Capacitance and electric fields question:

Discussion in 'Homework Help' started by daviddeakin, Mar 31, 2012.

  1. daviddeakin

    Thread Starter Active Member

    Aug 6, 2009
    This is a question from a past exam, that I'm trying to figure out for the purposes of revision:

    Consider a thin parallel-plate capacitor of an area of 2x10^-4 metres squared, with a plate separation of 0.1 mm. The left quarter of the capacitor is filled with a dielectric material with a relative dielectric permittivity εr1=6, and the
    right three-quarters, with another material with εr2=3. The charge on the top and bottom plates is +120 pC and – 120 pC, respectively.

    Calculate the electric flux densities and the electric field in the right and left parts of the device. State any assumptions used.

    Answer: Assume the charge is distributed uniformly. The flux density is therefore uniform:
    D = Q/A = 120pC / 2x10^-4 = 0.6C/m^2.

    Dividing this by the permittivity gives the electric field:
    E= D/(εr x εo) = 0.6/(6 x 8.85x10^-12) = 11.3 GV/m (left quarter)
    E= D/(εr x εo) 0.6/(3 x 8.85x10^-12) = 22.6 GV/m (right quarter)

    (b) Calculate the potential difference between the plates and the capacitance of the device.

    Here is where I am stuck. I can figure out the capacitance easily enough from C = εA/d (treating the two parts of the capacitor as separate capacitors in parallel).
    But how do I find the PD? I would have expected just to multiply the number of volts-per-metre by the separation, but in this case the E-field is different in each part of the capacitor which would give different PDs, which is clearly wrong! :confused: