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A dielectric slab of thickness d is inse...

A dielectric slab of thickness `d` is inserted in a parallel plate capacitor whose negative plate is at `x=0` and positive plate is at `x = 3d`. The slab is equidistant from the plates. The capacitor is given some charge. As one goes from `0` to `3d(1998)`.

A

The magnitude of the electric field remains the same

B

the direction of the electric field remains the same

C

The electrical potential increases continuously

D

The electric potentical increases at first, then decreases and again increases

Text Solution

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The correct Answer is:
B, C
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Knowledge Check

  • A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance

    A
    becomes zero
    B
    remains the same
    C
    decreases
    D
    increases

  • A positive charge q is given to each plate of a parallel plate air capacitor having plate area A and plate separation d, then

    A
    since both the plates are identially charged, therefore, capacitance become equal to zero
    B
    energy stored in the space between the capacitor plates is equal to `(q^(2))/(epsilon_(0)A^(2))`
    C
    no charge appears on inner surface of the plates
    D
    potential difference between the plates is equal to `(2qd)/(epsilon_(0)A)`

  • A parallel plate capacitor is charged by a battery. The battery is removed and a thick glass slab is inserted between the plates. Now,

    A
    The capacity of the capacitor is increased
    B
    The electrical energy stored in the capacitor is decreased
    C
    The potential across the plate is decreased
    D
    The electric field between the plates is decreased.