Find the capacitance between the terminals `A` and `C_(3)` If `epsilon_(r) = 2`

Explanation

The give system is broken into three capacitors `C_(1),C_(2)` and `C_(3) as shown in

`C_(1)=(epsilonA//2)/(d//2)`
`C_(2)=(epsilon_(0)epsilonA//2)/(d//2)`
where `C_(3)=(epsilon_(0)A//2)/d`
`C_(1)` and `C_(2)` are in series, and this series combination is in parallel with `C_(3)`. Then the equivalent capacitance between the terminals `A` and `B` is
`C_(AB)=(C_(1)C_(2))/(C_(1)+C_(2))+C_(3)=(epsilon_(0)A)/d((epsilon_(r))/(epsilon+1)+1/2)` (where `epsilon_(r)=2`)
`:. C_(AB)=(7epsilon_(0)A)/(6d)`