Finding pressure on a dielectric between a capacitor

[Titan97]

Homework Statement

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.

Homework Equations

F=qE

The Attempt at a Solution

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?

 

[zabini]

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.