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Byju's Answer

Standard XIII

Physics

New Capacitance in Dielectric

A parallel- p...

Question

A parallel- plate capacitor is placed in such a way that its plates are horizontal and lower plate is dipped into a liquid of dielectric constant $k$ and density $ρ$ . Charge on plates of capacitor is $Q$ and $- Q$ . [ $+ Q$ is charge on upper plate]

Level of liquid in the space between plates rises.

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Level of liquid in the space between plates falls.

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Net charge on the lower plate is

\frac{- Q}{k}

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Charge induced on the upper surface of liquid is

- Q (1 - k)

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Solution

The correct option is C Net charge on the lower plate is $\frac{- Q}{k}$ .
Due to negative charge induced on upper surface of liquid level of fluid rise.
Induced charge is $- q (1 - \frac{1}{k})$

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Similar questions

Q. A parallel- plate capacitor is placed in such a way that its plates are horizontal and lower plate is dipped into a liquid of dielectric constant

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A parallel- plate capacitor is placed in such a way that its plates are horizontal and lower plate is dipped into a liquid of dielectric constant k and density ρ. Charge on plates of capacitor is Q and −Q. [+Q is charge on upper plate]

The correct option is C Net charge on the lower plate is −Qk.Due to negative charge induced on upper surface of liquid level of fluid rise. Induced charge is −q(1−1k)

A parallel- plate capacitor is placed in such a way that its plates are horizontal and lower plate is dipped into a liquid of dielectric constant $k$ and density $ρ$ . Charge on plates of capacitor is $Q$ and $- Q$ . [ $+ Q$ is charge on upper plate]

The correct option is C Net charge on the lower plate is $\frac{- Q}{k}$ .
Due to negative charge induced on upper surface of liquid level of fluid rise.
Induced charge is $- q (1 - \frac{1}{k})$