Question

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

Answer 1

$\text{Step 1: Calculation of Area}$

Capacitance of a parallel plate capacitor is given by the relation:

$C=\frac{{\in }_{0}A}{d}$ $....\left(1\right)$

$\Rightarrow A=\frac{Cd}{{\in }_0}$

Putting Values:

$\therefore \text{Area,}A=\frac{2F\times (0.5\times {10}^{-2}m)}{8.85\times {10}^{-12}}=1130k{m}^2$

$\text{Step 2: Conclusion}$

Here, the area of the plates is too large$\left(1130k{m}^2\right)$, which is not possible in practical situation.

Therefore, the ordinary capacitors of area few $c{m}^2$ are normally in range of $\mu F$.

However, electrolytic capacitors do have a much larger capacitance $0.1F$ because of very minute separation between the conductors. As on reducing separation, capacitance increases significantly from equation $\left(1\right)$