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For the congiuration of mede of permitti...

For the congiuration of mede of permittiveties `epslon_(0), epsilon,` and `epsilon_(0)` between parallel plates each of area `A`, as shown in the equivalent capacitance is .
.

A

`epsilon_(0)A/d`

B

`epsilonepsilon_(0)A//d`

C

` (epsilonepsilon_(0)A)/(d(epsilon+epsilon_(0)))`

D

`(epsilonepsilon_(0)A)/((2epsilon+epsilon_(0))d)`.

Text Solution

Verified by Experts

The correct Answer is:
D
`C_(eq)=(epsilon_(0))/(d/K_(1)+d/K_(2)+d/K_(3)`
Hence, `K_(1)=K_(3)=1, K_(2)epsilon_(0)`.
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