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  4. 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 thev dielectric slab is (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:

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Q. 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 thev 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:

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Solution:

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$C _{0}=\frac{ \in _{0} A }{ d }$
$C ^{\prime}= C _{1}$ and $C _{2}$ in series.
i.e. $\frac{1}{C^{\prime}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}$
$\frac{1}{ C ^{\prime}}=\frac{(3 d / 4)}{\epsilon_{0} KA }+\frac{ d / 4}{\epsilon_{0} A }$
$\frac{1}{ C ^{\prime}}=\frac{ d }{4 \in_{0} A }\left(\frac{3+ K }{ K }\right)$
$C ^{\prime}=\frac{4 K C _{0}}{(3+ K )}$