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Coulomb's Law Electric Fields

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Two point charges are placed at the opposite corners of a rectangle as shown. What is the Electrical Field magnitude at each point due to the charges?


    2. Relevant equations

    Pythagorean theorem, E=kq/r^2


    3. The attempt at a solution

    r= square root of .8^2 +.4^2 =.89
    E= 8.99x10^9 X 4X10^-6 divided by (.89^2)
    E=8.99x10^9 X 2X 10^-6 divided by (.89^2)
    should I not be dividing by .89^2 but rather the radius of an electron?

    I am not sure if A and B should be negative or positive or if I am doing this right..?
     

    Attached Files:

  2. jcsd
  3. May 11, 2013 #2

    Simon Bridge

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    A and B are points, they are neither negative nor positive.
    However, you do need to bear in mind that the electric field is a vector - you only have the equation for the magnitude.

    Note, charge 1 is twice the size of charge 2.
    For each point, one charge is twice the distance that the other charge is.
    It's worth thinking about this - if you double r, then E goes down by... but what if you also double q? ... halve q?
     
  4. May 11, 2013 #3
    I dont really understand because to find the vector i need to use square root of e in the x direction squared + e in the y direction squared...

    I dont get how to find the vector in the x direction ...I have E= kq/r^2 or 8.99X10^9 X 4X10^-6 / (.400^2)

    but my professor crossed off .89^2 so I think I did that wrong?
     
  5. May 11, 2013 #4
    im getting 226499 N/C for Point A and 125639 N/C for Point B...?
     
  6. May 12, 2013 #5

    Simon Bridge

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    What is the field at point A due to the 2μC charge?
    Which direction does it point?

    What is the field at point A due to the 4μC charge?
    Which direction does it point?

    What is the total field vector at point A? (use i-j-k notation if you like)

    What is the magnitude of the total field at point A?
     
  7. May 12, 2013 #6
    125639 nc

    ???
     
  8. May 12, 2013 #7
    The magnitude at A and B are the same.
     
  9. May 12, 2013 #8

    Simon Bridge

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    @barryj: we'll see...

    it's no good to have the answer - you need to have the physics.
    Please read the PF rules.

    @baird.lindsay:
    I think we've identified some confusion here ...
    ... etc... please show your working.
    1. electric field at A due to 2 microCoulomb charge:
    ##E_1= kq_1/r_1^2##
    ##k=8.99 \times 10^9 \text{N.m}^{-2} \text{C}^{-2}##
    ##q_1=2\times 10^{-6}\text{C}##
    ##r_1=4\text{m}##
    ##E_1=1123.8\text{N/C}## direction: "upwards" (note: axis not defined in problem)

    Want to try that again?

    2. the field at point A due to the 4 microCoulomb charge:
    ##E_2=kq_2/r_2^2##
    ... what are each of the values?

    3. E vector is

    4. magnitude of E ...
    note:
    if ##\vec{E}=E_x\hat{\imath}+ E_y\hat{\jmath}##
    then ##|\vec{E}|=\sqrt{E_x^2 + E_y^2}##
     
  10. May 12, 2013 #9
    therefore 1256.4 N/C
    is this correct?
     
    Last edited: May 12, 2013
  11. May 12, 2013 #10

    Simon Bridge

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    You can edit the quote by inserting extra close and open-quote tags.

    Note: the i direction is usually to the right horizontally - the direction of the field at A due to the 4mcC charge is to the left ... i.e. the -i direction. But it makes no difference to the magnitude.

    I don't check people's arithmetic - but your method is now correct.
    Can you see where your thinking was out before?

    Now you can repeat for point B and see if Barryj was right ;)
     
  12. May 12, 2013 #11
    no they are different...i get it now ..thanks for all the help..this was driving me crazy!
     
  13. May 12, 2013 #12

    Simon Bridge

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    Just take things a step at a time - but notice how you didn't really need anyone to tell you when you had it right?
    Did you figure out how you were going wrong the first time?
     
  14. May 12, 2013 #13
    I stand corrected. Humble pie tastes terrible.
     
  15. May 12, 2013 #14

    Simon Bridge

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    Don't beat yourself up.
    As you advance in science you will find you are wrong far more times than you are right - that's the only way you can be sure of anything. The trick is to be wrong in the right way... so you can learn from it ;)
    http://www.ted.com/talks/kathryn_schulz_on_being_wrong.html

    It is a very common tactic to get you familiar with how the equations work by giving you problems where things get doubled and halved. This is a case in point.

    Note, charge 1 is twice the size of charge 2.
    For each point, one charge is twice the distance that the other charge is.
    It's worth thinking about this - if you double r, then E goes down by... (how much?) but what if you also double q? ... halve q?

    The point of the exercise is to understand the inverse-square law, not just be able to do the math.
     
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