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(B) The initial capacitance of the parallel plate capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$

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$\frac{4\varepsilon_r}{3\varepsilon_r + 1}$

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(B) The initial capacitance of the parallel plate capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the separation between them. When a dielectric slab of thickness $t = \frac{d}{4}$ and relative permittivity $\varepsilon_r$ is inserted,the new capacitance $C$ is given by the formula: $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{\varepsilon_r}}$ Substituting $t = \frac{d}{4}$ into the equation: $C = \frac{\varepsilon_0 A}{d - \frac{d}{4} + \frac{d/4}{\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{d(3\varepsilon_r + 1)}{4\varepsilon_r}}$ $C = \frac{4\varepsilon_r \varepsilon_0 A}{d(3\varepsilon_r + 1)}$ Since $C_0 = \frac{\varepsilon_0 A}{d}$,we can write: $C = C_0 \left( \frac{4\varepsilon_r}{3\varepsilon_r + 1} \right)$ Therefore,the ratio $\frac{C}{C_0} = \frac{4\varepsilon_r}{3\varepsilon_r + 1}$.

$A$ parallel plate capacitor has a capacitance of $C_0$. If a dielectric slab of relative permittivity $\varepsilon_r$ and thickness equal to one-fourth of the distance between the plates is inserted,the new capacitance is $C$. Then,the ratio $\frac{C}{C_0}$ is:

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