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Electric Field

  1. Feb 12, 2007 #1
    Ok I have a somewhat easy problem im getting confused with.

    We were givin a problem in class like this.

    [​IMG]

    And told to find E at A I'm assuming...

    so E1 is K*(5*10^-5/.3m^2) = 5*10^6 in the positive jHat direction

    E2 is K*(-6*10^-5/.36m^2) = 4.1*10^4 and its direction is 304 degrees...


    I understand how to get all that but What I dont understand is this.

    The teacher has the magnitude being 2.77*10^6 and the direction as 34 degrees. Did I do the right thing by finding E1 and E2? If so are they correct? and the answere he gave is that E at A and if so how did he get it? Thanks for the help!
     
  2. jcsd
  3. Feb 12, 2007 #2

    hage567

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    You've started out OK. But remember that an electric field is a vector quantity. You have to break up E from each charge into its components to be able to find the resultant field at A. So you have to come up with the sum of the components in the x direction and the sum in the y direction, and then put them together to get the final result.
     
  4. Feb 12, 2007 #3
    Ok I understand E1 goes from q1 to A along the y-axis so its direction and magnitude are what I have and it doesnt do anything in the x-direction. I guess can you give me a quick crash course in breaking it down into components?
     
  5. Feb 12, 2007 #4

    hage567

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    "Ok I understand E1 goes from q1 to A along the y-axis so its direction and magnitude are what I have and it doesnt do anything in the x-direction."

    This is opposite to your diagram, which has the x-axis as the vertical one.

    "I guess can you give me a quick crash course in breaking it down into components?"

    Using trig, you can find the angle of E from q2 towards A. Then break this up into x and y components at A using that angle. Think of things like right-angled triangles and vector addition.
     
  6. Feb 12, 2007 #5
    Sorry I drew that diagram quickley and reversed the x and y axis.
    This is whatI have so far...

    [​IMG]

    Is this true?

    E1x = 0
    E1y = 5e6

    But for E2 I dont know how to get differnt number for E2x and E2y. I found the degree and I know I need to use trig. On top of all that I still dont see how I'd get a magnitude od 2.77e6.... I've dona all this before but I just resumed class after taking Phys 1 two years ago.
     
  7. Feb 12, 2007 #6

    hage567

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    Is this true?

    E1x = 0
    E1y = 5e6

    This is good so far.

    Try this: At A, put some lines in for the x and y axes. You have the vector from A to q2 drawn in your diagram. Do you see that the angle of E2 from the drawn in y-axis at A is the same as the one you calculated? (By the "z" rule)
    You want to find the components of that vector along each of the x and y axes so that you get a right-angled triangle that "adds" up to that E2 vector.
    ie [tex] E^2=E_x^2+E_y^2 [\tex]
    This is where the trig comes in. If you write out the sin and cos of that angle in terms of the components, you will get an expressions for each. You know the total magnitude of E2, you wrote it down in your first post.
     
  8. Feb 12, 2007 #7
    I'm still not understanding... I have the directions and magnitudes of both E1 and E2 so E at point a is just E = E1+E2....

    and I can't vizualise what you are saying. I really have no idea. here what I got.

    [​IMG]
     
  9. Feb 12, 2007 #8

    hage567

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    I'm sure somewhere in your text there is an explanation of how to resolve vectors into their components. It's the same idea as if you were doing a dynamics question with Newton's second law.

    In the picture at A there is a right triangle. If you do not know the trig ratios of a right triangle (like sin(theta) =E2x/E2 in this example), looking them up might help.
     
    Last edited: May 8, 2007
  10. Feb 12, 2007 #9
    So sin(56) = E2x / 4.1e6)

    =3.4e6

    ??
     
  11. Feb 12, 2007 #10

    hage567

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    Yes, but your diagram isn't quite clear in showing it. Hopefully when my attachment gets approved you will see what I mean.
     
  12. Feb 12, 2007 #11

    hage567

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  13. Feb 13, 2007 #12
    Oh can I see your diagram now.. but shouldn't that verticel component be E2x?? As I said I first drew the graph wrong.
     
  14. Feb 13, 2007 #13

    hage567

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    "Oh can I see your diagram now.. but shouldn't that verticel component be E2x?? As I said I first drew the graph wrong."

    Yes, Sorry. I just left it the way I originally did it.
     
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