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An air-filled parallel plate capacitor has a capacity $2 \mathrm{pF}$, The separation of the plates is doubled and the interspace between the plates is filled with dielectric material, then the capacity is increased to $6 \mathrm{pF}$, The dielectric constant of the material is
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6
$$
\begin{aligned}
& \mathrm{C}=\frac{\mathrm{kA} \varepsilon_0}{\mathrm{~d}} \\
& \therefore \frac{\mathrm{C}_2}{\mathrm{C}_1}=\frac{\mathrm{k}_2}{\mathrm{k}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2} \\
& \therefore \frac{6}{2}=\frac{\mathrm{k}_2}{1} \cdot \frac{1}{2} \\
& \therefore \mathrm{k}_2=6
\end{aligned}
$$
\begin{aligned}
& \mathrm{C}=\frac{\mathrm{kA} \varepsilon_0}{\mathrm{~d}} \\
& \therefore \frac{\mathrm{C}_2}{\mathrm{C}_1}=\frac{\mathrm{k}_2}{\mathrm{k}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2} \\
& \therefore \frac{6}{2}=\frac{\mathrm{k}_2}{1} \cdot \frac{1}{2} \\
& \therefore \mathrm{k}_2=6
\end{aligned}
$$
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