulombs, but I am not gonna plug in the negative sign. Oops, and I left off coulomb on the other one here, sorry. And then again, the distance I want is from the charge to the point where we want to find the field, and that again is three meters, and we can't forget to square it. So should this contribution be positive or negative? I can't rely on the negative sign to tell me that, I've gotta look at what direction it goes. Since it goes to the right, that's the positive direction, so this is gonna be plus, these add up, these both go the same direction, the positive direction, so the total net electric field is just gonna be both of these added up. So if I do this, if I square this three I'm getting nine, and nine divided by nine is just one, so I get eight newtons per coulomb, and then this term is really the same thing, nine is divided by nine so that goes away, 10 to the ninth cancels with 10 to the negative ninth and all I'm left with is this eight, so it'd be plus eight newtons per coulomb. So each charge is contributing eight newtons per coulomb of electric field at this point which means that the total net electric field would just be 16 newtons per coulomb at that point. That is the net electric field, that's the magnitude of the net electric field at that point between them. And which way does it go, what's the direction? It goes to the right cuz both of these vectors pointed to the right so the total is gonna be twice as big as one of them and also to the right. Now if you have a case like this and both terms, you know both terms are gonna be equal, you can just write one of them down and multiply by two, you don't have to just add them both up, but I wanted to show you this way so you could see how everything works out. And in the end we get 16 newtons per coulomb for the total field which points to the right. Now what if we changed this, what if we made this instead of a negative eight nanocoulomb charge we made this a positive eight nanocoulomb charge? Well it would no longer create an electric field that points to the right. Positive charges create fields that point radially away from them, so it would create its electric field to the left, which means down here when we find its contribution to the electric field we'd have to include it as a negative contribution cuz it's pointing in the negative direction. Even though it's a positive charge, the contribution it gives to the total electric field is negative cuz it points in the negative direction. And that would give me zero, so if I had this a positive this whole thing would add up to zero cuz I'd have eight and then minus eight and I'd get zero newtons per coulomb, so the electric field would completely cancel right in the middle. So what I'm saying is you have to be very careful with your negative signs, don't just assume these contributions are always gonna add up. You can find each one always plugging in the charges as positive even if they're negative and then decide should I add or subtract these contributions based on whether they go to the right or to the left. If they point to the right you'd choose a positive in front of this term since it points in the positive x direction. And if they point to the left you're gonna choose a negative in front of this term because it would point in the negative x direction. So recapping, to find the total electric field from multiple charges, draw the electric field each charge creates at the point where you want to determine the total electric field, use this formula to get the magnitude of the contribution from each charge, then decide whether those contributions should be positive or negative based not on the sign of the charge but the direction the field is pointing from that charge, add up the two contributions, and that'll give you the total electric field at that point.