For changing the capacitance of a given parallel plate capacitor,
Question:

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 :

1. (1) $\mathrm{C}^{\prime}=\frac{3+\mathrm{K}}{4 \mathrm{~K}} \mathrm{C}_{0}$

2. (2) $\mathrm{C}^{\prime}=\frac{4+\mathrm{K}}{3} \mathrm{C}_{0}$

3. (3) $\mathrm{C}^{\prime}=\frac{4 \mathrm{~K}}{\mathrm{~K}+3} \mathrm{C}_{0}$

4. (4) $\mathrm{C}^{\prime}=\frac{4}{3+\mathrm{K}} \mathrm{C}_{0}$

Correct Option: , 3

Solution:

(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. $\frac{1}{\mathrm{C}^{\prime}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$

$\frac{1}{C^{\prime}}=\frac{(3 \mathrm{~d} / 4)}{\epsilon_{0} \mathrm{KA}}+\frac{\mathrm{d} / 4}{\epsilon_{0} \mathrm{~A}}$

$\frac{1}{C^{\prime}}=\frac{d}{4 \epsilon_{0} A}\left(\frac{3+K}{K}\right)$

$\mathrm{C}^{\prime}=\frac{4 \mathrm{KC}_{0}}{(3+\mathrm{K})}$