We know that,
[tex]\huge ➸ C = ϵ₀ \frac{A}{d} [/tex]
- C⤏Capacitance
- ϵ₀⤏permittivity of free space constant
- A⤏Area of capacitor plates
- d ⤏distance
given ↷
- C⤏5.31 nF = 5.31 x 10^-9
- ϵ₀⤏8.85 x 10^-12 F/m
- A⤏Area of capacitor plates
- d ⤏5mm
to find ↷
solution ↷
[tex]\begin{gathered}➸ C = ϵ₀ \frac{A}{d} \\ \end{gathered} [/tex]
[tex]\begin{gathered}➸ A = \frac{Cd}{ϵ₀} \\ \end{gathered} \\ [/tex]
[tex]\begin{gathered}➸ A = \frac{5.31 \times {10}^{ - 9} \times 5 }{8.85 \times {10}^{ - 12} } \\ \end{gathered} \\ [/tex]
[tex]➸ A =3 \times {10}^{3} {mm}^{2} [/tex]
[tex]➸ A =3 {m}^{2} [/tex]
Hence, the area is 3m²
