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Electric Field Between two Charges Equals 0

  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    A 3.17 μC and a -2.19 μC charge are placed 3.65 cm apart. At what point along the line joining them is the electric field zero? Assume that the first charge is at the origin and the second charge is at +3.65 cm.


    2. Relevant equations

    E = K(q) / r2
    Ep = E1 + E2
    0 = E1 + E2
    -E1 = E2
    Since one is negative and the other is positive they would really work together making it E1 + E2, then -E1 = E2

    3. The attempt at a solution


    Ep = E1 + E2
    0 = E1 + E2
    -E1 = E2
    -k(q1) / x2 = k(q2) / (r - x)2
    -(q1) / x2 = (q2) / (r2 - 2rx + x2)
    -(q1)(r2 - 2rx + x2) = (q2)(x2)
    -(3.17e-6)(0.03652 - 2(0.0365)x + x2) = (-2.19e-6)(x2)
    -(3.17e-6)(1.33e-3) - (7.30e-2)x + (x2) = (-2.19e-6)(x2)
    -(3.17e-6)22 + (7.30e-2)x - (4.22e-9) + (2.19e-6)x2 = 0
    (-9.80e-7)22 + (7.30e-2)x - (4.22e-9) = 0
    x = -b +/- √[b2 - 4ac / 2a
    x = [-(7.30e-2) +/- √(7.30e-2)2 - (4)(-5.36e-6)(-4.22e-9)] / 2(-9.80e-7)
    x = [-(7.30e-2) +/- (5.33e-3)] / [-1.96e-6]

    Sorry if that's hard to read. But after i did all this, i got answers way to big to even be close. They were easily in the 10s of thousands. What did i do wrong?
     
  2. jcsd
  3. Oct 10, 2011 #2

    SammyS

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    You mentioned that (for a point between the charges) the E-field for both charges is in the same direction. Actually, it points left.

    A line is infinitely long, so find a place along the x-axis, that's not between the charges.

    Do you suppose that point is to the right, or is it to the left of the two charges?
     
  4. Oct 10, 2011 #3
    After i got an answer the two points where to the right of the two charges.
     
  5. Oct 10, 2011 #4

    SammyS

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    What two points?

    There is one point at which the E-field is zero.

    That point is to the right ot the two charges. Why?
     
  6. Oct 10, 2011 #5

    lightgrav

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    you should only have one place that their field contributions cancel.
    (the other "zero" is at infinity)
    but yes, it's farther from the big charge sqrt(3.19/2.17) times as far.
     
  7. Oct 10, 2011 #6
    I mean, after doing Quadratic i ended up with two values, both where larger than 3.65 cm. That's what i mean when i said i have two points.

    It's to the right because we the field is going to the right. We were taught that you always imagine what a proton would do should it enter the field, and in this case it would be repelled from the positive charge and attracted to the negative charge to the right of it. Thus the point where the field is zero is the right of the two charges.
     
  8. Oct 10, 2011 #7

    lightgrav

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    It depends on which place has more charge.
     
  9. Oct 10, 2011 #8
    Alright, so we know that it's to the right, I don't see how this is helping solving the problem i have.
     
  10. Oct 10, 2011 #9

    lightgrav

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    solve for q2/q3 = (x-d)^2 / (x)^2 ... then take the root of both sides.
     
  11. Oct 10, 2011 #10
    q1 / q2 = (x - d)2 / (x)2
    3.17 / -2.19 = (x - d)2 / (x)2
    √[3.17 / -2.19 = (x - d)2 / (x)2]
    1.20 = (x - d) / x
    1.20x = x - d
    1.20(0.0365) = (0.0365 - d)
    0.0438 = 0.0365 - d
    -0.0438 = -0.0365 + d
    d = -0.0073 m => -0.73 cm Something like that then?
     
  12. Oct 10, 2011 #11

    lightgrav

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    the big charge has to be farther away, so its influence is spread out more.
    I thought "x" was the variable , and "d" was the constant.
    (I would've done 0.4617 = 1 - d/x)
     
  13. Oct 10, 2011 #12
    Where did you get the 0.4617 From? Everything else i can see.
     
  14. Oct 10, 2011 #13

    lightgrav

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    E3 = 3/(x)^2 . . . E2 = 2/(x-d)^2
    => sqrt(2/3) = (x-d)/x = 1 - d/x
    oops. I needed parentheses in my calculator!
     
  15. Oct 10, 2011 #14
    I'm assuming that the 3 and the 2 that you're using are the charges? Then why not add the Negative to the two? or is it irrelevant in this case

    E3 = 3/(x)^2
    E2 = 2/(x-d)^2

    3.17 / x2 = 2.19 / (x - d)2
    √(2.19 / 3.17) = (x - d) / x
    √(2.19 / 3.17) = (1) - (d / x)
    0.8311 = 1 - (d / x)
    0.8311 - 1 = -d / x
    -0.1688 = -d / x
    -0.1688x = -d
    -0.1688x = -(0.0365)
    x = -0.0365 / -0.1688
    x = 0.216m => 21.6 cm
     
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