A parallel plate capacitor having a separation between the plates $d$ , plate area $A$ and material with dielectric constant $K$ has capacitance $C_0$. Now one-third of the material is replaced by another material with dielectric constant $2K$, so that effectively there are two capacitors one with area $\frac{1}{3}\,A$ , dielectric constant $2K$ and another with area $\frac{2}{3}\,A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$ then $\frac{C}{{{C_0}}}$ is
  • A$1$
  • B$\frac{4}{3}$
  • C$\frac{2}{3}$
  • D$\frac{1}{3}$
JEE MAIN 2013, Diffcult
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