Question

Please, the answer.... Thanks  The area of each plate of a 6.8 pF capacitor is 4.8 × 10–6 m2. The plates are separated by a distance of 8.2 × 10–5 m. What is the dielectric constant of the dielectric used in the capacitor?

Answer 1

Capacitance (C) = 6.8 pF or 6.8 x 10^-12 Farad
Plate Area (A) = 4.8 x 10^-6 m^2
Distance (d) = 8.2 x 10^-5 m
Dielectric Constant (ε) = ?

Dielectric Constant Unit:
ε = C x d/a - > ε = farad x meter / meter² - > ε = farad/meter


C = ε x A/d

-> 6.8 x 10^-12 = ε x 4.8 x 10^-6/8.2 x 10^-5

-> 6.8 x 8.2 x 10^ (-12+(-5)) = ε x 4.8 x 10^-6

-> 55.76 x 10^-17 = ε x 4.8 x 10^-6

-> ε = 55.76 x 10^-17/4.8 x 10^-6

-> ε = 11.6 x 10^ (-17 - (-6))

-> ε = 11.6 x 10^ (-17 +6)

-> ε = 11.6 x 10^-11 F/m

Answer 2

Answer is 1.3 × 10^1. Just took on edg.