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 $\frac{3}{4} \mathrm{~d}$, where ' $\mathrm{d}$ ' is the separation
between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (C0 ) is given by the following relation :
Correct Option: 3,
$C_{0}=\frac{\epsilon_{0} A}{d}$
$\mathrm{C}^{\prime}=\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ in series.
i.e. $\frac{1}{C^{\prime}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}$
$\frac{1}{\mathrm{C}^{\prime}}=\frac{(3 \mathrm{~d} / 4)}{\epsilon_{0} \mathrm{KA}}+\frac{\mathrm{d} / 4}{\epsilon_{0} \mathrm{~A}}$
$\frac{1}{C^{\prime}}=\frac{d}{4 \in_{0} A}\left(\frac{3+K}{K}\right)$
$C^{\prime}=\frac{4 K C_{0}}{(3+K)}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.