Respuesta :
Answer:
a) flow must be doubled , b) Ф₂ =∛2² Ф₀, c) Ф does not change , d) flow is constant , e) flow is constant
Explanation:
For this exercise let's use Gauss's law, which states that the flow is equal to the waxed load divided by the permissiveness of the vacuum
Ф = E. dA = [tex]q_{int}[/tex] / ε₀
Let's apply this equation
a) if the charge inside the Gaussian surface is doubled by the equation the flow must be doubled
Ф₀ = [tex]q_{int}[/tex] / ε₀
Ф₂ = 2 [tex]q_{int}[/tex] / ε₀ = 2 Ф₀
b) the volume is V₂ = 2V₀. Area is the volume divided by the length
V = LLL = L³
A = LL = L²
A = V / L
V₂ = 2V₀
L³ = 2 L₀³
L = ∛2 L₀
V₂ = A₂ L
A₂ = V₂ / L
A₂ = 2V₀ / L = 2 L₀³ / (∛2 L₀)
A₂ = 2 /∛2 V₀ / L₀
A₂ = ∛2² A₀
Ф₀ = E A₀
Ф₂ = E A₂
Ф₂ = E ∛2² A₀
Ф₂ =∛2² Ф₀
The flow doubles when the volume is increased by the factor [tex]2^{2/3}[/tex]
c) the flow does not depend on the shape of the surface, so Ф does not change
d) the flow does not depend on the position of the load, while inside the surface the flow is constant
e) when the load leaves the surface, the load inside drops to ero, so the flow changes to zero
Ф = 0
