Capacitor partially dipped in water

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Homework Help Overview

The problem involves a parallel plate capacitor partially submerged in a liquid with a specified relative permittivity. The capacitor is charged and then disconnected from the power source, leading to questions about the change in liquid height between the plates and the forces acting on the liquid.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the heights of the liquid before and after charging the capacitor, questioning the clarity of the problem statement and the implications of the energy changes involved.

Discussion Status

Some participants are seeking clarification on the problem's details and the physical principles at play, particularly regarding the forces acting on the liquid and the reasons for changes in energy and height. There is an ongoing exploration of interpretations and assumptions without a clear consensus.

Contextual Notes

Participants note potential ambiguities in the problem statement, such as the definition of the ratio of heights and whether the liquid is confined. These aspects are under discussion but not resolved.

rohith291991
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Homework Statement


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...

Homework Equations


Adgh=F...for equilibrium...

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... won't 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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please someone help me out...
 
rohith291991 said:
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...


I think no one has dealt with this problem because it is unclear what the question is. What ratio is being taken here? Is it the ratio of the heights of the liquid between the plates before and after the capacitor has been charged? Is the liquid confined in any way? It reads as if either you omitted some details or the problem isn't well-defined.
 
no the capacitor is charged and the key remains closed...let the height of the water be h1...then the key is opened(the capacitor is no longer connected in the circuit)...let the height now be h2... the ratio h2/h1 is asked... i guess this is quitw clear...my doubt is y wud the heights change at all??
 
also i don't know what u mean by confined...there is a liquid surface that's all...it doesn't matter whether its the surface of a sea or water in a cup...i don't understand y it shud matter...the question seems to be clear enough...
 

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