Question

A parallel plate capacitor of capacitance $100\, pF$ is to be constructed by using paper sheets of $1\, mm$ thickness as dielectric. If the dielectric constant of paper is $4$, the number of circular metal foils of diameter $2\, cm$ each required for the purpose is

• A. 40
• B. 20
• C. 30
• D. 10

Answer 1

Answer: (D)

The arrangement of $n$ metal plates separated by dielectric acts as parallel combination of $(n-1)$ capacitors.
Therefore, $C=\frac{(n-1)\, K \varepsilon_{0} A}{d}$
Here, $ C =100 \,pF$
$=100 \times 10^{-12} F $
$ K=4, \varepsilon_{0}=8.85 \times 10^{-12} C ^{2} / Nm ^{2} $
$ A=\pi r^{2} =3.14 \times\left(1 \times 10^{-2}\right)^{2} $
$ d=1\, mm =1 \times 10^{-3} $
$(n-1) \times 4 \times 8.85 \times 10^{-12}$
$ \therefore 100 \times 10^{-12} =\frac{\times 3.14 \times\left(1 \times 10^{-2}\right)^{2}}{1 \times 10^{-3}}$
or$ n= \frac{1111.156}{111.156} $
$= 9.99 $
$ \approx 10 $