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Where does electric field go to zero?

  1. Jun 8, 2014 #1
    Hello
    Here is an example.we want to calculate value of x where Electric filed going to be zero.i want to use vector form of coulomb equation
    4340101300_1402253250.jpg
    So as we know
    8648817800_1402254057.png
    for 1c:
    6134598200_1402254056.png
    for -9C:
    8295098200_1402254056.png
    so as we do superposition we obtain
    7117543900_1402254057.png
    So
    8132467900_1402254057.png
    In this problem x doesn't have any real number.what is my problem?

    Any help appreciate
     
  2. jcsd
  3. Jun 8, 2014 #2
    Why does your second equation have a -ax, but your third equation doesn't?
     
  4. Jun 8, 2014 #3

    CAF123

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    The electric field due to the two charges should be pointing in different directions at some position x to the left of the 1C charge. This condition is not met in the equation where you superimpose the fields.
     
  5. Jun 8, 2014 #4
    Hello
    Dear Jilang
    Electric filed in equation 2 is in (-ax) direction but the third equation is in (+ax) direction it is because of charges positive or negative value.


    dear CAF123
    as i check the book answer that use only coulomb scale form,EF pointing and where electric filed going to be zero is in near 1c charge but outside of two charges distribution.
    8x+10%29%5E2%7D%3D%3E%5Cfrac%7B1%7D%7Bx%7D%3D%5Cfrac%7B3%7D%7B%28x+10%29%7D%3D%3Ex%3D5.gif
    but i want to use vector form of coulomb equation.
     
    Last edited: Jun 8, 2014
  6. Jun 9, 2014 #5

    CAF123

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    In your coordinate system, the E field from the 1C charge at a distance x defined in the sketch is ##E_{1C} = - kq_1/x^2 \underline{e}_x ##. That of the -9C charge is ##E_{-9C} = kq_2/(10+x)^2 \underline{e}_x##. ##q_1, q_2## here are the magnitudes of the charges and their differing sign is manifest in the reversal of the unit vector.
     
  7. Jun 9, 2014 #6
    Hello
    Dear CAF123
    Thanks for your response
    however i don't agree with your explanation it is because of charges act.for example we know that the electric filed line of a dot positive charge is :
    positive.gif
    and EF equation is :
    gif.latex?%5Cvec%7BE%7D%3D%5Cfrac%7BQ%7D%7B4%5Cpi%20%5Cvarepsilon%20r%5E2%7D.%5Chat%7Br%7D.gif
    but if we put a dot negative charge that EF line is same as here
    negative.gif
    we can only change the sign of charge and have a correct form.
    gif.latex?%5Cvec%7BE%7D%3D%5Cfrac%7B-Q%7D%7B4%5Cpi%20%5Cvarepsilon%20r%5E2%7D.%5Chat%7Br%7D.gif

    now why the vector form of coulomb equation doesn't work here?
     
  8. Jun 10, 2014 #7

    CAF123

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    That is the same as $$\vec E = \frac{Q}{4 \pi \epsilon_o r^2} (- \hat r) = -\frac{Q}{4 \pi \epsilon_o r^2} \hat r$$ If you look at the picture, the electric field due to a point charge is spherically symmetric and the radial vector is defined to be positive radially outwards (spherical coordinates convention). In the case of a negative charge, the field lines move in the radially inwards direction, which is exactly what ##-\hat r## in the above equation is telling you.
     
  9. Jun 10, 2014 #8
    Hi Baby_1 , your post #6 is correct. But the r vectors both go the same way, pointing to the left of both charges in the negative x direction.
     
  10. Jun 10, 2014 #9
    Thanks dear CAF123 & jilang for your response.
    I found my problem.i wrote wrong the distance vector.here is an Electromagnetic book that explain better
    8633574300_1402386266.png
    here is the correct answer :
    1031429900_1402426382.png
     
  11. Jun 10, 2014 #10

    BiGyElLoWhAt

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    Is that the correct answer or correct equation?

    Try this:

    You know the electric field is 0 at some point.

    This means that ##\sum E_{q} = ?##

    Expand the sum and you have an equality to work with that's fairly easy to solve, and you're "using" the vector form of columbs law. Really, in this situation, the vectors are only useful for telling you the direction of the contribution to the field at a point by the respective point charge.
     
  12. Jun 10, 2014 #11

    BiGyElLoWhAt

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    also I think you meant ##K\frac{9}{(x+10)^2}a\hat{x}##
     
  13. Jun 10, 2014 #12
    Well done Baby_1!
     
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