Question

What is the area of the plates of a $2F$ parallel plate capacitor, given that the separation between the plates is $0.5cm$? (You will realize from your answer why ordinary capacitors are in the range of $F$ or less.) However, electrolytic capacitors do have a much larger capacitance $\left( {0.1F} \right)$ because of the very minute separation between the conductors.)

Answer 1

Hint: Two conducting plates separated by a small distance constitute a parallel plate capacitor. The capacitance of the parallel plate capacitor is given in the question. The separation between the plates of the capacitor is also given. We have to find the area of the plates of the capacitor.

Formula used:
$C = \dfrac{{{\varepsilon _0}A}}{d}$
where $C$ stands for the capacitance of the capacitor,
${\varepsilon _0}$ is the dielectric constant of free space,
$A$ stands for the area of the plates of the capacitor, and
$d$ stands for the separation between the plates of the parallel plate capacitor.

Complete step by step answer:
We know that the capacitance of a parallel plate capacitor can be written as,
$C = \dfrac{{{\varepsilon _0}A}}{d}$
From this, we can write the area of the parallel plate capacitor as,
$A = \dfrac{{Cd}}{{{\varepsilon _0}}}$
The capacitance of the parallel plate capacitor is given as,
$C = 2F$
The distance between the two plates of the capacitor is given as,
$d = 0.5cm = 0.5 \times {10^{ - 2}}m$
The value of the dielectric constant of free space is given by
${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$
Substituting these values in the equation,
$A = \dfrac{{Cd}}{{{\varepsilon _0}}}$
$ \Rightarrow A = \dfrac{{2 \times 0.5 \times {{10}^{ - 2}}}}{{8.85 \times {{10}^{ - 12}}}} = 1130k{m^2}$
This area is too large and is unattainable. To reduce the area of the plates of the capacitor we take the capacitance in the range of $\mu F$.

Hence, the correct answer is option (B).

Additional information:
The capacitance can be defined as the amount of charge required to raise the potential by $1$ volt. The unit of capacitance is farad. The capacitance of a capacitor is $1$ farad if $1$ coulomb of charge given to the capacitor increases its potential by $1$ volt.

Note: A capacitor is an arrangement of two conductors to store electrical charges. The capacitance of a capacitor is the measure of its ability to store electrical charges. The capacity will increase if there is a dielectric medium in between the parallel plates.