The capacitance of a parallel plate capacitor | vedclass

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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon A}{d}$.For air as the medium,the capacitance is $C_{0} = \frac{\v

Vedclass · Accepted Answer

$0.4425 	imes 10^{-10}$

Vedclass · Answer

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon A}{d}$.For air as the medium,the capacitance is $C_{0} = \frac{\varepsilon_{0} A}{d} = 3 \mu F$.When a dielectric medium is introduced,the new capacitance is $C = \frac{\varepsilon A}{d} = 15 \mu F$.Taking the ratio of the two capacitances:$\frac{C}{C_{0}} = \frac{\varepsilon}{\varepsilon_{0}} = K$ (where $K$ is the dielectric constant).$\frac{15}{3} = \frac{\varepsilon}{\varepsilon_{0}} \implies 5 = \frac{\varepsilon}{\varepsilon_{0}}$.Therefore,the permittivity of the medium is $\varepsilon = 5 	imes \varepsilon_{0}$.$\varepsilon = 5 	imes 8.85 	imes 10^{-12} = 44.25 	imes 10^{-12} 	ext{ F/m}$.Converting to scientific notation: $\varepsilon = 0.4425 	imes 10^{-10} 	ext{ F/m}$.

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

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