Fig. 25-35 shows a variable “air gap” capacitor for manual tuning. Alternate plates are connected together; one group of plates is fixed in position, and the other group is capable of rotation. Consider a capacitor of$\mathit{n}\mathbf{=}\mathbf{8}$plates of alternating polarity, each plate having area$\mathit{A}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{}\mathit{c}\mathit{m}\mathbf{2}$ and separated from adjacent plates by distance $\mathit{d}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{40}\mathit{m}\mathit{m}$What is the maximum capacitance of the device?

$$\begin{gathered} \mathrm{n}=8 \\ \mathrm{A}=1.25{\mathrm{cm}}^2\left(\frac{{10}^{-4}{\mathrm{m}}^2}{1{\mathrm{cm}}^2}\right)=1.25\times {10}^{-4}{\mathrm{m}}^2 \\ \mathrm{d}=3.40\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=3.40\times {10}^{-3}\mathrm{m} \end{gathered}$$

Capacitance is directly proportional to area. So the surface area of the capacitor will be the capacitance. The whole capacitor consists of (n-1)identical single capacitors connected in parallel. Each capacitor has a surface areaand plate separation d. The maximum capacitance can be calculated by using the equation 25-9

Formulae are as follows:

$\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$

Where C is capacitance, A is the area, and d is distance.

But for (n-1) capacitors,

$\mathrm{C}=\frac{\left(\mathrm{n}-1\right){\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$

By substituting the given values,

$$\begin{gathered} \mathrm{C}=\frac{\left(8-1\right)\times 8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\times 1.25\times {10}^{-4}{\mathrm{m}}^2}{3.40\times {10}^{-3}\mathrm{m}} \\ =\frac{7\times 8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\times 1.25\times {10}^{-4}{\mathrm{m}}^2}{3.40\times {10}^{-3}\mathrm{m}} \\ =2.28\times {10}^{-12}\mathrm{F} \end{gathered}$$

Hence, the maximum capacitance of the device is $\mathrm{C}=2.28\times {10}^{-12}\mathrm{F}$.

Therefore, by using the formula of capacitance in terms of area, find the maximum capacitance.