A parallel-plate capacitor was lowered into water in a horizontal position, with water filling up the gap between the plates `d = 1.0 mm` wide. Then a constant voltage `V = 500 V` was applied to the capacitor. Find the water pressure increment in the gap.

Explanation

when the capacitor which is immersed in water is connected to a constant voltage source, it gets chagred. Suppose `sigma_(0)` is the free charge density on the condenser plates. Because water is a dielectric, bound charges also appear in it. Let `sigma'` be the surface dnesity of bound charges. (Because of homogenelt of the medium and uniformly fo the field when we ignore edge effects no volume density of bound charges exists.) The electric fiedl due to free charges only `(sigma_(0))/(epsilon_(0))`, that due to bound charges is `(sigma')/(epsilon_(0))` and the total electric field is `(sigma_(0))/(epsilon epsilon_(0))`, Recalling that the sign of bound charges is opposite of the free charges, we have
`(sigma_(0))/(epsilon epsilon_(0)) = (sigma_(0))/(epsilon_(0)) - (sigma')/(epsilon_(0))` or, `sigma' = ((epsilon - 1)/(epsilon)) sigma_(0)`
Because of the field that exists due to the free charges (not the total field, the field due to the bound charges must be excluded for this purpose as they only give rise to self energy effects), there is a force attractaing the bound charges to the near by plates. This force is
`(1)/(2) sigma' (sigma_(0))/(epsilon_(0)) = ((epsilon - 1))/(2 epsilon epsilon_(0)) sigma_(0)^(2)` per unit area.
The factor `(1)/(2)` needs an explantantion. Normally the force on a test charge is `qE` in an electric field `E`. But if the charge itself is produced by the electric field then the force must be constructed bit by bit and is
`F = int_(0)^(E) q(E') dE'`
if `q (E') prop E'` then we get
`F = (1)/(2) q (E) E`
This factor of `(1)/(2)` is well knwon. For example the energy of a dipole of momet `vec(p)` in an electric firld `vec(E_(0))` is `-vec(p). vec(E_(0))` while the energy per unit volume of a linear dielectric in an electric field is `- (1)/(2) vec(p). vec(E_(0))` where `vec(p)` is the plarization vector (i.e., dipole moment per unit volume). Now the force per unit area manifests itself as excess pressure of the liquid.
Nothing that `(V)/(d) = (sigma_(0))/(epsilon epsilon_(0))`
We get `Delta p = (epsilon_(0) epsilon (epsilon - 1) V^(2))/(2a^(2))`
Substituing, using `epsilon = 81` for water, gives `Delta p = 7.17 k Pa = 0.07 atm`.