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, Diffcult
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