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Math Help - Coulombs law and Electric Fields

  1. #1
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    Smile Coulombs law and Electric Fields

    HI there. I am revising for my A level Physics exam in January. I have printed out past paper questions, but have no qnswers to them! I have worked through this question and was wandering if anyone could check to see I have done it correctly.

    Any help greatly appreciated.

    Two point charges, each of magnitude 5x10^-6C, are placed horizontally 50mm apart at points A and B respectively.

    Calculate the force exerted on a point charge of 3x10^-6C at a point P, which is 20mm from A and 30mm from B, in the cases when:

    i) both charges A and B are positive (my answer 11.25N)
    ii) A is positive and the charge at B is negative (my answer 2.25N)

    Thank you very much for assistance!

    Mike
    Last edited by mikewhant; December 30th 2009 at 08:19 AM.
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  2. #2
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    Quote Originally Posted by mikewhant View Post
    HI there. I am revising for my A level Physics exam in January. I have printed out past paper questions, but have no qnswers to them! I have worked through this question and was wandering if anyone could check to see I have done it correctly.

    Any help greatly appreciated.

    Two point charges, each of magnitude 5x10^-6C, are placed horizontally 50mm apart at points A and B respectively.

    Calculate the force exerted on a point charge of 3x10^-6C at a point P, which is 200mm from A and 30mm from B, in the cases when:

    i) both charges A and B are positive (my answer 11.25N)
    ii) A is positive and the charge at B is negative (my answer 2.25N)

    Thank you very much for assistance!

    Mike
    if P is 30mm from B, and A and B are 50mm apart, how can P be 200mm from A?

    did you mean to type 20mm?
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  3. #3
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    sorry, yes, it is 20mm! thanks for pointing that out!
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  4. #4
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    I'm afraid my calculations differ from yours.

    How did you attempt to find the net force on charge P ?
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  5. #5
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    Ok, well I worked out the electric field strength due to A (2.25x10^6N/C) and then the same for B (1.5x10^6N/C)

    Then I worked out the force due to A (6.75N) and then for B (4.5N) and added them together giving 11.25N for part i

    For part ii, I said that since B was negative, it should be subtracted from A giving 2.25N for part ii.

    Thanks again for your help.

    Mike
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  6. #6
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    Quote Originally Posted by mikewhant View Post
    Ok, well I worked out the electric field strength due to A (2.25x10^6N/C) and then the same for B (1.5x10^6N/C)

    Then I worked out the force due to A (6.75N) and then for B (4.5N) and added them together giving 11.25N for part i

    For part ii, I said that since B was negative, it should be subtracted from A giving 2.25N for part ii.

    Thanks again for your help.

    Mike
    coulomb constant ... k = 8.99 \times 10^9

    q_A = 5 \times 10^{-6} C

    q_P = 3 \times 10^{-6} C

    r = 2 \times 10^{-2} m

    force exerted on charge P by charge A ...

    F = \frac{kq_Aq_P}{r^2} \approx 337 N
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  7. #7
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    Got it!

    Not too sure what I was trying to work out! I was using the equation

    E=kq/r^2 to work out the elctric field strength. Didnt need to do that!

    Thanks alot for that, take care

    Mike
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