A parallel plate capacitor is located horizontally, so that one of its plates is submerged into liquid while the other is over the surface. The dielectric constant of the liquid is equal to `k`. Its density is equal to `rho`. To what height will the level of the liquid in the capacitor rise after its plates gets a charge of surface density `sigma` ?
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Explanation

One way of doing this problem will be exactly as in previous case so let us try an alternative method based on energy. Suppose the liquid rises by a distance `h`. Then let us calculate the extra energy of the liquid as a sum of polarization energy and the ordinary gravitional energy. Teh latter is
`(1)/(2) h . rho g . Sh = (1)/(2) rho gS h^(2)`
If `sigma` is the free charge surface density on the plane, the bound charge density is, from the previous problem,
`sigma' = (epsilon - 1)/(epsilon) sigma`
This is also the vol,ume density of induced dipole moment i.e, Polarization Then the energy is, as before
`- (1)/(2) . sigma' E_(0) = (-1)/(2) sigma' (sigma)/(epsilon_(0)) = (-(epsilon - 1) sigma^(2))/(2epsilon_(0) epsilon)`
and the total polarization energy is
`-S (a + h) ((epsilon - 1) sigma^(2))/(2epsilon_(0) epsilon)`
Then, total energy is
`U (h) = -S (a + h) ((epsilon - 1) sigma^(2))/(2 epsilon_(0) epsilon) + (1)/(2) rho g Sh^(2)`
The actual height to which the liquid risea is detemined from the formula
`(dU)/(dh) = U' (h) = 0`
This gives `h = ((epsilon - 1) sigma^(2))/(2 epsilon_(0) epsilon rho g)`.