If ${q}_{{f}}$ is the free charge on the capacitor plates and ${q}_{{b}}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates, then bound charge $q_{b}$ can be expressed as 
  • A${q}_{{b}}={q}_{{f}}\left(1-\frac{1}{{k}}\right)$
  • B${q}_{{b}}={q}_{{f}}\left(1-\frac{1}{\sqrt{{k}}}\right)$
  • C${q}_{{b}}={q}_{{f}}\left(1+\frac{1}{\sqrt{{k}}}\right)$
  • D${q}_{{b}}={q}_{{f}}\left(1+\frac{1}{{k}}\right)$
JEE MAIN 2021, Diffcult
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