You have two identical capacitors and an external potential source. a) Compare the total energy stored in the capacitors when they ar

Question

You have two identical capacitors and an external potential source.
a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel.
U(parallel)/U(series) =?
b) Compare the maximum amount of charge stored in each case.
Q(parallel)/Q(series) =?
c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?
E(parallel)/E(series) =?

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Cherry 5 years 2021-08-23T05:47:38+00:00 1 Answers 88 views 0

Answers ( )

  1. thucuc
    0
    2021-08-23T05:49:14+00:00
    Reply

    Answer:

    Explanation:

    Given two identical capacitor

    i.e they have the same capacitance C

    Also let say we have external potential V

    a. Now for series connection,

    1/Ceq=1/C+1/C

    1/Ceq=(1+1)/C

    1/Ceq=2/C

    Then, Ceq=C/2

    Ceq=½C

    So the energy stored in a capacitor is given as

    Us= ½CeqV²

    Us=½×½CV²

    Us=¼CV²

    Now, for parallel connection

    Ceq=C+C

    Ceq=2C

    Then,

    Energy is given as

    Up=½CeqV²

    Up=½×2CV²

    Up=CV²

    Comparing this to series

    Us=¼CV², since CV²=Up

    Then, Us=¼Up

    Up=4Us

    Up/Us=4

    So the energy stored in the parallel capacitor connection is 4 times the energy stored in the series capacitor connection.

    b. Charges

    For series connection

    We know that same charge pass through series connection. Now, for series the potential difference is V/2 since they have the same capacitance, they will share the voltage

    And the charge is given as

    Q=CeqV

    Qs=½CV

    Qs=½CV for each capacitor

    Then the total charge will be

    Qs=½CV×2.

    Qs=CV

    Now, for parallel connection

    The have the same voltage but different charge,

    The charge for parallel is given as

    Q=CeqV

    Qp=2CV

    Now comparing

    Qp/Qc=2CV / CV

    Qp/Qc=2

    The maximum charge of the parallel is four times that of the series again

    c. Electric field is given as

    E=V/d

    The potential difference in parallel is V same voltage.

    Ep=V/d

    Now for series the potential difference is V/2 since they have the same capacitance, they will share the voltage

    Es=V/2d.

    Then,

    Ep/Es=V/d ÷ V/2d

    Ep/Es=V/d × 2d/V

    Ep/Es=2

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