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Homework Help: Coulomb triangle

  1. May 17, 2006 #1
    I've been trying to figure this problem out for quite awhile but am really struggling, help would be much appreciated. Thank you

    In the diagram to the right, the net force on the 1.0 mC charge is zero. What is the sign and magnitude of the unknown charge q?

    So the diagram is a triangle, which has two charges on the bottom, each 2.0mC, and 8 cm apart. The top is the unknown q, which from the middle of the triangle is 6cm up. And there is a 1.0mC charge in the centre of the triangle which is 3 cm up. (I have attached the diagram)

    Based on the measurements given, I'm first trying to come up with the x and y components for each of the forces. And this is where I'm already having difficulties. Q1 is bottom left, Q2 unknown top, Q3 bottom right, Q4 centre

    F12 = -F12xcos(theta)- F12ysin(theta)
    F13 = -F13x
    F14 = -F14xcos(theta)+ F14ysin(theta)

    x: -F12cos(theta)- F13 + F14cos(theta) = 0

    y: -F12sin(theta) + F14sin(theta) = 0

    Can anyone tell me if this looks remotely ok so far?

    Attached Files:

  2. jcsd
  3. May 18, 2006 #2
    Well generally a good procedure (well what I was taught and it is straight forward) is to first draw the force vectors. That is, draw the force vector of the force of Q1 on Q4. Then draw the force vector of Q3 on Q4.
    Just from doing this you see that the resultant force vector (excluding Q2 at the moment) is upwards. So you can see that Q2 will have to be positive to provide a repulsive force in the opposite direction to the resultant force from Q1 and Q3 on Q4.

    Then calculate the magnitude of the force vectors for Q1 and Q3 on Q4.

    F14=(kq1q4)/r^2 = (kq1q4)/(0.05)^2

    Where r was calculated using Pythagoras' Theorem.

    Do this for Q3 on Q4 as well. Remembering that you are only interested in the magnitude at this stage (not direction).

    You don't need to worry about calculating the x-component 'cause from symmetry you can see that the x-components of Q1 and Q3 will cancel each other and Q2 doesn't contribute any force in the x direction.

    So all you need to worry about is the y-component of the force. Since you have the magnitude of the forces from Q1 and Q3 you can find the y-components by multiplying the magnitude of the forces with sin(theta).

    y: F14y + F34y - F24y = 0

    From this you can see that the y components of F14 and F34 are added since I've taken the upwards force to be positive. Solve for F24 and you have your solution.
  4. May 18, 2006 #3
    Ok so for F14 & F34 = 7.2 x 10^8

    I found the angle for each of these to be 36.9 degrees, therefore F14 & F34 x sin(36.9) both = 4.32 x 10^8

    Based on big man's y component F14y and F34y add together to get F24 = 8.65 x 10^8

    So based on a top angle with Q24 of 33.7 I found Q2 to be 0.16 C.

    To me this doesn't seem right as its so much larger than any of the other charges?
  5. May 18, 2006 #4
    It's not quite right.

    With F14 and F34 you should have a magnitude of the order of 10^6 rather than of 10^8.

    F14 = (k*q1*q4)/0.05^2 = (8.99*10^9)*(0.002*0.001)/0.0025 = 7.19*10^6

    The magnitude of F34 will obviously be the same.

    Now you got the angle right and so F14y and F34y will be the same and will be:


    Then solving for q2 you should get a number that is slightly less than q4.

    I don't know what the "top angle with Q24" means 'cause q2 is lying on the same line as q4, meaning that all of it's force is directed along the y direction.

    ie y component= F14y + F34y -F24 = 0
    ie 2*F14*sin(36.9) - k*(q2*q4)/r1^2 = 0

    Where r1 is the distance between q2 and q4 (0.03 m).
    So yeah you were right that 0.16 C didn't make sense :)
  6. May 18, 2006 #5
    Ok so based on this I found Q2 to be +0.865 C which seems to be a more appropriate number. Am I correct in saying that it is positive?

    But I'm not sure that I understand why there is no angle, is it because the vector has only a y-component and so it just equals the force?
  7. May 18, 2006 #6
    I assume you meant that it was +0.865 mC or around there.

    Yes you are right in saying that it is positive because it has to provide a repulsion force so that it 'cancels' out the y-component forces from q1 and q3.

    There is no angle because q2 and q4 are lying on the same axis. This means that the force of q2 on q4 will only be directed along that axis, which means that there will be no x-component. So reiterating the point...there is no force in the x direction, all the force from q2 is being directed in the y direction only.
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