Q94 / 120 · VITEEE 2015 · Physics

The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. As a dielectric is introduced between the plates, the capacitance becomes 15 $\mu \mathrm{F}$. The permittivity of the medium in $\mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$ is

  • A
    $8.15 \times 10^{-11}$
  • Correct answer
    B
    $0.44 \times 10^{-10}$
  • C
    $15.2 \times 10^{12}$
  • D
    $1.6 \times 10^{-14}$

Solution

Capacitance of air capacitor $\mathrm{C}_{0}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=3 \mu \mathrm{F}$ When a dielectric of permittivity $\varepsilon_{\mathrm{r}}$ and dielectric constant $\mathrm{K}$ is introduced between the plates, then Capacitance, $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=15 \mu \mathrm{F}$ Dividing eq. (ii) by (i), we get $\begin{aligned} & \frac{\mathrm{C}}{\mathrm{C}_{0}}=\frac{\mathrm{d}}{\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}}=\frac{15}{3} \\ \Rightarrow \quad & \mathrm{K}=5 \\ \therefore \quad & \text { permittivity of the medium } \\ & \varepsilon_{\mathrm{r}}=\varepsilon_{0} \mathrm{~K} \\ &=8.85 \times 10^{-12} \times 5=0.44 \times 10^{-10} \end{aligned}$

Hence, the correct answer is option: 'B'.

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