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What is the voltage across the capacitor plates?

  1. Aug 10, 2012 #1
    A 15μF capacitor is connected to a 55V battery and becomes fully charged. The battery is removed and the circuit is left open. A slab of dielectric material is inserted to completely fill the space between the plates. It has a dielectric constant of 4.8. What is the voltage across the capacitor plates after the slab is in place?

    I used the formulas:

    Q = CV

    and

    V = Q / C*k

    Am I in the right direction? =/
     
  2. jcsd
  3. Aug 10, 2012 #2

    CWatters

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    Science Advisor
    Homework Helper

    Yes but would be better to show more of the working. For example by quoting the equation for a flat plate capacitor

    C=E0Era/d

    and explaining that the charge is constant.
     
  4. Aug 10, 2012 #3

    gneill

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    Staff: Mentor

    It depends upon how and when you apply them :wink: Why not take a shot at a solution so we can see what you have in mind?

    Also, try to get in the habit of using parentheses to clarify ambiguous orders of operations in your ascii formulas. For example, is the second equation V = (Q/C)*k, or V = Q/(C*k) ?
     
  5. Aug 10, 2012 #4
    Thanks for the replies!
    Please correct me if I'm wrong

    So first, I used the formula Q=CV to look for the charge of the capacitor connected to the 55V battery. Since the battery is removed, the charge Q remains the same right? Now when the dielectric material is inserted to fill the space, capacitance increases and with the equation Cinitial = Q/V, voltage should decrease ...?

    So, basically, I did C*K because capacitance increases by a factor K, then looked for the new voltage: V = Q/Cfinal

    Am I making any sense? lol
     
  6. Aug 10, 2012 #5

    gneill

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    Staff: Mentor

    It looks like you've got the right idea. What is your numerical result?
     
  7. Aug 10, 2012 #6
    mhmm, I have around 11V ... I hope that's correct =/
     
  8. Aug 10, 2012 #7

    gneill

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    Staff: Mentor

    It's in the right neighborhood, yes.
     
  9. Aug 10, 2012 #8
    Thanks for your help!
     
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