A slab of dielectric constant $K$ has the same crosssectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}\,d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be.(Given $C _{0}=$ capacitance of capacitor with air as medium between plates.)
JEE MAIN 2022, Diffcult
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$x + y +\frac{3 d }{4}= d$
$x + y =\frac{ d }{4}$
$\frac{ A \epsilon_{0}}{ d }= C _{0}$
$\Delta V = Ex +\frac{ E }{ k } \times \frac{3 d }{4}+ Ey$
$=\frac{3 Ed }{4 k }+ E ( x + y )$
$\Delta V = E \left[\frac{3 d }{4 k }+\frac{ d }{4}\right]$
$\Delta V =\frac{\sigma}{\epsilon_{0}}\left[\frac{3 d + dk }{4 k }\right]=\frac{ Qd }{ A \epsilon_{0}}\left[\frac{3+ k }{4 k }\right]$
$\frac{ Q }{\Delta V }= C =\frac{ A \epsilon_{0}}{ d }\left[\frac{4 k }{3+ k }\right]=\frac{4 k C _{0}}{ k +3}$
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