Respuesta :
the total electric flux through the two surfaces is equal.
Gauss law! Flux is proportional to the interior charge, the shape of the closed surface or its size is irrelevant as far as it encloses the same charge.
Gauss law! Flux is proportional to the interior charge, the shape of the closed surface or its size is irrelevant as far as it encloses the same charge.
According to the Gauss law, the electric flux through the closed surface is [tex]$\frac{1}{{{\varepsilon }_{0}}}$[/tex] times the charge enclosed by the surface.
[tex]$\Delta \phi =\frac{q}{{{\varepsilon }_{0}}}$[/tex]
Here, [tex]$\Delta \phi $[/tex] is the electric flux.
Gaussian surface a and b encloses the same positive point charges. So, the electric flux through surface a is four times larger than that through surface b is incorrect.
The total electric flux through surface b is eight times larger than that through surface a is incorrect because the electric flux is [tex]$\frac{1}{{{\varepsilon }_{0}}}$[/tex] times the total charge enclosed by the surface.
As one is aware that the electric flux is independent of the area of the Gaussian surface.
The total electric flux through surface a is eight times larger than that through surface b is also incorrect because the electric flux is independent of the area of the Gaussian surface.
Explanation:
Electric flux is independent of the area of the Gaussian surface. Since the charges enclosed by the surfaces are equal, then the electric flux through the surface will be equal.
Therefore, the total electric flux through the two surfaces is equal.
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