Question 620: Electrostats & Capacitance

Question

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 $$15 \mu \mathrm{F}$$. The permittivity of the medium in $$\mathrm{SI}$$ unit is $$[\varepsilon_0=8.85 	imes 10^{-12} \mathrm{SI}$$ unit]

Accepted Answer

The capacitance of a parallel plate capacitor with air is
$$C=\frac{\varepsilon_0 A}{d}\quad 	ext{.... (i)}$$
where, $$\varepsilon_0=$$ permittivity of free space or air. The capacitance with a dielectric medium between the plates is
$$C^{\prime}=\frac{\varepsilon_m A}{d} \quad 	ext{.... (ii)}$$
where, $$\varepsilon_m=$$ permittivity of medium.
Here, $$C^{\prime}=15 \mu \mathrm{F}, C=3 \mu \mathrm{F}$$
and $$\varepsilon_0=8.85 	imes 10^{-12}$$ SI unit
From Eqs. (i) and (ii), we get
$$\begin{aligned}
\frac{C^{\prime}}{C} & =\frac{\varepsilon_m A}{d} 	imes \frac{d}{\varepsilon_0 A}=\frac{\varepsilon_m}{\varepsilon_0} \
\varepsilon_m & =\frac{\varepsilon_0 C^{\prime}}{C}=\frac{8.85 	imes 10^{-12} 	imes 15 	imes 10^{-6}}{3 	imes 10^{-6}} \
& =44.25 	imes 10^{-12} 	ext { SI unit } \
& =0.4425 	imes 10^{-10} 	ext { SI unit }
\end{aligned}$$

Answer

$$8.845 	imes 10^{-11}$$

Answer

$$0.4425 	imes 10^{-10}$$

Answer

15

Answer

5

Electrostats & Capacitance

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