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Capacitance Equations

  1. Aug 7, 2009 #1
    Problem 1
    Two parallel metal plates are connected to a voltage source so that their potential difference is maintained at 350 V. The two plates are separated by 1.00 cm. An electron is released from rest at the negative plate and accelerates toward the positive plate. (a) Determine the electric field strength between the two plates. (b) Determine the amount of kinetic energy gained by the electron.

    a.) [tex]V=\int \vec{E} \cdot d\vec{S}[/tex]
    [tex]V=Ed[/tex]
    [tex]E=\frac{V}{d}[/tex]
    [tex]E=\frac{350}{.01}=35kV/m[/tex]

    b.)[tex]U_i+K_i=U_f+K_f[/tex]
    [tex]qV + 0 = 0 + \frac{1}{2}mv^2[/tex]
    [tex]qV = \frac{1}{2}mv^2[/tex]
    the KE gained by the moving electron is the PE lost, or qV.
    [tex]qV= (1.6x10^{-19})(350)=5.6x10^{-17}[/tex]

    Problem 2
    A 12.0 V battery is connected to a 6.0 nF parallel plate capacitor. (a) What amount of charge will exist on each plate once electrostatic equilibrium is reached? (b) What amount of energy is “stored” in the capacitor in this state?

    a.)[tex]C=\frac{Q}{V}[/tex]
    [tex]Q=VC[/tex]
    [tex]Q=(12)(6x10^{-9})=72x10^{-9} Coulombs[/tex]

    b.)[tex]U=\frac{1}{2}QV[/tex]
    [tex]U=\frac{1}{2}(72x10^{-17})(12)=4.3x10^{-7} Joules[/tex]

    Now, my question is this. How can the potential energy lost in the first problem (qv) not be the same amount of charge stored in the second problem (1/2 qV)? I've had both equations thrown at me where Electric Potential is equal to Electrostatic Potential Energy per charge (qV), and also where it's (1/2 qV).
     
    Last edited: Aug 7, 2009
  2. jcsd
  3. Aug 7, 2009 #2

    berkeman

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    Staff: Mentor

    In the first problem, the units of electric field are V/m, so your answer should be 35kV/m.

    Not sure I understand your question at the end...
     
  4. Aug 7, 2009 #3
    You're right. Edited.
     
  5. Aug 7, 2009 #4

    berkeman

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    Staff: Mentor

    So I'm not tracking this yet. Could you show the math where you get the two different results. The fundamental equations should be:

    E = V/d
    Q = CV
    F = qE
    Work = Force * Distance...
     
  6. Aug 7, 2009 #5
    If the Voltage is defined as the Potential Energy per unit charge, V=U/q. Therefore, U=qV, or the amount of work is equal to the charge times the voltage difference. However, when it's read this way, "The energy stored in a capacitor", it is U=1/2qV.

    see: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html#c1

    So I'm confused. How can the potential energy be calculated as qV when talking about Voltage but be 1/2qV when talking about stored energy?

    See: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1
     
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