A parallel plate capacitor having crosssectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of same area but thickness $d/2$ is inserted between the plates as shown in figure having dielectric constant $K (=4) .$ The ratio of new capacitance to its original capacitance will be,
  • A$4:1$
  • B$2:1$
  • C$8:5$
  • D$6:5$
NEET 2020, Medium
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