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Electric field of a plane

  1. Jan 28, 2007 #1
    Two particles are fixed to an x-y coordinate system:
    particle 1 of charge -5.00 microCoulombs lies on the x axis at x = + 6.00 cm and particle 2 of charge +5.00 microCoulombs lies on the y axis at y = + 8.00 cm. Midway between the particles, what is their net electric field, in unit vector notation?

    So I got:
    1/4piEo x q/r^2 - 1/4piEo x q/r^2

    (1/(4pi(8.85x10^-12)) x (-5 x 10^-6)/(.06)^2)i - (1/(4pi(8.85x10^-12)) x (5 x 10^-6)/(.08)^2)j

    -1.248e7i + 7024847j

    Is this anywhere near right?
  2. jcsd
  3. Jan 28, 2007 #2


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    Staff Emeritus
    Science Advisor

    Not quite.

    The electric field vector must point along the line from the positive charge to the negative charge, so one needs to determine the unit vector along that line.

    The problem asks for the field at the midpoint between the charges, and so one needs the distance between the positive charge and the midpoint, and the negative charge and the midpoint. For one point at (x, 0) and the other at (0, y), think of (x/2, y/2).
  4. Jan 28, 2007 #3
    so can anyone give me a better idea?
    Last edited: Jan 29, 2007
  5. Jan 29, 2007 #4
    pretty please?
    im a bit confused
  6. Jan 29, 2007 #5
    So is it more like:

    (1/(4*π*ε_0)*(5.00 μC / (3.00 cm)^2) for particle 1 and give it an 'i' and -1/(4*π*ε_0)*(5.00 μC / (4.00 cm)^2) for particle 2 and give it a j?

    Since thats r/2?
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