(II) To get an idea how big a farad is, suppose you want to make a 1-F air-filled parallel-plate capacitor for a circuit you are building. To make it a reasonable size, suppose you limit the plate area to . What would the gap have to be between the plates? Is this practically achievable?

The capacitor is a charge storage device. The capacitance of a capacitor relies on the area of plates and the gap between the plates.

The capacitance is given by,

\(C = \frac{{{\varepsilon _0}A}}{d}\) … (i)

Here,\({\varepsilon _0}\)is the permittivity of free space, A is the area of plates and dis the separation between the plates.

The capacitance is,\(C = 1\;{\rm{F}}\)

The area of each plate is,\(A = 1\;{\rm{c}}{{\rm{m}}^2}\)

From equation (i), the separation between the plates is given as:

\(d = \frac{{{\varepsilon _0}A}}{C}\)

Substitute the values in the above expression.

\(\begin{aligned}d &= \frac{{\left( {8.85 \times {{10}^{ - 12}}\;{\rm{F/m}}} \right)\left( {1\;{\rm{c}}{{\rm{m}}^2} \times \frac{{1\;{{\rm{m}}^2}}}{{{{10}^4}\;{\rm{c}}{{\rm{m}}^2}}}} \right)}}{{\left( {1\;{\rm{F}}} \right)}}\\d &= 8.85 \times {10^{ - 16}}\;{\rm{m}}\end{aligned}\)

Thus, thegap between the plates is\(8.85 \times {10^{ - 16}}\;{\rm{m}}\).

It is practically impossible to achieve this because it is less than the radius of a proton.