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Location of a charge

  1. Jan 20, 2008 #1
    A 28.1 nC charge is located at position ( x,y ) = ( 1.0\;{\rm cm},\;2.0\;{\rm cm}). At what ( x, y) position(s) is the electric field?

    a) - 225,000\hat{ i } {\rm N/C} ? , x=?
    b) (161,000\hat{ i }+80,500\hat{ j }) {\rm N/C} ? x=?
    c) same as b , y=?
    d) (21,600\hat{ i }-28,800\hat{ j }) {\rm N/C} ? , x=?
    e) same as d, y=?

    i have no idea how to go about this.
     
  2. jcsd
  3. Jan 20, 2008 #2

    CompuChip

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    You could start by calculating the electric field of the charge, and writing it in the form
    [tex]\vec E = E_1(x, y) \hat{i} + E_2(x, y) \hat{j} [/tex]

    Then you can solve all the questions by equating the components.
     
  4. Jan 20, 2008 #3
    the thing i am confused abt is that .. what am i gonna figure out .. E is given .. its 28.1nC
     
  5. Jan 20, 2008 #4

    CompuChip

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    E is the electric field, not the charge Q which is given. What electric field does a point charge "generate"?
     
  6. Jan 20, 2008 #5
    right ok ..
    so E = kq/r^3
    = (9x10^9)(28.1x10^-9)/ (25)

    r= sqrt(5) ... r^3 = .0224
    11290.178 N/C

    now .. if i take x-y-component i get ..

    theta = 60 deg

    y = 9777.58
    x = 5645.1 ..

    does that seem right?
     
    Last edited: Jan 20, 2008
  7. Jan 20, 2008 #6
    i am still going in circles with this .. i am no sure where am i mucking up
     
  8. Jan 20, 2008 #7
    can someone pls guide me through this, i have 2 hours to complete this assignment and this is one of the question i am stuck with ..

    any help is appreciated
    Thank you
     
  9. Jan 20, 2008 #8

    hage567

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    To begin with, you're not using the proper equation (it should be an r^2 term in the denominator not r^3) and you keep changing things in the middle of your work. I don't understand what you're trying to do here at all.

    For (a), look at the electric field that is described. It only has a horizontal component. So if the charge is sitting at (1,2), what direction (and orientation) do you think you should draw the electric field vector?
     
  10. Jan 20, 2008 #9

    well the E vector would be directed 60 deg from (0,0) .. thats what i thought .. but i am probably way off base
     
  11. Jan 20, 2008 #10
    ok so far i got:

    E - kQ/r^2 = (9x10^9)(28.1x10^-9)/.0005 = 5.058x10^5 N/C .. does that look good and sensible?
     
  12. Jan 20, 2008 #11

    hage567

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    But you already know the strength of the electric field. It's given to you. You are trying to find the position from the charge that the electric field takes on the value given in (a). How did you get 0.0005?
     
  13. Jan 20, 2008 #12

    hage567

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    The E vector does not come from (0,0), it is coming from the charge. You are just trying to describe it in terms of your coordinate system. If there is only a horizontal component (so no change in the vertical direction) to the electric field vector, which way will it point? How far away from the charge do you need to be to get an electric field strength of 225,000 N/C? Worry about that first and when you have this distance you can express it in (x,y) form since you know the position of the charge.
     
  14. Jan 20, 2008 #13
    0.0005 is from putting (1,2) in Pythagorean and then i get sqrt ((0.02)^2 + (0.01)^1)
    and when i did sqr of that i get 0.005

    and i was given charge 28.1nC ., i figured out the E field .. now the question is if the new E vector is 225,000 in the x direction ... how do i go back .. i cant seem to re-trace my steps
     
  15. Jan 20, 2008 #14

    hage567

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    So that means you've found the electric field strength at the origin. That's not relevant to this question as I understand it. Forget about the x,y coordinates for a minute. Can you tell me how far away from a point charge of magnitude 28.1 nC you have to be to get an electric field strength of 225,000 N/C? There is no trig or angles involved to find this part out. Just use the electric field equation for a point charge.

    You don't need to figure out the E field! It's given to you! You want to find where the electric field is 225,000 N/C. I don't know what else to say about this.

    Perhaps I'm not interpreting this problem correctly.
     
  16. Jan 20, 2008 #15

    rl.bhat

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    Go through the quote by Goldenwind dated 1-18-08
     
  17. Jan 20, 2008 #16

    hage567

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    Do you mean a thread by Goldenwind? If so, can you post the link for it?
     
  18. Jan 20, 2008 #17
    i gave it another shot .. and used
    225000 = kQ/r^2 .. and found out r^2 .. and that worked

    for the second one (part B) i did the same calc but from my calculations, i am not getting the right answer .. its close but not exact ...

    for this part (b) i got 4.96 and the answer is 4.35 .. i guess i got the concept but not hitting bulls eye ..
     
  19. Jan 20, 2008 #18
    i checked it .. thats for another problem

    my calculations are good for one part and now for another .. which seems strange
     
  20. Jan 20, 2008 #19

    hage567

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    OK. We'll leave it at that. But you should have had to take into account the x=1.

    This one is a bit trickier. This one you have to use trig. You know the components of the electric field, so you can find the angle between them. You can also work out the magnitude of the E field vector.

    Just draw a right angle triangle, and solve for the length of the x edge. You know what the hypotenuse of the triangle is (it's r = sqrt(kQ/E), where E is the magnitude of the total vector). You know the angle (from above).

    You do have to remember to add/subtract the distance charge is displaced from the origin. So if you find the x length to be a certain value, you must add the x=1 onto it to find the x position from the origin. Does that make sense? (Sorry, I'm getting tired.)
     
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