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Finding an electric field magnitude with tension and weight of hanging ball

  1. Jan 27, 2010 #1
    1. The problem statement, all variables and given/known data
    As shown in the figure above, a ball of mass 0.720 g and positive charge q =39.4microC is suspended on a string of negligible mass in a uniform electric field. We observe that the ball hangs at an angle of theta=15.0o from the vertical. What is the magnitude of the electric field?

    (The figure shows the ball hanging 15 degrees to the left of the y-axis, and the electric field points to the left)


    2. Relevant equations
    F = ma
    E = F/q
    Sum of forces = 0
    Sum of forces = tension + e-field - weight = 0


    3. The attempt at a solution
    I drew a free-body diagram, and found the forces due to the tension and weight to be
    F = mg*cos(15) or F = 6.816
    Then I used E = F/q, to get 1.73*10^5 N/C
    I think I'm missing a step? Or calculating the force incorrectly? Thank you!
     
  2. jcsd
  3. Jan 27, 2010 #2
    Without delving into the math to heavily you found the tension on the string, now what component of that tension is exactly opposite of the force due to the electric field?

    I think this should help if you have not got it yet.
     
  4. Jan 27, 2010 #3
    Hi Pdargn, thank you for the response!
    I think the weight (mg) is the component of the tension that is opposite the force due to the electric field? I'm still not sure how to find the overall force to use in the E = F/q equation?
     
  5. Jan 27, 2010 #4
    I thought I saw you had that the tension was equal to cos(theta)mg...

    It should be.

    So now what component of the tension is parallel to the electric field thus the electric force?
     
  6. Jan 27, 2010 #5
    I tried doing mg*cos(theta) = Fsin(theta), but the F I got gave me the wrong answer when I put it in E = F/q. I'm still not sure what I'm doing wrong with the force. I thought if the tension of the string (mg*cos(theta) = force e-field (F*sin(theta) I would get the overall force, can you help me to figure out where I'm going wrong?
     
  7. Jan 27, 2010 #6
    If you draw a free body diagram of all the forces acting on the charged mass, you get mg acting straight down, T acting up and to the right, and F(electric) acting to the left... yes? or is this not in concert with your diagram?
     
  8. Jan 27, 2010 #7
    Yes, that is exactly what my diagram looks like. So the force of the E-field must equal the force of the tension minus the weight?
    I tried F(electric) = mg*cos(theta) - mg
    But it's still wrong.
    I'm sorry, I appreciate your help but I really don't understand how to correctly put the forces in the equation to find the F(electric)???
     
  9. Jan 27, 2010 #8
    So do we agree that some component (the horizontal component) of tension must be supplying the force that must be equal to the force due to the electric field if the charged mass stays suspended (not accelerating) at this angle?

    And we see there is no way the force of gravity (mg) could supply the force necessary because it is perpendicular to the electric force?

    But we do see that we can find the tension, using cos(theta)mg and that we have to find the horizontal component of this tension?

    I shall return.
     
  10. Jan 27, 2010 #9
    I guess I don't understand how to find the horizontal component of the tension. I would assume it involves sin(theta) as that is the horizontal side, and parallel to the electric force, but I still don't know how to set up the equation.
    I tried mg*cos(theta) - sin(theta) = electric force, but that is also wrong.
     
  11. Jan 27, 2010 #10
    Why did you subtract and why did you use sin(theta)?

    The horizontal component of the tension would be T*cos(90-theta).
    And T is equal to mgcos(theta).

    Does this make sense? Look at your free body diagram and the angles in those right triangles.
     
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