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The capacitance of a parallel plate cond...

The capacitance of a parallel plate condenser is `C_(0)` If a dielectric of relative permittivity `epsilon_(r)` and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of `(C)/(C_(0))` will be -

A

`(5epsi_(r))/(4epsi_(r)+1)`

B

`(4epsi_(r))/(3epsi_(r)+1)`

C

`(3epsi_(r))/(2epsi_(r)+1)`

D

`(2epsi_(r))/(2epsi_(r)+1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`C_(o)=(epsi_(o)A)/(d),C=(epsi_(o)A)/(d-(d)/(4)+(d)/(4epsi_(r)))=(epsi_(0)A)/((3d)/(4)-(d)/(4epsi_(r)))`
`=((4epsi_(r))/(3epsi_(r))+1)C_(0)`
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Knowledge Check

  • To reduce the capacity of a parallel plate condenser, separation between the plates is

    A
    reduced and area of the plates decreased
    B
    decreased and area of the plates increased
    C
    increased and area of the plates decreased
    D
    increased and area of the plates increased

  • The capacitance of a parallel plate condenser with a separation of 4 mm between the plates is 7muF . If a mica sheet (K=6) of thickness 2 mm and of the same area is introduced betwenn the plates, its capacitance will be

    A
    `6muF`
    B
    `9muF`
    C
    `12muF`
    D
    `15muF`

  • The capacity of a parallel plate condenser is C. When the distance between the plates is halved, its capacity is

    A
    2 C
    B
    C
    C
    0.25 C
    D
    0.2 C

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