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Capacitors and Potential Difference

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data
    (Note: This is the fourth part of a single question, but it's the only part I'm having trouble with. Don't worry, it's not as simple as Q/V. :rofl:)

    "What is the potential difference between a point midway between the [circular, parallel] plates [which are 1.43 x 10-4 m apart] and a point that is 1.23 x 10-4 m from one of the plates? Answer in units of V."

    Givens: (Some of these I have gotten from the (proven correct) solutions from other parts of the question.)

    [tex]\Delta V_{0} = 0.148 V[/tex]
    [tex]\Delta d_{1} = 7.15 \times 10^{-5} m[/tex]
    [tex]\Delta d_{2} = 1.23 \times 10^{-4} m[/tex]
    [tex]C_{0} = 5.264 \times 10^{-15} F[/tex]
    [tex]Q_{0} = 7.79 \times 10^{-14} C[/tex]

    [tex]\Delta V_{N} = ?[/tex]

    2. Relevant equations
    [tex]C = \frac{Q}{\Delta V}[/tex]
    [tex]\Delta V = E\Delta d = \frac{\Delta PE_{e}}{q}[/tex]
    [tex]PE_{electric} = \frac{1}{2}C\Delta V^{2} = \frac{Q^{2}}{2C}[/tex]

    3. The attempt at a solution

    Alright. My problem with this question isn't so much mathematical as conceptual. If I could figure out some things, I could definitely apply some equations to solve it.

    -Should I use point-charge equations for this?
    -Since we're talking about voltage between points and not charges, can I use the charge capacity of the capacitor in my equations?
    -Can I add up the distances between the points, or do I need to work them separately because they are on different sides of the source of the field?
    -If I change the voltage, wouldn't I be unable to use my values for [tex]C[/tex] and [tex]Q[/tex], since they rely on a certain voltage?
    Last edited: Feb 16, 2007
  2. jcsd
  3. Feb 17, 2007 #2
    Since in a parallel plate capacitor, the electric field is constant at all points, the potential gradient must also be a constant. Therefore, the potential at a point midway will be q/2v.
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