Physics, 07.04.2020 16:00 lizzyhearts

Each plate of a parallel-plate air-filled capacitor has an area of 0.0020 m2, and the separation of the plates is An electric field of is present between the plates. What is the surface charge density on the plates

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Answer from: veah808

Answer

Each plate of a parallel-plate air-filled capacitor has an area of 2×10^−3m², and the separation of the plates is 2×10^-²mm. An electric field of 9.7×10^5V/m is present between the plates. What is the surface charge density on the plates? (ϵo=8.85×10^−12C²/N⋅m²)

Explanation

Given that,

Area of the plate is 2×10^-3m²

A = 2×10^-3m²

The separation between plate is 2×10^-2mm

d = 2×10^-2mm = 2×10^-2 × 10^-3m

d = 2 × 10^-5 m

The electric field present is 9.7 × 10^5 V/m

E = 9.7 × 10^5 V/m

ϵo = 8.85×10^−12C²/N⋅m²

We want to find the charge density?

The charge density of a capacitor can be determine using the charge density formula

σ = ϵo • E

Where E is the electric field

σ is the surface charge density

σ = 8.85 × 10^-12 × 9.7 × 10^5

σ = 8.58 × 10^-6 C/m²

Another method

The capacitance of the capacitor can be calculated using

C = kϵoA/d

The dielectric is air filled and it value is

k = 1.00059

C = 1 × 8.85 × 10^-12 × 2×10^-3/2×10^-5

C = 8.85 × 10^-10 F

The charge on a capacitor can be calculated using

q = CV

Where V = Ed

Then, q = C•E•d

q = 8.85× 10^-10×9.7×10^5 × 2×10^-5

q = 1.7169 × 10^-8 C.

Then, using charge density formula

σ = q / A

σ = 1.7169×10^-8/2×10^-3

σ = 8.58 × 10^-6 C/m²

So the charge density is σ = 8.58 × 10^-6 C/m²

Both are correct.

Answer from: Quest

The masses of the objects and their densities

Answer from: Quest

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