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Capacitance Problem

  1. Jul 16, 2012 #1
    The Problem: An empty capacitor is connected to a 12.0 V battery and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material (k=2.8) is inserted between the plates. Find the amount by which the potential difference changes, and state whether this change is an increase or a decrease.

    My attempt at a solution:

    Using the equations Q=CV and C=(k*A*ε)/d

    And considering that A, ε, d, and Q stay constant,
    And k is 1 to begin with since the space is empty, and 2.8 after the material is inserted,

    I get
    1*12(volts)=2.8V
    Thus, V=4.3 Volts.

    However, the book gives the answer as 7.7 V. Help?
     
  2. jcsd
  3. Jul 16, 2012 #2

    TSny

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    Gold Member

    Everything's ok except you didn't quite answer the question as stated. You want to find how much V changes.
     
    Last edited: Jul 16, 2012
  4. Jul 18, 2012 #3
    Oh, I see!

    ΔV=Vf-Vi=4.3V-12V=-7.7 V

    Thus, the voltage decreases by 7.7 V.

    Thanks for your help!
     
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