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A parallel plate condenser is filled wit...

A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is `A "metre"^(2)` and the separtion is `t` metre. The dielectric constants are `K_(1)` and `K_(2)`, respectively. Its capacitance in farad will be

A

`(epsilon_(0)A)/(t) (K_(1)+K_(2))`

B

`(2epsilon_(0)A)/(t) (K_(1)+K_(2))`

C

`(epsilon_(0)A)/(t).(K_(1)+K_(2))/(2)`

D

`(epsilon_(0)A)/(t).(K_(1)+K_(2))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
The given arrangement is equivalent to two capacitors in parallel. In each capacitor the area of the plate will
`:.` Equivalent capacitance `C' = C_(1) + C_(2)`
`= (K_(1) epsilon_(0)A//2)/(t) + (K_(2) epsilon_(0)A//2)/(t) = ((K_(1)+K_(2))epsilon_(0)A)/(2t)`
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Knowledge Check

  • A parallel plate capacitor is filled with dielectrics. What is capacitance?

    A
    `(2K_(1)K_(2))/(K_(1) + K_(2))`
    B
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  • A capacitor of plate area A and separation d is filled with two dielectrics of dielectric constant K_(1) = 6 and K_(2) = 4 . New capacitance will be

    A
    `4 (A epsilon_(0))/(d )`
    B
    ` 4.8 (A epsilon_(0))/(d )`
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    ` 5 (A epsilon_(0))/(d )`
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    D
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