Force problem! What is the electric force on q3 due to the other 2 charges?

  1. 1. The problem statement, all variables and given/known data

    Three point charges are fixed in place in the right triangle shown below, in which q1 = 0.71 µC and q2 = -0.67 µC. What is the electric force on the +1.0-µC (let's call this q3) charge due to the other two charges?


    magnitude ____ N
    direction ______°, measured counterclockwise from the +x-axis

    2. Relevant equations


    3. The attempt at a solution

    The only thing I don't know how to do is calculate the x and y components of the charge on q3 due to q1.

    Here's what I did:

    First I found the missing side, 6 cm. Then I converted all the cm to m by multiplying it by .1 m and then I converted all the uC to C by multiplying it by X10^-6.

    q1: Fx= k(q1)(q3)/r^2
    = 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 6/10
    = 8.99e9 (0.71e-6) (1.0e-6)/(0.10)^2 * 8/10
    q2: Fx=k(q2)(q3)/r^2
    =8.99e9 (-0.67e-6) (1.0e-6)/(0.10)^2

    Total force= Fx added together. And then sqrt. (Fx^2 + Fy^2)

    As for the direction, tan theta= (Fy/Fx) and then add 180 to it?

    Please help me, it's due tonight, thank you so much in advance.
  2. jcsd
  3. berkeman

    Staff: Mentor

    Welcome to the PF.

    You should convert the polar form of the vectors to rectangular form, add the forces in rectangular form, and then convert back to polar form. Otherwise, it's too easy to make mistakes.

    The polar --> rectangular conversion for the force due to q2 is easy, since all of the force is in the x direction, right? What about the conversion for the force due to q2? Can you show us the x and y components of that force? And then add the forces component-wise, and convert back to the polar answer format that they are asking for...
  4. Thank you so much, this forum is awesome, you replied so quick!

    I don't know what polar or rectangular form mean, I'm sorry, I'm in Intro physics.
    I think the x-component of the force q2 is (0.71e-6 cos 30) and the y-component is (0.71e-6 sin 30)?
    Can you please let me know if I'm on the right track?
  5. berkeman

    Staff: Mentor

    Good, except watch your signs. q1 and q3 are both +, so the force is repulsive or attractive? And given the answer to that, you need to call the component of the force in the + direction positive if it points in the direction of the + x axis, and negative it if points in the direction of the - x axis. And the same thing for the y component of the force.

    Once you have the x and y components of the forces (making sure the signs are right), add the x components to get the resultant x force, and add the y components. Then covert back to the polar form that the question asks for.
  6. Since they're both +, the force is repulsive.

    So I have:
    q2: Fx= - k(0.71e-6)(1e-6)/(0.71e-6 cos30)^2
    Fy= - k(0.71e-6)(1e-6)/(0.71e-6 sin30)^2

    (both components are negative, because the force points in the negative x direction, since they're repulsive to the positive charge of q3.)

    q3: Fx= k(-0.67e-6)(1e-6)/(0.6)^2

    So, Fx= 532160742267 + -0.0167313889 = 5.32160742e11
    Fy= -1.29705876e10

    What does "polar form" mean?
    Last edited: Jan 23, 2011
  7. berkeman

    Staff: Mentor

    Sorry, Q2 is in line with Q3 on the x-axis, so I don't understand how you have x and y components to that force. I could be spacing something, though.

    Polar/Rectangular component conversions:

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