**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 realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

**Solution:**

Capacitance of a parallel capacitor, *V* = 2 F

Distance between the two plates, *d* = 0.5 cm = 0.5 × 10−2 m

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

$C=\frac{\in_{0} A}{d}$

$A=\frac{C d}{\epsilon_{0}}$

Where,

$\epsilon_{0}=$ Permittivity of free space $=8.85 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$

$\therefore A=\frac{2 \times 0.5 \times 10^{-2}}{8.85 \times 10^{-12}}=1130 \mathrm{~km}^{2}$

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of µF.