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Capacitance when dielectric is introduced

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data
    a dielectric of dielectric constant 3 fills three fourth of the space between the plates of a parallel plate capacitor. What percentage of the energy is stored in the dielectric.


    2. Relevant equations

    U=CV2/2
    = epsilon0AV2/d.2

    3. The attempt at a solution
    When dielectric is introduced the above equation changes to epsilon0AV2/(d-t+t/epsilonr)
    I substituted 3d/4 for t since dielectric fills three fourth of space between the plates of capacitor and 3 for relative permeability.
    When simplified i got the answer epsilon0AV2/d
    which is twice the value of U that is 2U
    In the book the answer is given as 50% which i don't understand. How this 50% comes? Thanks in advance, revered members
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 14, 2011 #2
    how did you take V to be constant?
    is it given that capacitor is connected to a battery?
     
  4. Jul 14, 2011 #3
    no sir. its not given. please guide me to solve this problem
     
  5. Jul 14, 2011 #4
    is that the exact question given to you?
    because i think something else should also be given.
    ans also, the language of question is new to me
     
  6. Jul 14, 2011 #5
    yes sir i have the typed the exact question. no other details are given
     
  7. Jul 14, 2011 #6
    there is only 1 way i got 50% energy increase due to introduction of dielectric.
    first let capacitor is isolated (safe to assume)
    let it has plate area A and separation d.
    Energy, U = Q^2/2C

    find energy when there is no dielectric, in terms of epsilon,A,d not in C (pretty easy :) )

    then find net capacitance when dielectric is inserted.

    now find energy,
    *note: now Q will remain same as capacitor is separated.

    you'll see that increase in energy is 1/2 of initial energy ... which is stored due to (or in other words, stored inside) capacitor.
     
  8. Jul 20, 2011 #7
    sir, can i proceed in this way as given in my attachment? Of course i dont know how 50% comes as given in the book? thanks in advance,sir
     

    Attached Files:

    Last edited: Jul 20, 2011
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