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Finding the position of an electric field.

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A -10.0 nC charge is located at position (x,y) = (2cm, 1cm).

    At what (x,y) position(s) is the electric field:

    Here is where they give me the different electric fields, one of which is: 161,000[tex]\hat{i}[/tex] - 80,500[tex]\hat{j}[/tex] N/C

    The answers must be in cm.


    2. Relevant equations
    [tex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}[/tex]


    3. The attempt at a solution
    What I first did was plug in the the numbers I'm given:

    E = 161000
    epsilon_0 = 8.85x10^-12
    q = 10x10^-9

    Then I try to get r^2 by itself by multiplying it with both sides and then dividing both sides by 161000.

    I now have: [tex]r^2 = \frac{10\times10^{-9}}{4\pi(8.85\times10^{-12})(161000)}[/tex]

    Which gives me r^2 = 5.85*10^-4

    Squarerooting the answer gives me 0.0236 meters which converted to cm is 2.36 rounded to 2.4.

    Taking 2.4, I have added to the initial 2cm in the beginning of the problem to get 4.4 and have tried subtracting from 2 to get -0.4 but both are wrong. I did the first problem (E = -225,000 i-hat) the exact same way and I got the correct answer. My friend tried it out and also got the same answers as me.

    http://i385.photobucket.com/albums/oo297/jonyamikun/2664.jpg
     
    Last edited: Jan 21, 2009
  2. jcsd
  3. Jan 21, 2009 #2

    LowlyPion

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    I don't understand your problem statement.

    You have a charge.

    What is the other electric field? What is the field you cite? What is the 225,000 from you mention at the end?

    What exactly are you trying to find?
     
  4. Jan 21, 2009 #3
    I'm sorry if my description was too vague. Here is a picture if it helps.

    http://i385.photobucket.com/albums/oo297/jonyamikun/2664.jpg
     
    Last edited: Jan 21, 2009
  5. Jan 21, 2009 #4

    LowlyPion

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    Ohhh. I see. You have a point charge at 2,1. If the field strength is xxx, where are you measuring the field from? Got it.

    What you have then is an r-hat. The direction is given by the ratio of the i,j components passing through 2,1 and the magnitude of r or |r| is given by the sum of the squares away from 2,1.
     
  6. Jan 21, 2009 #5
    Sorry but I can't seem to figure out how this will help me solve the problem.

    I clicked on "show answer" for parts C and D and the answers are x=0, and y=2. This just doesn't make sense to me. Maybe I'm interpreting something wrong.
     
  7. Jan 21, 2009 #6

    LowlyPion

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    Your direction is determined by 1) the sign of the charge. It will be pointing toward 2,1 at all points. And 2) in the case of the 160i and -80.5j it lays on a line sloped 2x/1y pointing in from the top left. (Draw a diagram in the x,y plane.) This means that it intersects the y axis at (0,2).

    Now calculate the magnitude along this line as (160,000)2 + (80,500)2 with your -10nC charge and you will have your distance.
     
  8. Jan 21, 2009 #7

    LowlyPion

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    The reason you got the first one right was because it was on a line || to the x-axis passing through 2,1 and you didn't need to account for the distance as the sum of the squares.
     
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