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  4. A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness (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 C0= capacitance of capacitor with air as medium between plates.)

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Q. A slab of dielectric constant $K$ has the same cross-sectional 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 MainJEE Main 2022Electrostatic Potential and Capacitance

Solution:

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$ x+y+\frac{3 d}{4}=d $
$ x + y =\frac{ d }{4} $
$ \frac{ A \in_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 \in_0}\left[\frac{3+ k }{4 k }\right]$
$ \frac{ Q }{\Delta V }= C =\frac{ A \in_0}{ d }\left[\frac{4 k }{3+ k }\right]=\frac{4 kC _0}{ k +3}$