A large vessel is filled with a liquid. Two vertical plates touch the surlace of the liquid (Fig. 25.7). The dimensions of the plates are a and b, the distance between them is d. The plates have been charged by applying a voltage Po and then disconnected from the voltage source. To what height will the liquid rise? Ignore capillary effects.

Explanation

When the liquid flows through a strongly non-uniform electric field close to the edges of the plates, it is polarized and drawn into the space between the plates. Since the charge on the plates remains unchanged in the process, and the capacitance of the capacitor increases, this is accompanied by a decrease in the energy of the field. This decrease is compensated by the increase in the potential energy of the column of líquid held between the plates. From the law of conservation of energy
`(q^(2))/(2C_(0))=(q^(2))/(2C)+(mgh)/(2)`
Here `C_(0)=(epsi_(0ab))/(d),C=(epsi_(0)b)/(d)[a+(epsi-1)h]` (see Problem 25.5) `m= rho bhd`. Substituting these values into the first formula we obtain, after some simplifications,
`(epsi-1)q^(2)=epsi _(0)rho g a b^(2) [a+(epsi-1)h]`
Express the charge on the plates in terms of the potential,
`q=varphi_(0)C_(0)= epsi_(0)varphi_(0)ab//d`
After simplifications we obtain a quadratic equation
`h^(2)+(a)/(epsi-1)h-(epsi_(0)a varphi_(0)^(2))/(rho g d^(2))=0`
Solving it we obtain the height the liquid rises.