Respuesta :
Answer:
E = ½ σ₀/ 2ε₀ = ½ E₀
Explanation:
The charge in the initial metal wall is in the right part, but in metals the electrons are mobile, so that the charge is divided into two parts of the wall, therefore the charge density is
σ₂ = σ₀ / 2
the zero index is for the insulating wall and the index 2 for the metal wall
To find the field you can use Gauss's law,
Ф = E. dA = [tex]q_{int}[/tex] /ε₀
If we use a cylinder as a Gaussian surface, the sides of the cylinder do not contribute to the flow and the bases give a scalar producer that is reduced to the algebraic product
E A = q_{int} /ε₀
The surface charge density is
σ = Q / A
q_{int} = Q = σ A
E A = σ₂ A /ε₀
E = σ₂ /ε₀
As the electric field from the wall extends to both sides the value on one side is
E = σ₂ / 2 ε₀
E = σ₀/2 1/2ε₀
E = ½ σ₀/ 2ε₀
E = ½ E₀
The field is half of the field created by the insulating wall.
- E = ½ σ₀/ 2ε₀
- E= ½ E₀
Charge density
The charge in the initial metal wall is in the right part, but in metals the electrons are mobile, so that the charge is divided into two parts of the wall, therefore the charge density is
σ₂ = σ₀ / 2
The zero index is for the insulating wall and the index 2 for the metal wall
Gauss's Law is used for calculating field,
[tex]\phi = E. dA \\\\\phi = q_{int} / E_0[/tex]
If we use a cylinder as a Gaussian surface, the sides of the cylinder do not contribute to the flow and the bases give a scalar producer that is reduced to the algebraic product.
[tex]E A = q_{int} /E_0[/tex]
The surface charge density is
[tex]\sigma = Q / A\\\\q_{int} = Q = \sigma A\\\\E A = \sigma_2 A /E_0\\\\E = \sigma_2 /E_0[/tex]
As the electric field from the wall extends to both sides the value on one side is
[tex]E = \sigma_2 / 2 E_0\\\\E = \sigma_0/2 *1/2E_0\\\\E = 1/2 \sigma_0/ 2E_0\\\\E =1/2 E_0[/tex]
The field is half of the field created by the insulating wall.
Note: ε₀ is denoted by E₀.
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