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Capacitance pdf crosscheck

  1. Feb 23, 2016 #1
    A capacitor has a capacitance C and is charged using a battery. After disconnecting the battery the two plates are separated until they are twice as far apart as originally. What happens to the voltage across the capacitor?
    I think answer should be voltage increases . As battery has been disconnected q will remain same . Distance between the plates is increasing , capacitance should decrease.
    ##Q##= CΔV
    As Q is constant and C is decreasing , V should increase.
    But according to http://www.bama.ua.edu/~tmewes/PH-106/2008/clicker%20questions/CH26%20Clicker.pdf [Broken]
    Go to page 10
    Voltage goes down . It is wrong . And formula is also wrong. Is that correct?

    I have gone through all 14 pages of this pdf and found them right except this one , but I want to confirm are all other 13 pages correct?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 23, 2016 #2

    davenn

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    I agree with you ... it may have been a genuine highlighting mistake ?


    it increases because you do work on the system to spread the plates, this results in an increase in potential energy


    Dave
     
    Last edited by a moderator: May 7, 2017
  4. Feb 23, 2016 #3
     
  5. Feb 23, 2016 #4
    Yes, the formula is wrong on that page.(10)
     
  6. Feb 24, 2016 #5

    SammyS

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    You are right.

    Solving ##\ Q= C\,ΔV\ ## for ΔV gives ##\displaystyle\ \Delta V = \frac QC \ ##.

    That pdf / PowerPoint has the incorrect ##\displaystyle\ \Delta V = \frac CQ \ ##.
     
    Last edited by a moderator: May 7, 2017
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