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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.
The surface charge density on the plates is [tex]\sigma = 8.58 \times10^{-6} \;C/m^2[/tex]
Given information:
The area of the plate is [tex]A = 2\times10^{-3}m^2[/tex]
The separation between plates is [tex]d = 2\times10^{-5}m[/tex]
The electric field present is [tex]E = 9.7 \times10^5 V/m[/tex]
Parallel plate capacitor:
The electric field inside a parallel plate capacitor with surface charge density σ is given by:
E = σ / ϵ₀
The surface charge density of the capacitor can be determined by:
σ = ϵ₀E
Where E is the electric field
σ is the surface charge density
[tex]\sigma = 8.85 \times 10^-{12} \times 9.7 \times10^5\\\\\sigma = 8.58 \times10^{-6} \;C/m^2[/tex]
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