Q103 / 120 · VITEEE 2007 · Physics
A parallel plate capacitor of capacitance $100 \mathrm{pF}$ is to be constructed by using paper sheets of $1 \mathrm{~mm}$ thickness as dielectric. If the dielectric constant of paper is 4 , the number of circular metal foils of diameter $2 \mathrm{~cm}$ each required for the purpose is
- A40
- B20
- C30
- Correct answerD10
Solution
The capacitance (C) of a parallel plate capacitor with dielectric is given by
$\begin{array}{l}
C=\frac{\epsilon_{0} A}{d-t\left(1-\frac{1}{K}\right)} \\
C=100 p F=100 \times 10^{-12} F, \in_{0}=8.85 \times 10^{-12} \\
A=\pi \mathrm{I}^{2}=3.14 \times\left(10^{-2}\right)^{2}(\mathrm{r}=\mathrm{lcm}) \\
\mathrm{t}=10^{-3} \mathrm{~m}, \mathrm{~d}=\mathrm{t}, \mathrm{K}=4
\end{array}$
$\mathrm{C}_{1}=\frac{8.85 \times 10^{-12} \times 3.14 \times 10^{-4}}{10^{-3}-10^{-3}\left(1-\frac{1}{4}\right)}$
$=\frac{27.79 \times 10^{-16}}{10^{-3}-0.75 \times 10^{-3}}$
$=\frac{27.79 \times 10^{-13}}{0.25}$
$=\frac{27.79 \times 10^{-13}}{0.25}=111.16 \times 10^{-13} \mathrm{~F}$ (for 1 set)
Required $\mathrm{C}=100 \times 10^{-12} \mathrm{~F}$
$\Rightarrow \mathrm{n}=\frac{\mathrm{C}}{\mathrm{C}_{1}}=\frac{100 \times 10^{-12}}{111.16 \times 10^{-13}} \simeq 10$
Hence, the correct answer is option: 'D'.