Two square plates `(lxxl)` and dielectric `(l/2xxt/2xxl)` are arranged as shown in. Find the equivalent capacitance of the structrue.
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Options

A. `(epsilon_(0)A)/t((3K+1)/(2(K+1)))`
B. `(2epsilon_(0)A)/t((K+1)/(K+3))`
C. `(epsilon_(0)A)/t((K+1)/(K+3))`
D. `(epsilon_(0)A)/t((2K+1)/(2K+3))`

Answer

A

Explanation

We can make equivalent circuit of given system in two ways as in (a) and (b). But (b) is wrong and (a) correct.
Here `C_(1)=(epsilon_(0)A//2)/(t//2)=(epsilon_(0)A)/t`
`C_(2)=(Kepsilon(0)A//2)/(t/2)=(KepsilonA)/t` and `C_(3)=(epsilon_(0)A//2)/t=(epsilon_(0)A)/(2t)`

Now, `C_(eq)=(C_(1)C_(2))/(C_(1)+C_(2))+C_(3)`
`=(epsilon_(0)A)/t[(3K+1)/(2(K+1))]`.