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Net force on a charge in a square

  1. Feb 15, 2014 #1
    1. The problem statement, all variables and given/known data
    There are 4 charges in the corner of a square with the lengths of the sides measuring 18.2cm
    q1= 2.34μC (top left) q2=-1.58μC (top right) q3= 2.03μC (bottom left) q4= -3.00μC (bottom right)

    →Calculate net force on q4
    →Calculate net force at the center of the square


    2. Relevant equations
    Fnet=|KQ|/r^2
    K constant = 8.99x10^9

    3. The attempt at a solution

    q1= [(8.99x10^9)(2.34x10^-6)]/(0.182)^2 = 635086.3422 @ 315°
    q2= [(8.99x10^9)(1.58x10^-6)]/(0.182)^2 = 428818.9832 @270°
    q3= [(8.99x10^9)(2.03x10^-6)]/(0.182)^2 = 550950.9721 @ 0°

    for my next step I was going to calculate x and y components and then plug it into the
    d=*sqrt [(x^2)+(y^2)]

    to calculate net force on q4... but I'm very uncertain about how to continue or if my previous calculations are even right.
     
  2. jcsd
  3. Feb 15, 2014 #2
    You're on the right track. If you calculate the components for each, then you can find the magnitude from the addition of all those components.

    Remember to watch your signs on the components too. Also, for the calculation of the force of q1, there's an error in the value for the separation -- you want to use the length of a diagonal instead of a side length.
     
  4. Feb 15, 2014 #3
    do I square the distance anyway?
     
  5. Feb 15, 2014 #4
    Yes, you want to square the distance, but use the length of the diagonal for the force from q1.
     
  6. Feb 15, 2014 #5
    Solution??

    Ok so this is what I did
    → Using pythagoreon theorem I found the length of the diagonal line toward q4 which = 025738684 m

    I applied this to the previous calculation that I did wrong. But also, wouldn't the angle of q2 be 90° because it is a negative force and therefore repels away from the negative force of q4?

    Anyway this is what I did
    q1= [(8.99x10^9)(2.34x10^-6)]/(0.25738684)^2 = 317543.2411 @ 315°
    q2= [(8.99x10^9)(1.58x10^-6)]/(0.182)^2 = 428818.9832 @90°
    q3= [(8.99x10^9)(2.03x10^-6)]/(0.182)^2 = 550950.9721 @ 0°

    X and Y components:
    E1 -- x: 224536.9791 y: -224536.9791
    E2 -- x: 0 y: 428818.9832
    E3 -- x: 55090.9721 y: 0

    I then added the x components and the y components to calculate E4
    x: 775487.9512 y: 204282.0041

    I then used the distance formula to calculate the net field on E4:
    FnetE4: *sqrt [(775487.9512^2)+(204282.0041^2)]
    FnetE4: 801943.0776C ???


    Do I still need to calculate the angle?
    I did but it doesn't seem right.

    Angle tan -1:y/x
    =204282.0041/775487.9512
    =14.8°
     
  7. Feb 15, 2014 #6
    That all looks right for the most part, but because you calculated the net field due to all the charges, you have to multiply by q4 to get the net force.
     
  8. Feb 15, 2014 #7

    haruspex

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    You mean, the field at q4 due to the other three charges, right?
    Distance formula? C?
    You are asked for the force on q4, but it looks to me that you have only got as far as the field.
    Why does that seem wrong?
     
  9. Feb 15, 2014 #8
    Yes the charge on q4 due to the other 3.
    So I would use F=qE to calculate the net force on it right?
    It just seems wrong because the direction would be north east?
     
  10. Feb 15, 2014 #9
    So then to calculate the net force at the center of the square I've taken half of the distance I had for the charge of q1 -> q4 = 25.738684 / 2 = 12.869342cm

    So to calculate the center force I used the formula

    Fnet= |Kq1q2|/r2 once for the left side and once for the right side of the square

    so Fnet on the left side was =0.042704298 and =.0426126

    I am a little lost how to continue or if that is even right... Any help is appreciated!
     
  11. Feb 15, 2014 #10
    Yes, you have to do F = qE to get the net force. Also keep in mind that since q4 is negative, the force will be in the opposite direction as the field, so the force will actually be southeast.

    And for the center of the square, it doesn't make since to calculate the force at the center of it, since there's nothing in the center to have a force exerted upon it. I assume it means field, so use the field equation. And I don't really understand your process. What do you mean by the left half and the right half of the square? Shouldn't you calculate the fields from each of the four charges and then add them up?
     
  12. Feb 15, 2014 #11

    haruspex

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    Yes, but that's the field. The charge q4 is negative, so the force will point...?
     
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