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

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$ε_{0} h ρ g$

$\frac{ε_{0}}{2} h ρ g$

$2 ε_{0} h ρ g$

$2 ε_{0} ρ g$

answer is C.

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Detailed Solution

Indenced surface charge $σ_{i} = σ (1 - \frac{1}{k})$
If liquid level used up by h

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$\Rightarrow F_{u p} = m g = (h s p) g$
$q E = h s p g$

$σ_{i} s [2 x \frac{σ}{2 \in_{0}} - \frac{σ_{i}}{2 \in_{0}}] = h s p g$

$σ (1 - \frac{1}{K}) [2 σ - σ (1 - \frac{1}{K})] = 2 \in_{0} h p g$

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