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Finding pressure on a dielectric between a capacitor

  1. Dec 4, 2015 #1

    Titan97

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    Gold Member

    1. The problem statement, all variables and given/known data
    A parallel plate capacitor was lowered into water in a horizontal position, with water filling up the gap between the plates (gap d=1.0mm). Then, a constant voltage V=200volt is applied to the capacitor. Find the increment of pressure in the water between the gap.

    2. Relevant equations
    F=qE

    3. The attempt at a solution
    IMG_20151204_150222_024.JPG

    Let the slab represent water between the capacitor plates. On the left surface, the field due to the left plate is ##\frac{\sigma}{2\epsilon_0}## along negative x-axis and that due to the right plate is also ##\frac{\sigma}{2\epsilon_0}## along the same direction. Total field on one surface is ##\frac{\sigma}{\epsilon_0}##.
    Same force acts in opposite direction on the surface on the right.
    This tends to expand the dielectric.

    Let induced charge density on water be ##\sigma'##

    Therefore, $$F=qE=\sigma'A\frac{\sigma}{\epsilon_0}$$
    And $$P=\sigma'\frac{\sigma}{\epsilon_0}$$
    But answer given is $$P=\frac{1}{2}\sigma'\frac{\sigma}{\epsilon_0}$$

    Where does the 1/2 come from?
     
  2. jcsd
  3. Dec 7, 2015 #2
    The contribution to the total electric field on the surface due to the plate dipped in water is reduced k times as a dielectric that is water is present in between.
     
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