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Homework Help: Finding the magnitude and direction in a electric field.

  1. Feb 12, 2013 #1
    Three point charges are placed at the vertices of an equilateral triangle (of side .2m). The triangle is setup on an X-Y plain. On the top of the triangle on the Y-axis the charge is q7= -7μC, the positive X-axis the charge is q3= 3μC, and on the -X-axis the charge is q2= 2μC.

    Part A).
    What is the magnitude and the direction of the electric field at the origin?

    Part B).
    What would be the acceleration (magnitude and direction) of an electron placed at the origin?

    Part C).
    Would the acceleration (magnitude and direction) of the proton be different than that of the electron? Explain.
  2. jcsd
  3. Feb 12, 2013 #2


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    As per forum rules, you must post any equations you have that you believe may be relevant and your own attempt at solution, as far as you got.
  4. Feb 12, 2013 #3
    I'm going to parasite on this thread, because it deals with a similar topic.

    the way i approached it was use the electric field equation,

    E = [itex]\frac{1}{4\piε}[/itex] [itex]\int[/itex] [itex]\frac{1}{|r|^2}[/itex] [itex]\widehat{r}[/itex]dq

    for which i substituted [itex]\widehat{r} = cosψ = \frac{z - Rcos\theta}{r}[/itex] and for [itex] |r| = √R^2sin^2θ + (z - Rcosθ)^2[/itex] and [itex] dq = δda = δr^2sinθdθd\phi[/itex]

    where R is the radius of the sphere upon whos surface the charge is distributed.

    Now i substitute all of these wonderful thing into the electric field equation, all giddy to finally solve it :D Integrating over the entire surface of the sphere, that is from 0 to [itex]\pi [/itex] for [itex]d\theta[/itex] and 0 to [itex]2\pi [/itex] for [itex]d\phi[/itex]

    E = [itex]\frac{1}{4\piε} \int \frac{1}{ (√R^2sin^2θ + (z - Rcosθ)^2)^2 }\frac{z - Rcos\theta}{(√R^2sin^2θ + (z - Rcosθ)^2)}δr^2sinθdθd\phi[/itex]

    [itex]\int d\phi = 2\pi[/itex] nothing exciting there, parametrize [itex] u = cosθ, du = -sinθ dθ [/itex]

    E = [itex]\frac{1}{4\piε} \int \frac{δ(2\pi r^2sinθ)(z - Rcos\theta)}{ (√R^2sin^2θ + (z - Rcosθ)^2)^3 }dθ \Rightarrow E = \frac{1}{4\piε} \int \frac{δ(2\pi r^2sinθ)(z - Rcos\theta)}{ (√R^2 + z^2 - 2Rzcosθ)^3 }dθ \Rightarrow E = -\frac{1}{4\piε} \int \frac{δ(2\pi r^2)(z - Ru)}{ (√R^2 + z^2 - 2Rzu)^3 }du[/itex]

    at this point i pretty much just hur dur, try to integrate by partial fractions, and get nowhere because its like no other i've met before.

    I realize this is a pretty standard integral, being an inverse cosine law and all. Any online resources?

    Would someone please hint as to how to solve this, have i made any mistakes.


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    Last edited: Feb 12, 2013
  5. Feb 12, 2013 #4
    step one, draw the question. apply principle of superposition; consider one charge at a time, then sum individual forces.

    Part A
    think about which force vectors are going to cancel at the origin.

    Part B
    use the result from part A :D

    Part C

    is pretty vanilla. opposites attract.
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