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For changing the capacitance of a given parallel 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 $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation
$C ^{\prime}=\frac{3+ K }{4 K } C _{0}$
$C ^{\prime}=\frac{4+ K }{3} C _{0}$
$C ^{\prime}=\frac{4 K }{ K +3} C _{0}$
$C ^{\prime}=\frac{4}{3+ K } C _{0}$
Solution
$C _{0}=\frac{\epsilon_{0} A }{ d }$
$C ^{\prime}= C _{1}$ and $C _{2}$ in series.
i.e. $\frac{1}{ C ^{\prime}}=\frac{1}{ C _{1}}+\frac{1}{ C _{2}}$
$\frac{1}{ C ^{\prime}}=\frac{(3 d / 4)}{\epsilon_{0} KA }+\frac{ d / 4}{\epsilon_{0} A }$
$\frac{1}{ C ^{\prime}}=\frac{ d }{4 \epsilon_{0} A }\left(\frac{3+ K }{ K }\right)$
$C ^{\prime}=\frac{4 K C _{0}}{(3+ K )}$981-s542
