Question 102 of 180

For changing the capacitance of a given paralle plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\dfrac{3}{4}d$, where $d$ is the separation between the plates of parallel plate capacitor. The new capacitance ${C}^{'}$ in terms of original capacitance ${C}_{0}$ is given by the following relation :

A${C}^{'}=\dfrac{3+K}{4K}{C}_{0}$B${C}^{'}=\dfrac{4+K}{3}{C}_{0}$C${C}^{'}=\dfrac{4K}{K+3}{C}_{0}$D${C}^{'}=\dfrac{4}{3+K}{C}_{0}$

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Answer: (C) ${C}^{'}=\dfrac{4K}{K+3}{C}_{0}$

${C}_{0}=\dfrac{{ϵ}_{0}A}{d}$
${C}^{'}={C}_{1}$ and ${C}_{2}$ in series.
i.e. $\dfrac{1}{{C}^{'}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}$
$\dfrac{1}{{C}^{'}}=\dfrac{\left(3d/4\right)}{{ϵ}_{0}KA}+\dfrac{d/4}{{ϵ}_{0}A}$
$\dfrac{1}{{C}^{'}}=\dfrac{d}{4{ϵ}_{0}A}\left(\dfrac{3+K}{K}\right)$
${C}^{'}=\dfrac{4K{C}_{0}}{\left(3+K\right)}$