A parallel plate capacitor of capacitance $100 \mathrm{pF}$ is to be constructed by using paper sheets of $1.0 \mathrm{~mm}$ thickness as dielectric. If the dielectric constant of paper is $4.0$, the number of circular metal foils of diameter $2.0 \mathrm{~cm}$ each required for this…

Question

A parallel plate capacitor of capacitance $100 \mathrm{pF}$ is to be constructed by using paper sheets of $1.0 \mathrm{~mm}$ thickness as dielectric. If the dielectric constant of paper is $4.0$, the number of circular metal foils of diameter $2.0 \mathrm{~cm}$ each required for this purpose is

Aicharya · Accepted Answer

$C=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}=10^{-10} \mathrm{~F}$ Let the number of sheets of foils required be $n$. They will form $(n-1)$ capacitors. If $K$ is the dielectric constant of the dielectric, the capacitance is given by $C=\frac{K \varepsilon_{0}(n-1) A}{d}$ or $n-1 =\frac{C d}{K \varepsilon_{0} A}=\frac{C d}{K \cdot 4 \pi \varepsilon_{0}} \cdot \frac{4 \pi}{\pi r^{2}}$ $=\frac{4 C d}{K \cdot 4 \pi \varepsilon_{0} r^{2}}$ $=\frac{4 \times 10^{-10} \times 1 \times 10^{-3} \times 9 \times 10^{9}}{4 \times\left(1.0 \times 10^{-2}\right)^{2}}=9$ or $n=10$ Hence the correct choice is (a).

A parallel plate capacitor of capacitance \(100 \mathrm{pF}\) is to be constructed by using paper sheets of…

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