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Electric Field Due to Point Charges?

  1. Dec 4, 2005 #1
    Hello!

    Just hoping for a bit of help on a homework problem I'm doing at the moment...

    Basically this is the question:

    Three point charges are placed along the y axis: a charge q at y=a, a charge -2q at the origin and a charge q at q=-a. Such an arrangement is called an electric quadrupole.
    a. Find the magnitude and direction of the electric quadrupole at points along the positive x-axis.
    b. Use the binomial expansion to find an approximate expression for the electric field valid for x>>a. Contrast this behaviour to that of the electric field of a dipole.


    I've tried to do this loads of different ways...

    I've basically used E = 1/4pi...q/r^2 (didn't know how do get a symbol for the permittivity of free space!) and then for the vector have put (xi - aj) for q at a and so on... But then when I add them up to come up with a total E I just get something that looks really messy - although it is to the power of a half and could be expanded using the binomial, it just looks wrong...

    Any input on how I could try to solve the question would be greatly appreciated!

    Thanks, Sam(antha)
     
  2. jcsd
  3. Dec 4, 2005 #2

    siddharth

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    What you did seems to be right. Just find the vector sum of the Electric field due to the three charges. The y-component of the field due to the charge at y=a and y=-a should cancel. So along the x-axis, your net Field should also point along the x direction.
    If you post and show exactly where you are having difficulties (ie, which equations) it will be easier to help.
     
    Last edited: Dec 4, 2005
  4. Dec 4, 2005 #3
    Thanks!

    This is where I'm getting confused though, because if the y components cancel then when I find the magnitude of the field I will just be squaring the x components, so I won't have anything to the power of a half to expand using the Binomial? :confused:
     
  5. Dec 5, 2005 #4

    siddharth

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    You will have a power of 3/2. When you take the x-component due to the 2 charges at y=a and y=-a, the field will be
    [tex] \frac{q \cos\theta }{4 \pi \epsilon (a^2 + x^2)} [/tex]
    So find [tex] \cos \theta [/tex] and substiute to find the Field.
    So you will be able to use your approximation when x>>a
     
    Last edited: Dec 5, 2005
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