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The separation between two plates of a p...

The separation between two plates of a parallel plate condenser is filled with two dielectric media as shown in figure . The ratio of its capacities , with and without dielectric , is

A

`K_(1)(K_(1)+K_(2))`

B

`2K_(1)K_(2)//(K_(1)+K_(2))`

C

`(K_(1)+K_(2))//2`

D

`K_(2)(K_(1)+K_(2))`

Text Solution

Verified by Experts

The correct Answer is:
b

In this arrangement , separation/thickness of `K_(1)`= thickness of `K_(2)=(d)/(2)`
Without dielectric , `C_(0)=(epsilon_(0)A)/(d)`
The two capacitors formed with `K_(1)andK_(2)` are in series
`C_(1)=(K_(1)epsilon_(0)A)/(d//2)=2K_(1)*(epsilon_(0)A)/(d)=2K_(1)C_(0)`
`C_(2)=(K_(2)epsilon_(0)A)/(d//2)=2K_(2)*(epsilon_(0)A)/(d)=2K_(2)C_(0)`
In series combination,
`C_(eq)=(C_(1)C_(2))/((C_(1)+C_(2)))=((2K_(1)C_(0))(2K_(2)C_(0)))/(2K_(1)C_(0)+2K_(2)C_(0))`
`C_(eq)=(4K_(1)K_(2)C_(0)^(2))/(2C_(0)(K_(1)+K_(2)))=(2K_(1)K_(2)*C_(0))/((K_(1)+K_(2)))`
`(C_(eq))/(C_(0))=(2K_(1)K_(2))/((K_(1)+K_(2)))`
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Knowledge Check

  • A dielectric introduced between the plates of parallel plate condenser

    A
    decreases the electric field between the plates
    B
    decreases the capacity of the condenser
    C
    increases the charge stored in the condenser
    D
    increases the capacity of the condenser
  • A dielectric introduced between the plates of parallel plate condenser

    A
    decreases the electric field between the plates
    B
    decreases the capacity of the condenser
    C
    increases the charge stored in the condenser
    D
    increases the capacity of the condenser
  • The capacity of a parallel plate condenser filled with material of dielelctric constant 8 is 16 muF . Its capacity, if the dielectric is removed , will be

    A
    `1 muF`
    B
    `2 muF`
    C
    `4 muF`
    D
    `0.5 muF`
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