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(A) The original capacitance of the air-filled parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of constant $K$ is i

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$(K+3) \frac{C}{4}$

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(A) The original capacitance of the air-filled parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.When a dielectric slab of constant $K$ is introduced to fill one-fourth of the area,the capacitor can be considered as two capacitors in parallel.One capacitor has air as the dielectric with area $\frac{3A}{4}$ and plate separation $d$. Its capacitance is $C_1 = \frac{\varepsilon_0 (3A/4)}{d} = \frac{3}{4} \frac{\varepsilon_0 A}{d} = \frac{3C}{4}$.The other capacitor has a dielectric $K$ with area $\frac{A}{4}$ and plate separation $d$. Its capacitance is $C_2 = \frac{K \varepsilon_0 (A/4)}{d} = \frac{K}{4} \frac{\varepsilon_0 A}{d} = \frac{KC}{4}$.Since these two capacitors are in parallel,the equivalent capacitance is $C_{net} = C_1 + C_2$.$C_{net} = \frac{3C}{4} + \frac{KC}{4} = \frac{C}{4}(K+3)$.

$A$ parallel plate capacitor with air as the dielectric has capacitance $C$. $A$ slab of dielectric constant $K$ and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

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