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Conceptual question about coulomb's law

  1. Aug 18, 2011 #1
    I know that for electric field you always have to break E up into the x and y components for each charge---do you have to do the same thing when finding electric force using Coulombs law? I get the correct answer when i do NOT break F up into components (ie when finding F between 2 charges with one on the x-axis and the other on the y axis)...can someone please explain why I do not need to break up F into F_x and F_y?
     
  2. jcsd
  3. Aug 18, 2011 #2

    Pengwuino

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    Why wouldn't you need to break up your electric force by components too? The electric force is, in the absence of magnetic effects, simply the charge (which is a scalar) times the electric field. Anything you do with the field you'd do with the force. You probably should tell us specifically what question you're looking at that is confusing you.
     
  4. Aug 18, 2011 #3
    I'm actually looking at a bunch of different questions and comparing how I solved each one. Someone just told me that whenever the answer asks only for the magnitude (of either F or E) then I do not ever use components. Is this true? I'm a little hesitant because I know that using sine or cosine seriously manipulates the numbers, and when I think of "magnitude" I really only think of getting a positive result. (I mean it DOES make sense that I wouldn't use components if I'm only looking for magnitude, I'd just like this to be verified before I commit it to memory.)
     
  5. Aug 18, 2011 #4

    Pengwuino

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    It depends what they meant.

    The magnitude of the field is given by [itex]|F| =\sqrt{F_x^2 + F_y^2 + F_z^2}[/itex] so you do need the components to calculate this. And actually this is true of ANY vector quantity, not just the electric force or field or what have you.
     
  6. Aug 18, 2011 #5
  7. Aug 18, 2011 #6

    Pengwuino

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    You do have to have to solve for everything component-wise.

    So say you have A at the upper tip. You compute the magnitude between, for example, charge A and C (not the C you just used, but the identifier for what sphere we're talking about). The magnitude is given by [itex]F = {{kq_Aq_C}\over{d^2}}[/itex]. Now, if that were the end of the story, this is all you would need. However, you have other charges that have to be taken into account so you need to break that magnitude up into x/y components. Then you have to add in the other charges (whose forces you break up until x/y components as well). You do your vector addition and find the total force acting on A, broken up into components, and at the very end you can use the equation I put up earlier.
     
  8. Aug 18, 2011 #7
    I'm not sure you answered my question; *just* for finding the magnitude of F between A and C, why do I not need to solve as such:

    F_x = k(q_A)(q_c)cos(theta)/(d^2)
    F_y = k(q_A)(q_c)sin(theta)/(d^2)

    and then obtain lFl by sqrt[(F_x)^2 + (F_y)^2]

    hopefully that makes more sense.
     
  9. Aug 18, 2011 #8

    Pengwuino

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    No, you don't need to use the components if it's JUST 2 charges. If there is more than 1 charge, however, you MUST use components.
     
  10. Aug 18, 2011 #9
    Ok but why? haha sorry but I really need to understand the reasoning behind what I'm doing or else I see it starting a downward spiral as the course progresses. is it maybe because if you dont have to add any like-components together then the sin^2 + cos^2 would just equal 1 and make converting into components arbitrary?
     
  11. Aug 18, 2011 #10

    gneill

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    If there is only one force acting then the magnitude of the resultant force is equal to the magnitude of that single force; no vector components are required because there's only a single force acting in a single direction. When there are two or more forces acting and they are acting along different directions, then you must decompose each force into components, add the like components, then determine the overall magnitude from the composition of those components. It boils down to vector addition.
     
  12. Aug 18, 2011 #11
    ok thanks---its been like 2 years since i had physics 1 unfortunately :(
     
  13. Aug 18, 2011 #12

    Pengwuino

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    Another thing to realize is that if you say you have 2 charges only, you can arbitrarily establish a coordinate system so that the charges lie on the same axis and it's exactly as if you started with 2 charges on say, the x-axis.
     
  14. Aug 18, 2011 #13
    ok that makes sense, and i thought about that i just didnt know if it was allowed lol thanks again! :-)
     
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