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Question

A parallel-plate capacitor is located horizontally so that one of its plates is submerged into a liquid while the other is over its surface. The permittivity of the liquid is equal to ϵ. Its density is equal to ρ. To what height will the level of the liquid in the capacitor rise after its plate get a charge of surface density σ ?

A
No change in liquid level
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B
ρ2(21)0ρg
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C
ρ2(21)20ρg2
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D
ρ2(1)0ρg
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Solution

The correct option is C ρ2(21)20ρg2

When the capacitor gets charged electric field will be induced between the plates. Since water is dielectric, charge will gets distributedd in it.
As we know the total charge induced in the dielectric is
σ=σ(11)
Assuming the top plate is positively charged, there will be electric field from top to bottom. Simillarly electric field in water will be from bottom to top.
The electic field due to water will pull the water surface upward due to which the water level rises to some height h
Now calculating net electric field ,
E1=σ20 (downward)
E3=σ20=σ(11)20 (upward)
E2=σ20 (downward)
ENet=(E1+E2)E3=σ20+σ20=σ(11)20
=σ0[112+12]=σ0[12+12]
ENet=σ20[1+1] (downward)
For calculating force,
F=qE As, q=σA
FA=σE
=σ(11)σ2[1+10]
FA=σ220(112)=P (upward)
As this in equilibrium then,
ρgh=σ220(112)
h=σ22ρg0(112)=σ22ρg0(212)
For detailed solution watch the next video.

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