The capacitance of a parallel plate capacitor with air as a medium is $6 \, \mu ext{F}$. With the introductio

Question

The capacitance of a parallel plate capacitor with air as a medium is $6 \, \mu	ext{F}$. With the introduction of a dielectric medium, the capacitance becomes $30 \, \mu	ext{F}$. The permittivity of the medium is: $(\varepsilon_0=8.85 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2})$

Accepted Answer

1. **Calculate Dielectric Constant (K):** The capacitance of a parallel plate capacitor with a dielectric medium ($C$) is related to its capacitance with air ($C_0$) by the formula $C = KC_0$, where $K$ is the dielectric constant of the medium [NCERT Class 12, Physics Part I, Sec 2.13, Eq 2.54].
   $$K = \frac{C}{C_0} = \frac{30 \, \mu	ext{F}}{6 \, \mu	ext{F}} = 5$$
2. **Calculate Permittivity of the Medium ($\varepsilon$):** The permittivity of the medium $\varepsilon$ is related to the permittivity of free space $\varepsilon_0$ by the relation $\varepsilon = K \varepsilon_0$ [NCERT Class 12, Physics Part I, Sec 2.13, Eq 2.52].
   $$\varepsilon = 5 	imes (8.85 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2})$$
   $$\varepsilon = 44.25 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}$$
   $$\varepsilon = 0.4425 	imes 10^{-10} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}$$
   Rounding to two decimal places, $\varepsilon \approx 0.44 	imes 10^{-10} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}$.

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