Question

The rectangular plates in a parallel-plate capacitor are 0.063 m × 5.4 m. A distance of 3.5 × 10–5 m separates the plates. The plates are separated by a dielectric made of Teflon, which has a dielectric constant of 2.1. What is the capacitance of the capacitor?

Answer 1

Answer:

capacitance of the capacitor = 0.18 μ  F

Explanation:

Area of the plate A = 0.063 m x 5.4 m = 0.3402 m²

distance between the plate d = 3.5 × 10–5 m

dielectric value for Teflon K = 2.1

capacitance of capacitor = ?

Formula for capacitance of parallel plate is as follows ,

$C= \frac{K\epsilon_0 A}{d}$ (Where $\epsilon_0 = 8.82 \times 10^-12\ \frac{F}{m}$ )

putting the values in the equation,

C = $\frac{2.1\times8.82\times10^-12\times0.3402 }{3.5\times10^-5}$ $=0.18\times 10^-6$ =0.18 μ  F