If ${\mathrm{q}}_{\mathrm{f}}$ is the free charge on the capacitor plates and ${\mathrm{q}}_{\mathrm{b}}$ is the bound charge on the dielectric slab of dielectric constant $\mathrm{k}$ placed between the capacitor plates, then bound charge ${\mathrm{q}}_{\mathrm{b}}$ can be expressed as :

• A

  ${\mathrm{q}}_{\mathrm{b}}={\mathrm{q}}_{\mathrm{f}}\left(1-\frac{1}{\sqrt{\mathrm{k}}}\right)$
• Correct answer
  B

  ${\mathrm{q}}_{\mathrm{b}}={\mathrm{q}}_{\mathrm{i}}\left(1-\frac{1}{\mathrm{k}}\right)$
• C

  ${\mathrm{q}}_{\mathrm{b}}={\mathrm{q}}_{\mathrm{f}}\left(1+\frac{1}{\sqrt{\mathrm{k}}}\right)$
• D

  ${\mathrm{q}}_{\mathrm{b}}={\mathrm{q}}_{\mathrm{f}}\left(1+\frac{1}{\mathrm{k}}\right)$

Hence, the correct answer is option: 'B'.