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What is the force of attraction between ...

What is the force of attraction between the two plates of a parallel plate capacitor? Assume that, area of each plate of the capacitor is A and one plate is charged with +Q and the other with -Q.

Text Solution

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Intensity of electric field at a point near a charged plate having surface density of charge `sigma` is,
`E = (sigma)/(2in_0)`
Now, `sigma = (Q)/(A)`
`:. E = (Q)/(2A in_0)`
Magnitude of the charge on the other plate of the capacitor = Q.
So, the force experienced by the plate is,
`F = QE = Q*(Q)/(2A in_0)= (Q^2)/(aA in_0)`.
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Knowledge Check

  • The capacitance of a parallel plate capacitor depends on

    A
    the thickness of the plates
    B
    the charge accumulated on the plates
    C
    the potential difference between the two plates
    D
    the distance between the two plates

  • If the distance between the plates of a parallel plate capacitor is doubled while the battery is kept connected, then

    A
    the charge stored in the capacitor will be doubled
    B
    the battery will absorb some amount of energy
    C
    the electric field between the two plates will be halved
    D
    work will be done by an external agent on the two plates

  • The electrostatic force between the metal plates of an isotated parallel plate capacitor C having a charge Q and area A, is

    A
    proportional to the square root of the distance between the plates
    B
    linearly proportional to the distance between the plates
    C
    independent of the distance between the plates
    D
    inversely proportional to the distance between the plates

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