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Q. The capacitance of a parallel plate capacitor with air as 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 is

NTA AbhyasNTA Abhyas 2022Electrostatic Potential and Capacitance

Solution:

The capacitance of air capacitor
$C_0=\frac{A \varepsilon_0}{d}=3 \mu F \text {...(i) }$
When a dielectric of permittivity $\varepsilon_r$ or dielectric constant $K$ is introduced between the plates of the capacitor, then its capacitance
$C=\frac{K A \varepsilon_0}{d}=15 \mu F \ldots \text { (ii) }$
Dividing Eq. (ii) by Eq. (i)
$\frac{C}{C_0}=\frac{\frac{K A \varepsilon_0}{d}}{\frac{A \varepsilon_0}{d}}=\frac{15}{3}$
$\therefore K=5$
Permittivity of the medium
$\varepsilon =\varepsilon_0 K $
$=8.854 \times 10^{-12} \times 5$
$=44.27 \times 10^{-12} $
$=0.44 \times 10^{-10} C ^2 N ^{-1} m ^{-2}$