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Liquid in a capacitor as dielectric (IE Irodov 3.144)

1 vote

A parallel plate capacitor is located horizontally so that one of its plates is submerged into the liquid while the other is over its surface. The permittivity of the liquid is equal to $\epsilon$, its density is equal to $\rho$. To what height will the level of the liquid in the capacitor rise after its plates get a charge of surface charge density $\sigma$?

This is a type of controversial problem, as it contains a different answer by the different author, two different answers provided for this are

$$1) \ h=\dfrac{(\epsilon-1)\sigma^2}{2\epsilon_o\epsilon\rho g}$$ $$2) \ h=\dfrac{(\epsilon^2-1)\sigma^2}{2\epsilon_o\epsilon^2\rho g}$$

Solution for 1 and solution for 2

Please help!

asked Oct 30 in Physics Problems by n3 (318 points)
edited Oct 31 by sammy gerbil
Professor Arora explains both at the  start and end of his video why the method in solution 1 is incorrect.
But book IE Irodov gave the answer for solution 1. Can IE Irodov go wrong!!!
It is difficult to decide between the two methods. I am not entirely convinced by either one.

Probably the answer in Irodov is correct. However, solution 1 provides no explanation to justify why it works. It is quite easy to write down formulas and get the required answer. It is more difficult to show why it must be the correct answer.

Professor Arora provides a clear explanation for his answer (solution 2), which is quite convincing. He also gives a reason why the other solution is incorrect. However, I think he has used the energy method to solve a similar problem in which the capacitor is vertical. If the energy method gives the wrong answer for the horizontal capacitor, why should it give the correct answer when the capacitor is vertical? For the vertical capacitor he explains that the force on the dielectric arises from the fringe field, but for the horizontal capacitor he ignores the fringe field.

It will probably take me a few days more to do enough research and thinking to decide which answer is correct.
Thanks, Sir, for still working on my problem.

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