Question

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.

Answer 1

When the capactior which is immersed in water is connected to a constant voltage source, it gets charged. Suppose σ₀ is the free charge density on the condenser plates. Because water is a dielectric, bound charges also appear in it. Let σ' be the surface density of bound charges. (Because of homogeneity of the medium and uniformity of the field when we ignore edge effects no volume density of bound charges exists.) The electric field due to free charges only  that due to bound charges is σ'/ε₀ and the total electric field is Recalling that the sign of bound charges is opposite o f the free charges, we have 

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 attracting the bound charges to the near by plates. This force is

The factor 1/2 needs an explanation. Normally the force on a test charge is qE in an electric field E. But if the charge itself is produced by the electric filed then the force must be constructed bit by bit and is

This factor of 1/2 is well known. For example the energy of a dipole of moment  in an electric field vector E₀ is  while the energy per unit volume o f a linear dielectric in an electric field is is  the polarization vector (i.e. dipole moment per unit volume). Now the force per unit area manifests itself as excess pressure of the liquid.