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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

k

and density

ρ

. Charge on plates of capacitor is

Q

and

- Q

. [

+ Q

is charge on upper plate]

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

K

and density

ρ

. Each plate has an area

A

. The plates are now connected to a battery which supplies a positive charge of magnitude

+ Q

to the upper plate. Find the rise in the level of the liquid in the space between the plates.

Q. Assertion: Charges are given to plates of two plane parallel plate capacitors

C_{1}

and

C_{2}

(such that

C_{2} = 2 C_{1}

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K

is pressed to complete the circuit. Finally the net charge on upper plate and net charge the circuit. Finally the net charge on upper plate and net charge on lower plate of capacitor

C

is positive.
Reason : In a parallel plate capacitor both plates always carry equal and opposite charge.

Q. On your first day as an electrical technician you are asked to determine the leakage current in a capacitor. The capacitor consists of two parallel-plates. The plates have equal and opposite charge

Q

. The dielectric has dielectric constant

K

and a resistivity

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. The plates have area

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and separation between the two plates is

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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 $- q (1 - \frac{1}{k})$