Separating capcitor plates

Discussion in 'Physics' started by poopscoop, Oct 15, 2013.

  1. poopscoop

    Thread Starter Member

    Dec 12, 2012
    I just walked out of a physics test, and one of the questions was to the effect of:

    You charge a parallel plate capacitor to 'Q', disconnect the battery, and double the distance between the plates. How does the potential difference (Voltage) change?

    After some thought, I deduced that if Q remains the same and Q=CV (C=capacitance in farads), then V must double to maintain the same charge Q.

    My reasoning is that by separating the plates, you are doing work on them. This energy must be stored somewhere, and since C cannot increase (In fact it is dropping) then V must increase by doubling.

    The overall energy in the capacitor will increase according to U=C(V^2)/2, so the energy will double after accounting for the drop in capacitance.

    Sound correct?
    If I miss this question I don't have the luxury of an explanation as to why I missed it without going to some inconvenient office hours.
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
    You have done well here and deserve a pint.

    If you wish to do it mathematically , you have not used the fact that the capacitance between two plates is proportional to the area and inversely proportional to the distance apart, the constant being the permittivity.

    C = \frac{{\varepsilon A}}{d}

    Using this to obtain the ratio of capacitance as distance, d doubles

    \frac{{{C_2}}}{{{C_1}}} = \frac{{\frac{{\varepsilon A}}{{2{d_1}}}}}{{\frac{{\varepsilon A}}{{{d_1}}}}} = \frac{1}{2}


    {C_1} = 2{C_2}

    Using energy as an intermediary as you have done, at constant charge, Q

    U = \frac{1}{2}\frac{{{Q^2}}}{C}

    So the ratio of energies is using

    U = \frac{1}{2}\frac{{{Q^2}}}{C}

    \frac{{{U_2}}}{{{U_1}}} = \frac{{\frac{{{Q^2}}}{{2{C_2}}}}}{{\frac{{{Q^2}}}{{4{C_2}}}}} = 2


    {U_2} = 2{U_1}

    Inserting the ratio of energies and using your different energy relation

    \frac{{{U_2}}}{{{U_1}}} = \frac{{\frac{{Q{V_2}}}{2}}}{{\frac{{Q{V_1}}}{2}}} = 2


    {V_2} = 2{V_1}

    This confirms all the ratios you deduced.
  3. WBahn


    Mar 31, 2012
    Very good job!

    Your explanation is well-reasoned and easy to follow.

    About all that could be done more would be to show it mathematically (which studiot has already done).
  4. alfacliff

    Well-Known Member

    Dec 13, 2013
    I believe this is included in the Kasmir (spelling) effect.
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
    No this is just bog standard electric circuit theory in classical physics.

    The casimir effect is a quantum effect.
  6. BillO

    Distinguished Member

    Nov 24, 2008
    An real life application of this result is the condenser microphone.