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.)

• Correct answer
  A

  $\frac{4K{C}_0}{3+K}$
• B

  $\frac{3K{C}_0}{3+K}$
• C

  $\frac{3+K}{4K{C}_0}$
• D

  $\frac{K}{4+K}$

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