Question

calculate the area of a paper required to construct a parallel plate capacitor of 0.004 micro Farad if the dielectric constant of paper be 2.5 and its thickness 0.025 mm.​

Answer 1

Solution:

Capacitance of parallel plate capacitor, C = 0.004 μF = 0.004 * 10⁻⁶ F

Dielectric constant of paper, K = 2.5

Thickness of paper, d = 0.025 mm = 0.025 * 10⁻³ m

Permittivity of free space, ε₀ = 8.854 * 10⁻¹² F/m

$C = (K \epsilon_{0} ) \frac{A}{d}$

$0.004 * 10^{-6} = (2.5 * 8.854 * 10^{-12} ) \frac{A}{0.025 * 10^{-3} }$

$A = \frac{0.004 * 10^{-6} * 0.025 * 10^{-3}}{2.5 * 8.854 * 10^{-12} }$

$A = \frac{0.0001 * 10^{-6} * 10^{-3}}{22.135 * 10^{-12} }$

$A = \frac{0.0001 * 10^{-6-3+12} }{22.135 }$

$A = \frac{0.0001 * 10^{3} }{22.135 }$

$A = 0.00000451773 * 10^{3}$

$A = 0.00451773$

$A = 4.51773 * 10^{-3} m^{2} \approx 4.52 * 10^{-3} m^{2}$

The area of paper is 4.52 * 10⁻³ m²