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Homework Help: Finding the electric field at a point on the x axis due to two charges.

  1. Feb 3, 2010 #1
    Problem: a) 2 charges are located on the y axis at positions +2 m (-2 microC) and -2 m (-1 microC) respectively. No other charges are present. Each grid is 1 meter. Compute the x and y components of the electric field at the point indicated at the position x = +2 m. Give 2 answers, Ex and Ey.
    b) Draw electric field lines. Note the 2 charges are not the same magnitude. Note also there are certain points where the direction of the field can be determined easily.
    c) If an electron is placed at the point indicated by the open circle on the x axis, what is the magnitude of the force on the electron? An electron's charge is -1.6 x 10^-19 C.

    Solutions: a) I found that the Ex=0 N/C because of symmetry making the net field completely in the y direction.
    For the y direction I determined that the r would be 2.83 m by using pythagorean's theorem. But i was getting confused on how to add the components of this direction. Would it be something like this: Using E = kq/r^2
    E1 = (9x10^9)(2x10^-6)/2.83^2 = 2247.5 N/C E1y=E1sin(49.97 degrees)=1588.39 N/C
    E2=(9x10^9)(1x10^-6)/2.83^2=1123.75 N/C E2y=E2sin(49.97 degrees)=-794.19 N/C
    Ey=794.2 N/C

    b) I wasn't really sure how to go about this other than knowing that electric field lines point towards negative charges, and the 2 charges will repel each other.

    c) I wasn't sure how to go about this.
     
  2. jcsd
  3. Feb 4, 2010 #2

    rl.bhat

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    Since two charges are different, Ex is not equal to zero.
    From where did you get 49.97 degrees?
     
  4. Feb 4, 2010 #3
    The 49.97 was actually 44.97, i mistakenly typed it. I got from taking the inverse sin of (2/2.81). How would I go about calculating Ex then?
     
  5. Feb 4, 2010 #4
    I found Ex by doing E1x=E1cos(44.97) and E2x=E2cos(44.97). I found E1x=1590.05 N/C and E2x=795.05 N/C. So adding those together Ex=2385.08 N/C. Is this correct?
     
  6. Feb 4, 2010 #5

    rl.bhat

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    yes. It is correct.
     
  7. Feb 4, 2010 #6
    Is the value for the Ey the same as Ex?
     
  8. Feb 4, 2010 #7

    rl.bhat

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    No.
    Ey = E1y - E2y
     
  9. Feb 4, 2010 #8
    How would I go about finding c?
     
  10. Feb 4, 2010 #9

    rl.bhat

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    Force on an electron F = E*q = Sqrt(Ex^2 + Ey^2)*q
     
  11. Feb 4, 2010 #10
    Finally will the force be negative because of the negative charge of the electron? Thanks so much for all your help!
     
  12. Feb 4, 2010 #11

    rl.bhat

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    Usually we wont say positive or the negative force. Instead we mention the direction of the force.
     
  13. Feb 6, 2010 #12
    Hi, I got a really small, negative answer to this problem... does -6.41 E -32N sound right to anyone? Thanks!
     
  14. Feb 6, 2010 #13

    rl.bhat

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    I don't think it is correct. E^-32 is not possible. Show your calculations.
     
  15. Feb 6, 2010 #14
    I first calculated new E values for charges -2microC and -1microC due to the new electron on the x-axis and got: E-2= k*(2E-6)(1.6E-19)/8 and E-1=k(1.0E-6)(1.6E-19)/8. Then I calculated E values for the x-directions: Ex-2=E2*(2/sqrt8)=2.54E-16 and Ex-1=E-1*(2/sqrt8)... then I added them together and got an Ex total= 3.8E-16. Then I basically got the same values for E in the y-direction, but subtracted 1.27E-16 from 2.54E-16 to geth 1.27E-16 total E value in the y-direction. I then took the square root of the sum of the squares of both of these Ex and Ey total values and got 4.0E-16... then finally, to find the total Force i multiplied this number by the charge of the electron, -1.6E-19. ... maybe i did too much work to figure this out?
     
  16. Feb 6, 2010 #15

    rl.bhat

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    E-2= k*(2E-6)(1.6E-19)/8
    The above expression is of force F not the field E.
    So your answer may be
    4.0E-16.N
     
  17. Feb 6, 2010 #16
    oh, i thought F = E * q ? and electric field always equaled K Q1Q2/ r^2 ??
     
  18. Feb 6, 2010 #17

    rl.bhat

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    Electric at any point is
    E = k*q/r^2.
     
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