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Consider a pendulum of length d/2 calculate the electric field inside capacitor

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    consider a parallel-plate capacitor with square plates of side L and distance d (<<L) between them, charged with charges +Q and -Q. The plates of the capacitor are horizontal, with the lowest lying on the x-y plane, and the orientation is such that their sides are parallel to the x and y axis, respectively

    consider a simple pendulum of length d/2 and mass m, hanging vertically from the centre of the top plate, that can oscillate in the x-z plane. Calculate the electric field inside the capacitor, and the electrical force acting on the mass if it has charge q (the field produced by the charge q itself can be neglected, and the polarity of the capacitor is such that the force will act in the negative z direction).

    2. Relevant equations
    E=-grad V
    =[q/(4pi*epsilon0)][1/(r^2)]*(r-hat)

    3. The attempt at a solution

    should i just use the above equation, and sort of ignore the r-hat at the end? would the answer to what the electric field is be zero?

    then would the electrical force be F=Bqv?
    but what would i put in for B and v?
     
  2. jcsd
  3. Aug 3, 2011 #2
    Your attempt at a solution is almost completely wrong. Your second equation,

    E =[q/(4pi*epsilon0)][1/(r^2)]*(r-hat)

    is the equation for the electric field created by a single point charge. If you distribute many electric charges on the surface of an object, such as a capacitor, the electric field is different; not surprisingly, the formula for the electric field depends on the shape of the object. It is possible to use your formula, and some calculus, to calculate the electric field created by charges distributed on any shape. Your book should include an explanation of how to do this. Alternatively, the formula for an electric field inside a parallel-plate capacitor is used very often, so you can probably find the formula in your book or online.

    Now, this gives you the electric field, but you still need to find the force on the charged particle. The one you quote, F = Bqv, is not the one you need. That is actually the force from a magnetic field on a moving charged particle. There are many equations that start with "F;" you need to find the one that applies to this problem, which means you have to read the text around the equation, not just the equation itself. Again, the formula you need can be found in your book or Wikipedia.
     
  4. Aug 14, 2011 #3
    I found another equation:

    F(subscript 1 2)=[1/(4pi*epsilon)]*[[q(subscript1)*q(subscript2)]/[(r-subscript1 2)^2]]*(r-hat-subscript1 2)

    but I'm not sure how this can be used since there is only one charge mentioned in the question and the equation involves two charges
     
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