Physics - Capacitor Question with Solution | TestHub

Capacitance of air capacitor ${\mathrm{C}}_0=\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}}=3\mu \mathrm{F}$ When a dielectric of permittivity ${\epsilon }_{\mathrm{r}}$ and dielectric constant $\mathrm{K}$ is introduced between the plates, then Capacitance, $\mathrm{C}=\frac{\mathrm{K}{\epsilon }_0\mathrm{A}}{\mathrm{d}}=15\mu \mathrm{F}$ Dividing eq. (ii) by (i), we get $\begin{array}{rl}& \frac{\mathrm{C}}{{\mathrm{C}}_0}=\frac{\mathrm{d}}{\frac{{\epsilon }_0\mathrm{A}}{\mathrm{d}}}=\frac{15}{3}\\ \Rightarrow & \mathrm{K}=5\\ \therefore & \text{ permittivity of the medium }\\ & {\epsilon }_{\mathrm{r}}={\epsilon }_0\mathrm{K}\\ & =8.85\times 10^{-12}\times 5=0.44\times 10^{-10}\end{array}$