A slab of dielectric constant K has same crosssectional area as plates of a parallel plate capacitor

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2. Electric Potential and Capacitance

hard

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.)

A

$\frac{4 KC _{0}}{3+ K }$

B

$\frac{3 KC _{0}}{3+ K }$

C

$\frac{3+ K }{4 KC _{0}}$

D

$\frac{ K }{4+ K }$

(JEE MAIN-2022)

Solution

$x + y +\frac{3 d }{4}= d$

$x + y =\frac{ d }{4}$

$\frac{ A \epsilon_{0}}{ d }= C _{0}$

$\Delta V = Ex +\frac{ E }{ k } \times \frac{3 d }{4}+ Ey$

$=\frac{3 Ed }{4 k }+ E ( x + y )$

$\Delta V = E \left[\frac{3 d }{4 k }+\frac{ d }{4}\right]$

$\Delta V =\frac{\sigma}{\epsilon_{0}}\left[\frac{3 d + dk }{4 k }\right]=\frac{ Qd }{ A \epsilon_{0}}\left[\frac{3+ k }{4 k }\right]$

$\frac{ Q }{\Delta V }= C =\frac{ A \epsilon_{0}}{ d }\left[\frac{4 k }{3+ k }\right]=\frac{4 k C _{0}}{ k +3}$

Standard 12

Physics

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