The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the…
The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. With the introduction of a dielectric medium between the plates, the capacitance becomes 15 $\mu F$. The permittivity of the medium in SI unit is $\left[\epsilon_{0}=8 \cdot 85 \times 10^{-12}\right.$ SI unit]
15
$8.845 \times 10^{-11}$
$0.4425 \times 10^{-10}$
44. 5
Solution
Capacitance of a parallel plate capacitor $\mathrm{C}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}}$
here $\varepsilon$ is the permitivity of the medium
For air medium $\varepsilon=\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \quad \therefore \mathrm{C}^{\prime}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}} \ldots .1$
whearas for dielectric medium capacitance $\mathrm{C}=\frac{\varepsilon \mathrm{A}}{\mathrm{d}} \ldots 2$
From 1 and 2
$\begin{array}{l}
\frac{\mathrm{C}}{\mathrm{C}^{\prime}}=\frac{\varepsilon}{\varepsilon_{0}} \\
\therefore \varepsilon=\frac{15}{3} \times 8.85 \times 10^{-12}=\left(44.25 \times 10^{-12}\right)=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{2}
\end{array}$
Hence the permitivity of the medium $\varepsilon=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{2}$