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Homework Help: Magnitude of electric field at a point problem

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data

    If Q = 80 nC, a = 3.0 m, and b = 4.0 m in the figure, what is the magnitude of the electric field at point P?

    [PLAIN]http://img19.imageshack.us/img19/1596/picat.jpg [Broken]

    2. Relevant equations

    [tex]\vec{E} = k_e \frac{|q|}{r^2}[/tex]

    [tex]\left\{\begin{matrix}\vec{E}_x=k_e \frac{|q|}{a^2+b^2} \cos \ \phi
    \\ \vec{E}_y=k_e \frac{|q|}{a^2+b^2} \sin \ \phi\end{matrix}\right.[/tex]

    ke = 8.9 x 109 (Coulomb's constant)

    3. The attempt at a solution

    The charge -Q only has an x component, the electric field due to this charge is:

    [tex]\vec{E}_{(-Q),x}= (8.9 \times 10^9) \times \frac{-80 \times 10^{-9}}{4^2}=-44.5[/tex]

    The two "2Q" charges cancel out each other's y-components, so we need to only consider their x components:

    [tex]\vec{E}_{(2Q), x} = (8.9 \times 10^{9}) \times \frac{2 \times (80 \times 10^{-9})}{3^2+4^2} \cos 45 = 40.27[/tex]

    I got the 45 degree angle is from here:

    [PLAIN]http://img200.imageshack.us/img200/5499/pic2ae.jpg [Broken]

    So the total electric field is:

    [tex]\vec{E}_{(-Q), x}+2\vec{E}_{(2Q), x}=-44.5+ 2\times(40.27)=36.05 N/C[/tex]

    But this is wrong since the correct answer must be 47 N/C. What's wrong with my calculation? Any help is greatly appreciated.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 21, 2011 #2


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    Homework Helper

    why 45°?

  4. Jul 21, 2011 #3
    check angle again ...

    tan 45 = 1

    is a=b? no

    do it again
  5. Jul 21, 2011 #4
    Thank you very much guys, I got it. :smile:
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