A cylindrical capacitor conneced to a `dc` voltage source `V` touches the surface of water with its end (Fig) The sepration `d` between the capacitors electrodes is substantially less than their mean radius. Find a height `h` to which the water level in the gap will rise. The, capilary effects are to be negelected.

Explanation

If `C_(0)` is the inital capacitance of the condenser before water rises in it then
`U_(i) = (1)/(2) C_(0) V^(2)`, where `C_(0) = (epsilon_(0) 2l pi R)/(d)`
(`R` is the mean radius and `l` is the length of the capacitor plates.)
Suppose the liquid rises to a height `h` in it. Then the capacitance of the condenser is
`C = (epsilon epsilon_(0)h 2pi R)/(d) + (epsilon(l-h) 2pi R)/(d) = (epsilon_(0) 2pi R)/(d) (l + (epsilon - 1) h)`
and energy of the capacitor and the liquid (inculding both gravitational and electrostatic condtributions) is epsilon `(1)/(2) (epsilon_(0) 2pi R)/(d) (l + (epsilon - 1)h) V^(2) + rho g (2pi R hd) (h)/(2)`
If the capacitor were not connected to a battery this energy would have to be minimized. But the capacitor is connected to the battery and in effect the potential energy of the capacitor and the liquid increases by
`delta h ((epsilon_(0) 2pi R)/(2d) (epsilon - 1) V^(2) + rho g (2pi Rd) h)`
and that of the cell diminishes by the quantity `A_(cell)` which is the product of charge flown and `V`
`delta h (epsilon_(0) (2pi R))/(d) (epsilon - 1) V^(2)`
In equilbrium, the two must balance, so
`rho g dh = (epsilon_(0) (epsilon - 1) V^(2))/(2d)`
Hence `h = (epsilon_(0) (epsilon - 1) V^(2))/(2rho g d^(2))`