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Determine the total charge needed to apply to the electroscope.

  1. Sep 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A large electroscope is made with "leaves" that are [itex]78[/itex]-cm-long wires with tiny [itex]24[/itex]-g spheres at the ends. When charged, nearly all the charge resides on the spheres.

    If the wires each make a [itex]26^{\circ}[/itex] angle with the vertical, what total charge [itex]Q[/itex] must have been applied to the electroscope? Ignore the mass of the wires.

    2. Relevant equations

    [itex]F = mg[/itex]
    [itex]F = \dfrac{kq_1q_2}{r^2}[/itex]

    3. The attempt at a solution

    Tried this way:

    [itex]F_{\text{electrostatic}} = F_{\text{gravity}}[/itex]
    [itex]\dfrac{k(\frac{Q}{2}(\frac{Q}{2}))}{r^2} = mg\cos(\theta)[/itex]

    [itex]r = 2L\sin(\theta) [/itex] is the distance between the two spheres and the length of the string, so we have

    [itex]Q = 4L\sin(\theta)\sqrt{\dfrac{mg\cos(\theta)}{k}} \approx 6.63 \times 10^{-6} C[/itex]

    But the answer is wrong.
     
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  3. Sep 5, 2013 #2

    gneill

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    Why would the electrostatic force be equal to the gravitational force? That would only be the case if the angle of the wires was 45° from vertical...

    Why would the gravitational force depend upon the angle (mg cos(θ))?

    Why not draw the setup and the FBD for one of the spheres?
     
  4. Sep 7, 2013 #3
    So much stressed with this problem since it's pretty hard to work out. Also, there are so many wrong answers online.

    Working out with the free-body diagram from the paper I drew, I have

    [itex]\Sigma F_x = F_T \cos(\theta) - mg = 0 \rightarrow F_T = \dfrac{mg}{\cos(\theta)}[/itex]
    [itex]\Sigma F_y = F_T \sin(\theta) - F_E = 0 \rightarrow F_E = F_T\sin(\theta) = mg\tan(\theta)[/itex]
    [itex]F_E = k\dfrac{(\frac{Q}{2})^2}{d^2} = mg\tan(\theta) \rightarrow Q = 2d\sqrt{\dfrac{mgtan(\theta)}{k}}[/itex]

    I got [itex]5.6 \times 10^{-6} \mbox{C}[/itex] but still marked wrong.
     
    Last edited: Sep 7, 2013
  5. Sep 7, 2013 #4

    TSny

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    Your formula looks correct to me. But I get a different numerical answer.
     
    Last edited: Sep 7, 2013
  6. Sep 7, 2013 #5
    What did you get in two sig fig, TSny? I want to be sure that I get the correct answer for the problem. MasteringPhysics, the physic HW system, is a big problem for me since it has some problems with answer-checking.

    The actual solution is [itex]5.569457441... \times 10^{-6}[/itex] C. In terms of 2 sig. fig., I obtain [itex]5.6 \times 10^{-6}[/itex] C. I want to double-check the answer with yours.
     
    Last edited: Sep 7, 2013
  7. Sep 7, 2013 #6

    TSny

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    Show your calculation with numbers. Maybe we can spot an error.
     
  8. Sep 7, 2013 #7
    The post was edited while you made the post. Here is the solution I posted already.
     
  9. Sep 7, 2013 #8

    TSny

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    I meant to show the details of plugging the numbers into the formula. In particular, what value did you use for d? for m? I don't get an answer near 5.6 μC.
     
  10. Sep 7, 2013 #9
    Here is the actual set up.

    [itex] Q = 2(78 \times 10^{-2} \mbox{m})\sqrt{\dfrac{(24 \times 10^{-3} \mbox{kg})(9.8 \mbox{m}/\mbox{s}^2) \tan(26^{\circ})}{9.0 \times 10^{9} \mbox{N}\cdot \mbox{m}^2/\mbox{C}}} [/itex]

    where...

    [itex]d = 78 \times 10^{-2} \mbox{m}[/itex]
    [itex]m = 24 \times 10^{-3} \mbox{kg}[/itex]
     
  11. Sep 7, 2013 #10

    TSny

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    Think about your value of ##d##. Is it correct?
     
  12. Sep 7, 2013 #11
    To me, I believe it's correct since 1 m = 100 cm.
     
  13. Sep 7, 2013 #12

    TSny

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    What does ##d## represent physically in this equation?
     
  14. Sep 7, 2013 #13
    Oops. Forgot the "basics". Sorry about that.

    ##d## represents the distance between "charges". So with trig, I got [itex]d = 78 \times 10^{-2}\sin(26^{\circ} \mbox{m})[/itex] Thus, we have:

    [itex] Q = 2(78 \times 10^{-2} \times \sin(26^{\circ}) \mbox{m})\sqrt{\dfrac{(24 \times 10^{-3} \mbox{kg})(9.8 \mbox{m}/\mbox{s}^2) \tan(26^{\circ})}{9.0 \times 10^{9} \mbox{N}\cdot \mbox{m}^2/\mbox{C}}} [/itex]

    ...which is approximately 2.4 x 10-6 C.
     
  15. Sep 7, 2013 #14

    TSny

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    Still not quite right. You had it correct in your first post (where you used ##r## instead of ##d##).
     
  16. Sep 7, 2013 #15
    But then ##r = 2L\sin(\theta)##. Why is it that?
     
  17. Sep 7, 2013 #16

    TSny

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    I don't get an answer of 6.6 μC when I plug in the numbers.
     
  18. Sep 7, 2013 #17

    TSny

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    Note that your formula for Q was not correct in the first post.
     
  19. Sep 7, 2013 #18

    TSny

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    That's what you wrote for ##r##. How did you get that?
     
  20. Sep 7, 2013 #19
    That is by trig. The formula from the first post is wrong as you said.

    Let's not stick to that "first post" conversation too much since it seems much unnecessary to stress too much about the previous wrong answer. As you said before, the following equation is correct:

    ## Q = 2d\sqrt{\dfrac{mg\tan(\theta)}{k}} ##

    ##d## is the distance between the charges. With trig, the distance between the sphere and the vertical is ##(78 \times 10^{-2})\sin(26^{\circ})##. Multiply that by 2 to get ##2(78 \times 10^{-2})\sin(26^{\circ})##. Thus, d could be that value.
     
  21. Sep 7, 2013 #20

    TSny

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    Yes, that's correct. ##d = 2L\text{sin}\theta##. But note that in the formula for Q, d is multiplied by 2. So, there's a factor of 2 in calculating d and another factor of 2 in the formula for Q.
     
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