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 uF or less. However, electrolytic
capacitors do have a much larger capacitance
(0.1F) because of very minute separation
between the conductors.]​

Answer 1

Explanation :

capasitence of a parallel capacitor , V = 2F distance between the two plates , d = 0.5 multiple 10 power -2 m

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

C = E0 A upon d

A = Cd upon E0

where,

E0 = premittivity of free space = 8.85 ✖ 10 power -12 C power 2 N power -1 m power -2

therefore , A = 2 multiple 0.5 multiple 10 power -2 upon 8.85 multiple 10 power - 12

= 1130 kms

hence , the area of plates is too large . To avoid this situation , the capacitance is taken the range of uF.

Answer 2

Answer:

• 1130 km²

Explanation:

Given

• The capacitance is C = 2F.
• The plate separation is, d = 0.5 cm = 0.5 × 10$^{-2}$m.
• The capacitance of a parallel plate capacitor is given by the relation,

Tofind

• Area of capacitor

Solution

$C = \frac{ \varepsilon _{0} A}{d}$

$A = \frac{CV}{\varepsilon _{0}}$

$= \frac{ 2 \times 0.5 \times 10^{-2}}{8.85 \times 10^{-12}}$

= 1129943502 m²

= 1130 km²

Hence

The area of capacitor is 1130 km²