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Derive expression for the capacitance of...

Derive expression for the capacitance of the parallel plate capacitor,

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Explain the effect of dielectric capacitance of parallel plate capacitor and obtain the formula of dielectric constant.

Write the capacitance of parallel plate capacitor with medium of dielectric of dielectric constant K.

Knowledge Check

  • Which of the following one, the capacitance of parallel plate capacitor does not depend ?

    A
    On the area of the plate
    B
    On the distance between two plates
    C
    On the charge of the plate
    D
    On the shape of the plate

  • A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates force of attraction between the plates of the parallel plate air capacitor is :

    A
    `(C^(2)V^(2))/(2d^(2))`
    B
    `(C^(2)V^(2))/(2d)`
    C
    `(CV^(2))/(2d)`
    D
    `(CV^(2))/(d)`

  • In the beginning the space between the plates of a parallel plate capacitor contains air and thereafter it is filled up with a medium of dielectric constant K. Then ......

    A
    The electric field and the capacitance become K times.
    B
    The electric field becomes `(1)/(K)` times and the capacitance becomes K times.
    C
    The electric field becomes K times and the capacitance becomes `(1)/(K)` times.
    D
    The electric field and the capacitance become `(1)/(K)` times

    • Similar Questions

    Explore conceptually related problems

    An ideal cell is connected across a capacitor as shown in figure. The initial separation between the plates of a parallel plate capacitor is d. The lower plate is pulled down with a uniform velocity v. Neglect the resistance of the cirucit. Then the variation of charge on capacitor with time is given by

    Obtain the formula for the effective capacitance of the parallel combination of different n capacitors.

    What is the area of the plates of a 2 F parallel plate capacitor given that the separation between the plates is 0.5 cm ?

    Prove that the force acting on one plate due to the other in a parallel plate capacitor is F =(1)/(2) (CV^(2))/(d) .

    What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

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