Which electromagnetic radiation can be seen with the human eye?
1 answer:
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Answer:
The charge on plane A is 2σ
The charge plane B on the right is σ
Explanation:
To solve this exercise we can use Gauss's law, to find the electric field created by each plane, let's create a cylinder-shaped Gaussian surface, where the base is parallel to the plane.
Gauss law
Ф = ∫ E. dA = /ε₀
with the field it must be perpendicular to the plane and the normal one to the base of the cylinder is perpendicular, the dot product is reduced to the algebraic product.
The charge inside the cylinder
σ = q_{int} / A
q_{int} = σ A
let's replace
the flow is on both sides of the plane
2E A = σ_A /ε₀
E = σ_A / 2ε₀
Let's analyze plane A on the left, the electric field is salient so the charge must be positive
For plane B on the right we repeat the calculation, finding that the charge is positive
Now we can find the electrified field in each part
To the left of plate A
E = E_A + E_B
3σ/ 2ε₀ = σ_A/2ε₀ + σ_B / 2ε₀
3 σ = σ_A + σ_B
The electric field between the planes is the subtraction of each elective field
E = E_A -E_B
σ / 2ε₀ = σ_A / 2ε₀ - σ_B / 2ε₀
σ = σ_A -σB
Let's write our system of equations and solve
3 σ = σ_A + σ_B
σ = σ_A - σ_ B
let's add the two equations
4 σ = 2 σ_A
σ_ A = 2 σ
therefore the charge on plane A is 2σ
now let's subtract the equations
2 σ = 2 σ_B
σ_B = σ
The charge plane B on the right is σ