For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $\mathrm{K}$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} \mathrm{~d}$, where ' $\mathrm{d}$ ' is the separation between the plates of parallel plate capacitor. The new capacitance $\left(C^{\prime}\right)$ in terms of original capacitance $\left(\mathrm{C}_{0}\right)$ is given by the following relation :
Correct Option: , 3
(3)
$\mathrm{C}_{0}=\frac{\epsilon_{0} \mathrm{~A}}{\mathrm{~d}}$
$\mathrm{C}^{\prime}=\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ in series.
i.e. $\frac{1}{\mathrm{C}^{\prime}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$
$\frac{1}{C^{\prime}}=\frac{(3 \mathrm{~d} / 4)}{\epsilon_{0} \mathrm{KA}}+\frac{\mathrm{d} / 4}{\epsilon_{0} \mathrm{~A}}$
$\frac{1}{C^{\prime}}=\frac{d}{4 \epsilon_{0} A}\left(\frac{3+K}{K}\right)$
$\mathrm{C}^{\prime}=\frac{4 \mathrm{KC}_{0}}{(3+\mathrm{K})}$
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