A parallel-plate capacitor with plate separation of 4.0 cm has a plate area of 6.0 × $10 ^{- 2}$ $m ^ { 2 }$ What is the capacitance of this capacitor if a dielectric material with a dielectric constant of 2.4 is placed between the plates,completely filling the space? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor with air or vacuum between the plates is given by:
C = ε₀A/d
where ε₀ = 8.85 × 10^-12 N∙m^2/C^2 is the permittivity of free space, A is the area of each plate, and d is the separation between the plates.
When a dielectric material is placed between the plates, the capacitance increases by a factor κ, called the dielectric constant or relative permittivity:
C' = κC
The dielectric constant of a material is a measure of how much it reduces the electric field between the plates, and it is always greater than or equal to 1. In this case, κ = 2.4, so the capacitance becomes:
C' = κε₀A/d = 2.4 × 8.85 × 10^-12 × 6.0 × 10^-2 / 4.0 × 10^-2 = 32 × 10^-12 F
Therefore, the best choice is E, 32 × 10^-12 F.

A Parallel-Plate Capacitor with Plate Separation of 4 10−210 ^{- 2}10−2

A Parallel-Plate Capacitor with Plate Separation of 4 $10 ^{- 2}$