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Electric field calculation doubt (dipole)

  1. Apr 6, 2017 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Hello all, i am having a hard time with EM, specially with the most simple concepts that i dont seem to grasp at all and the simple exercises that should be quite simple to solve.

    So the question is. ( i will omit the constant (1/4pieo) for practice sake

    When i calculate the eletric field, first i determine the r (distance vector) just like i learned in this instructional video (). The example given is to calculate the electric field on the y axis due to two point charges on the x axis, with opposite signs.

    Then i do: E1= q(-xi+yj) / ((x2+y2))3/2
    E2=-q
    (xi+yj) /((x2+y2))3/2

    That works just fine and i get the resultant E from the vector sum of each E. But that method is not working when i have to calculate the eletric field on the x axis.

    In this exercise, that is giving me trouble. i have to determine E in a x point between a and -a ( positions of q and -x) respectively. I can see, quite easily, that the resultant vector should be pointing to -i axis, but i cant get at the teacher answer. My attempt at solving:

    E1=q*(x-a)i/(x-a)3
    E2=-q(x+a)i/(x+a)3

    Summing those two give a different result, that should be: -2q*(x2+a2) divided by ((x-a))4

    Where exactly is my mistake?

    Thanks in advance
     
  2. jcsd
  3. Apr 6, 2017 #2

    BvU

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    Hello rbsan, :welcome:

    Do not delete the template. Use it (it's compulsory -- and it's good for you too).
    Here at PF we try to let folks fence for themselves as much as possible -- that way they often manage subsequent exercises all on their own :smile: . So:

    Mistake must be there then. Could you show your working ?
     
  4. Apr 6, 2017 #3
    Sorry about not using the template! Next time i will use it.

    The calculations are fine.

    It seems that if i input both charge as if they were negative:
    E1=-q/(x-a)^2
    E2=-q/(x+a)^2
    i get the correct answer.

    That makes sense in a vector sense (both electric field lines are pointing along the -i axis) but solving this way i completely ignore the charge sign.
    And it is not compatible with the solution method taught by that instructional video.

    Is that video wrong?
    Should i always disregard charge sign when calculating electric field and only take into consideration the direction E is pointing to?
     
  5. Apr 6, 2017 #4
    If you are just approaching the subject I think it would be more meaningful to try derive the electric field in a generic point (x,y,z) and move from there, cause there's some interesting insights you can get by working from the general cases down to the specific ones.
    Once you have an expression for E(x,y,z), you can narrow it down to the x axis as E(x,0,0), and then restrict it to -a<x<a.

    Anyway, I didn't understand if this formula
    is what you actually get from your calculations or what you should get; because you can clearly say it's not correct just by looking at the denominator: you have [itex](x-a)^2[/itex] and [itex](x+a)^2[/itex] as denominators of your E fields, and there's no way you can turn it into [itex](x-a)^4[/itex].
    So, in the first case, you should just check your calculations; in the second one, your source made an error.

    Are you sure you are ignoring the charge sign? What is it that defines the direction of the electric field then? How can you say they are directed along the negative x axis if you ignore the charge? :biggrin:
    Trying to find a way around to get the correct result is a good thing I guess, but you should also be careful and understand why it works, otherwise its just cheating :rolleyes:

    By the way, if you do not show clearly what result you got, and how you got it, most people will have a hard time trying to help you in a meaningful way.
     
  6. Apr 6, 2017 #5
    Yes that was my input mistake, the correct input is (x^2-a^2)^2 ( sorry for bad input)

    Yes i know that in thesis i am not actually ignoring the charge sign when i draw the field lines, since the negative charges point towards themselves from the observational point;- and positive charges point towards the observational point.

    But what i am doubt with is that if i need to not only take that into consideration but also it's sign on the algebraic formula.

    ''Trying to find a way around to get the correct result is a good thing I guess, but you should also be careful and understand why it works, otherwise its just cheating ''

    That is true and i am guilty of that :(
    But unfortunately sometimes i first learn how to solve the exercises and then i actually understand the concept. I wish i would always grasp the concept at first, but some stuff take a while to digest. ( for me at least)
     
  7. Apr 6, 2017 #6
    Let's see the formula then: if you have N electric fields, generated by N charges, superposition principle allows you to sum them all up:
    $$
    \vec{E} = \sum_{i=1}^N\vec{E}_i
    $$
    Now you will just have to apply a basic vector sum. Just be careful when working with vectors and there will be no sign ambiguities.
    If you have doubts, forget the E fields for a moment and consider 2 vectors, $$\vec{a} = a\hat{i}$$ and $$\vec{b}=-b\hat{i}$$ where both a and b are positive. What's their vector sum?

    It's all normal, physics takes time. I am a student and I can definitely relate. Just be patient, take your time, read, practice, hit your head against the wall, and every piece will fit, eventually ^^
     
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