Find the capacitance between A and B if three dielectric slabs of dielectric constants `K_(1) area `A _(2)` and thickness `d_(2)`) are inserted between the plates of a parallel late capacitor of plate area A. (Given that the distance between the two plates is `d_(1)=d_(2)+d_(2)`.)
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Explanation

It is equivalent to `C=C_(1)+(C_(2)C_(3))/(C_(2)+C_(3))`
`C=(A_(1)K_(1)epsi_(0))/(d_(1)+d_(2))=((A_(2)K_(2)epsi_(0))/(d_(1)).(A_(2)K_(3)epsi_(0))/(d_(2)))/((A_(2)K_(2)epsi_(0))/d_(1)+(A_(2)K_(3)epsi_(0))/(d_(2)))`
`=(A_(1)K_(1)epsi_(0))/(d_(1)+d_(2))=(A_(2)^(2)K_(2)K_(3)epsi_(0)^(2))/(A_(2)K_(2)epsi_(0)d_(2)+A_(2)K_(3)epsi_(0)d_(1)))`
`=(A_(1)K_(1)epsi_(0))/(d_(1)+d_(2))+(A_(2)K_(2)K_(3)epsi_(0))/(K_(2)d_(2)+K_(3)d_(1))`