The capacitance of a parallel plate capacitor is \(2.5 \mu \mathrm{F}\). When it is half filled with a dielectric as shown in figure, its capacitance becomes \(5 \mu \mathrm{F}\). The dielectric constant of the dielectric is

Given \(\mathrm{C}=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=2.5 \mu \mathrm{F}\)
When half filled with air,
\(\mathrm{C}_1=\frac{\varepsilon_0(\mathrm{~A} / 2)}{\mathrm{d}}=\frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}\)
\(\left(\because \varepsilon_{\mathrm{r}}=1\right)\)
When half filled with a dielectric,
\(\mathrm{C}_2=\frac{\varepsilon_{\mathrm{r}} \varepsilon_0(\mathrm{~A} / 2)}{\mathrm{d}}=\frac{\varepsilon_{\mathrm{r}} \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}} \text {, }\)
From the figure, it can be seen that \(\mathrm{C}_1\) and \(\mathrm{C}_2\) are in parallel configuration.
\(\begin{array}{ll}
\therefore \quad & \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_1+\mathrm{C}_2 \\
& \text { Given, } \mathrm{C}_{\mathrm{eq}}=5 \mu \mathrm{F} \\
& \Rightarrow 5 \mu \mathrm{F}=\frac{\varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}+\frac{\varepsilon_{\mathrm{r}} \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}} \\
& 5=\frac{2.5}{2}+\varepsilon_{\mathrm{r}} \frac{2.5}{2} \\
& \frac{3.75}{1.25}=\varepsilon_{\mathrm{r}} \\
\therefore \quad & \varepsilon_{\mathrm{r}}=3
\end{array}\)