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
JEE MAIN 2013, Diffcult
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$C_{0}=\frac{k \epsilon_{0} A}{d}$
$C=\frac{k \epsilon_{0} 2}{3 d}+\frac{2 k \epsilon_{0} A}{3 d}=\frac{4}{3} \frac{k \epsilon_{0} A}{d}$
$\therefore$ $\frac{{\text{c}}}{{{{\text{C}}_0}}} = \frac{{\frac{{4k{\varepsilon _0}A}}{{3d}}}}{{\frac{{k{\varepsilon _0}A}}{d}}} = \frac{4}{3}$
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