(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A parallel plate capacitor is dipped in a liquid of relative permittivity k horizontally, so that the upper plate remains outside. A source of emf V volt is used to charge the capacitor through a key K. The key K is first closed and then opened. The ratio of respective heights of the level of the liquid from the lower plate is....

2. Relevant equations

Adgh=F....for equilibrium...

3. The attempt at a solution

i got the first height as h=(((K^2)-1)q^2)/(2(A^2)(K^2)epsilon0dg)

where q is the charge on the capacitor, A is the area of each plate, d is the density...

now my doubt is how to find the force acting on the water in the second case... wont it be the same as the first case???...why should the height change at all???

i saw the solution and it goes like this.. First the potential energy as a function of h was found and the negative of its derivative wrt h gave the force.....but why does the energy vary??/ in fact it was given that the energy in the second case is less than in the first case...y is this so?? how is the energy lost??

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# Homework Help: Capacitor partially dipped in water

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