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Magnitude of e field

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the magnitude of the E field at the origin due to a single charge (either the + or the -) located 5 cm above the horizontal axis?

    2. Relevant equations
    KQ/D^2



    3. The attempt at a solution
    in an earlier question it asked What is the direction and magnitude of the E field at the origin due to the single negative charge (ignore all others) on the horizontal axis?" SO i used columbs law "KQ/d^2, (8.99x10^9x30x10^-9)/.05^2 and i got 1.1x10^5 which was the correct answer, but this questions answer is supposed to be 5.4X10^4. I was wondering why this is the case as all the numbers are the same from an earlier question, exepct now its vertical.
     

    Attached Files:

  2. jcsd
  3. Feb 22, 2012 #2
    Please provide earlier question :)
     
  4. Feb 22, 2012 #3
    the earlier question was "What is the direction and magnitude of the E field at the origin due to the single negative charge (ignore all others) on the horizontal axis?"
     
  5. Feb 22, 2012 #4
    and the present question?
     
  6. Feb 22, 2012 #5
    I didn't get how all the numbers are same in both question , aren't they different
     
  7. Feb 22, 2012 #6
    Because in both questions you are still 5 cm away from the e field, and the charges are still 30 nc. The first question the negative charge is to the left of the center and in the second question the charge (the problem said could be + or -) is 5 cm above the center. So aren't the numbers the same if i were to use the same equation?
     
  8. Feb 22, 2012 #7
    Sorry in question 1 the negative charge is 5 cm to the left of the origin and question 2 the charge is 5 cm above the origin.
     
  9. Feb 22, 2012 #8
    but the number of charges are different , they are infinite
    and also they are distributed along lines not at a point
     
  10. Feb 22, 2012 #9
    just for the sake of imagination
    think of a person pushing a box with the help of a rope alone
    and think many people pushing a box with many ropes
     
  11. Feb 22, 2012 #10
    aren't there only 1 charge in both questions as question 1 says to ignore the other charges and question 2 says there is a single charge? would i have to use a different formula?
     
  12. Feb 22, 2012 #11
    i really wish if you could post both the question exactly separated by space , i am confused
     
  13. Feb 22, 2012 #12
    Sorry
    1."What is the direction and magnitude of the E field at the origin due to the single negative charge (ignore all others) on the horizontal axis?"

    2.What is the magnitude of the E field at the origin due to a single charge (either the + or the -) located 5 cm above the horizontal axis?

    I know how to do 1, its just question 2 thats tripping me up.
     
  14. Feb 22, 2012 #13
    where is the particle placed in Q1
     
  15. Feb 22, 2012 #14
    particle is at the origin, and the negative charge is on the left
     
  16. Feb 22, 2012 #15
    left ? what is the coordinate?
     
  17. Feb 22, 2012 #16
    If you look at the picture the particle would be at the center and the negative charge is on the x axis to the left of the origin. So it would be 5 cm away, as the entire horizantel axis is 10cm.
     
  18. Feb 22, 2012 #17
    see magnitude of electric field will be same for all points whose distance from the particle is same
     
  19. Feb 22, 2012 #18
    locus will be a sphere , where the whole surface will have same magnitude at all points on the surface
     
  20. Feb 22, 2012 #19
    See magnitude of electric field is same for all points which are equidistant to the particle ( which is a locus of a hollow sphere )
     
  21. Feb 22, 2012 #20
    Thats what i was thinking, but on the answer sheet the it said the magnitude of
    1. was 1.1x10^5 N/C to left and for 2 it said the magnitude was 5.4X10^4 N/C. Im stumped as to how the magnitude could be different when all the numbers used is essentially the same.
     
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