1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Capacitor and dielectric

  1. Apr 13, 2016 #1
    1. The problem statement, all variables and given/known data

    A ##12 \mu F## capacitor ##C_{1}## consists of plates ##A## and ##B##. An ##18 \mu F## capacitor ##C_{2}## consists of plates ##C## and ##D##.

    (a) Each are charged up by a ##10 V## battery such that plates ##A## and ##C## are positively charged. They are disconnected from the battery and connected into a series circuit with plate ##A## connected to plate ##D## and plate ##B## connected to plate ##C##, and the system allowed to come to equilibrium. What charge is now on plate ##D##?

    (b) Both capacitors ##C_{1}## and ##C_{2}## have square plates of side ##x_{1}## cm and plate separation ##d## cm. The difference is that ##C_1## has only air between its plates, but ##C_2## also has a square plate of dielectric of side ##x_1##, relative permittivity ##\epsilon_{r}## and thickness ##t##. In terms of ##\epsilon_{r}##, what fraction ##(t/d)## of the gap is occupied by the dielectric?

    (c) Capacitor ##C_2## is again charged up by a battery of voltage ##V##, and remains connected to it. What force is required to extract the plate of dielectric from the gap? Give answer in terms of ##V## and ##x_1##.

    2. Relevant equations

    3. The attempt at a solution

    (a) Charge on capacitor ##C_1## is ##Q=CV=120 \mu C##, and charge on capacitor ##C_2## is ##Q=CV=180 \mu C##

    Now the arrangement of the capacitors in the series circuit is as follows:

    Untitled.jpg

    Therefore, after redistribution of charges among plates ##D## and ##A##, an excess negative charge of ##60 \mu C## will remain on the wire connecting plates ##D## and ##A##. Therefore, the charge on plate ##D## is ##-30 \mu C##.

    Am I correct so far?
     
  2. jcsd
  3. Apr 13, 2016 #2

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    I haven't worked out the answer, but how do you know that the negative charge splits evenly between plates [itex]A[/itex] and [itex]D[/itex]?
     
  4. Apr 13, 2016 #3
    Oh wait! The voltage across the two capacitors must be the same in magnitude, so the charge of ##-60 \mu C## must divide between plates ##A## and ##D## in the ratio of their capacitances.

    ##C_{1}:C_{2} = 2:3##, so charge on plate ##D = -36 \mu C##, and charge on plate ##A = -24 \mu C##.

    Am I correct so far?
     
  5. Apr 14, 2016 #4

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    I think that's right.
     
  6. Apr 14, 2016 #5
    Ok, but I am confused about one aspect of my solution.

    The voltage drop across capacitor ##CD## going from ##C## to ##D## must be equal to the voltage drop across capacitor ##BA## going from ##B## to ##A##. Also, plate ##C## is at a higher potential than plate ##D##. Doesn't that mean that plate ##B## must be at a higher potential than plate ##A##?
     
  7. Apr 14, 2016 #6

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes. Any two points connected by a conductor (no resistance) are at the same potential, so B&C are the same potential, and A&D are at the same potential.
     
  8. Apr 14, 2016 #7
    In that case, isn't plate ##A## positively charged before it was connected to the capacitor ##CD##, but afterwards, plate ##A## became negatively charged due to distribution of charge between the plates ##A## and ##D##?

    And similarly, isn't plate ##B## negatively charged before it was connected to the capacitor ##CD##, but afterwards, plate ##B## became positively charged due to distribution of charge between the plates ##B## and ##C##?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted