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Electric Field at a distance

  1. Aug 26, 2009 #1
    1. The problem statement, all variables and given/known data

    disk.jpg

    2. Relevant equations

    E=kq/r^2

    3. The attempt at a solution

    The electric field due to 1 of the electrons is:

    E=kq/r^2
    E=(9e9)(1.6E-19) / (.0325^2)
    E=1.363E-6

    The electric field due to the other electron is the same, correct? They should be.

    So, the net electric field due to the 2 electrons is twice this, or 2.72E-6.

    When I submit this, it comes back incorrect. What am I doing wrong? It's only asking for magnitude, not direction, so I'm submitting the absolute value, disregarding the direction.
     
  2. jcsd
  3. Aug 26, 2009 #2

    Doc Al

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    Staff: Mentor

    What direction does it have?

    The magnitude is the same.

    An electric field is a vector, so you must add them like vectors.

    You can certainly disregard the direction of the net field when presenting your answer, but you must consider the direction of each contribution as you add them up.
     
  4. Aug 26, 2009 #3
    So since the x-components of the Efield vectors cancel, the resulting Efields are only in the y direction. So the net magnitude from the Es vectors is 1.36E-6(cos45)*(2)? And direction would be -y?
     
  5. Aug 27, 2009 #4

    Doc Al

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    Right!
     
  6. Aug 27, 2009 #5
    Okay I got the answer to (b) as 1.93E-6 and it's correct.

    However, moving on to (d), I don't think I'm doing something right, although I can't find any error.

    E=kQ/d^2 //Coulombs Law
    E=(9e9)(1.6e-19) / d^2 = 1.44E-9 / d^2

    d^2=z^2+r^2
    d^2=.023^2 + .0023^2 = 5.34e-4

    1.44E-9/5.34e-4 = 2.70E-6

    2.70e-6(cos45)=1.91e-6 //This is the force along the y-axis from one electron.

    1.91e-6*2=3.81e-6 //Multiply by two for two electrons. X-components cancel out.

    The final answer I get, 3.81e-6 does NOT correlate with the next question, however, in which it states that the field from (d) should be less than the field from (b). 3.81e-6 > 1.93e-6.

    What am I missing?

    (I absolutely HATE these online homework submissions.)
     
  7. Aug 27, 2009 #6

    Doc Al

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    Staff: Mentor

    Good.

    Better tell me what (d) is asking for. (May as well post the entire problem.)
     
  8. Aug 27, 2009 #7
    OH! HA. I forgot I cropped the image :)

    (a) At the proton's location, what is the magnitude of the electric field Ec due to electron Ec?

    Answer: 2.72E-6 which is correct.

    (b) At the proton's location, what is the magnitude of the net electric field Es,net due to electrons Es?

    Answer: 1.93E-6 which is the one we worked on and is correct.

    (c) The proton is then moved to a distance z = R/10. What then is the magnitude of Ec at the proton's location?

    Answer: 2.72E-4 which is correct.

    (d) What is the magnitude of Es,net at the proton's location after it has moved to a distance z = R/10?



    (e) From (a) and (c) we see that as the proton moves nearer to the disk, the magnitude of Ec increases. Why does the magnitude of Es,net decrease, as we see from (b) and (d)?
     
  9. Aug 27, 2009 #8
    This looks like a problem from Serway.

    Anyway you would have to calculate individual vector forces on the proton, resolve them into components and then add .
     
  10. Aug 27, 2009 #9

    Doc Al

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    Here's the problem:
    When the proton moves, the angle is no longer 45 degrees.
     
  11. Aug 27, 2009 #10
    Understood. But it still doesn't work out right:

    tan theta = .0023/.023
    theta = 5.71 degrees

    2.696E-6 (cos 5.71) * 2 = 5.36E-6

    According to that answer, it's INCREASING from the original 1.93E-6 which contradicts what (e) is asking.
     
  12. Aug 27, 2009 #11

    Doc Al

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    You have the wrong component: cos(theta) gives you the x-component. (When the angle was 45 it didn't matter, since sin(45) = cos(45) )
     
  13. Aug 27, 2009 #12
    Ooooookay. Again.

    tan theta = .0023/.023
    theta = 5.71 degrees
    90-5.71=84.29 //The other angle in the triangle. I HATE upside down triangles.

    2.696E-6 (cos 84.29) * 2 = 5.36E-6 \\I could have used (sin 5.71) too, right?

    The correct answer is: 5.37E-7, which indeed is smaller than the original.

    To answer (e) though, I'm not so sure why. The reason why (c) is bigger than (a) is because the distance is decreasing. The distance is limiting to zero as P-->Ec. As the denominator gets closer to zero, the value of the efield would be increase.

    However, I can't grasp why the efield due to Es got smaller.
     
  14. Aug 28, 2009 #13

    Doc Al

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    Yes, since sinθ = cos(90-θ).

    For Es, what matters is the y-component. As the angle with the horizontal gets smaller, so does the y-component.
     
  15. Aug 28, 2009 #14
    That makes sense now. I was thinking about it mathematically, and not visually.

    Thanks for all the help. I'm sure I'll need more.
     
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