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Poteintial difference to move mass

  1. Sep 27, 2007 #1
    There's a small sphere of mass m that hangs by a thread. the sphere is between two parallel plates L apart. The sphere has a charge Q. What is the potential difference that will make the plates to assume an angel of 20 degrees with the vertical?

    [tex]F=QE=mg \sin \theta [/tex]
    [tex]E=\frac{mg \sin 20}{Q} [/tex]
    [tex] V= - \int _0 ^L \frac{mg \sin 20}{Q}dx[/tex]
    [tex] V= - \frac{mgL \sin 20}{Q}d[/tex]

    i feel like this question should involve more work. was my thought process correct?
  2. jcsd
  3. Sep 27, 2007 #2


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    Yes, it's correct. (But the d in your last line is presumably a typo.) Also, since the question asks for "potential difference", the minus sign is not really relevant.
  4. Sep 27, 2007 #3
    yes it's a typo

    could you explain why the potential difference is not relevant?
  5. Sep 27, 2007 #4

    Andrew Mason

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    How do you get the first statement? Do a free body diagram of the forces on the suspended mass. There is the tension in the string, gravity and the electric force. From a free body vector diagram you should be able to get the expression for E and then V (=E/L) .

  6. Sep 28, 2007 #5

    Andrew Mason

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    If you do the free body diagram you will see that the gravitational and electric forces have to be balanced by the tension in the string.

    [tex]\vec{T} = q\vec{E} + m\vec{g}[/tex]

    This means that:

    [tex]T\sin{20} = qE[/tex]
    [tex]T\cos{20} = mg[/tex]


    [tex]qE/mg = \tan{20}[/tex]

    [tex]EL = V[/tex], so

    [tex]V = mgL \tan{20}/q[/tex]

    Last edited: Sep 28, 2007
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