B Why doesn't the flux through a Gaussian surface change with a change in shape?

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The flux through a Gaussian surface remains constant despite changes in shape, provided the net charge inside remains the same. This is because flux is the integral of the electric field and surface area, and when the surface area increases, the electric field decreases correspondingly. For a point charge, the electric field decreases with the square of the distance, while the surface area increases with the square of the radius, resulting in a constant product. Thus, any change in surface area is offset by a change in the electric field strength. Overall, the relationship between surface area and electric field ensures that the total flux remains unchanged.
Idyia
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Why doesn't the flux through a Gaussian surface change, when the shape is changed? (while keeping the net charge inside it the same)
Flux is the dot product of electric field and surface area, so wouldn't it change if surface area is changed?
 
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Idyia said:
Why doesn't the flux through a Gaussian surface change, when the shape is changed? (while keeping the net charge inside it the same)
Flux is the dot product of electric field and surface area, so wouldn't it change if surface area is changed?
True, flux is integral of dot product of electric field and \vec{da}, that's an infinitesimal element of a surface. But if you change shape of your surface, let's say you move some part of surface away from charges, the flux through the part of your surface that you didn't touch stays the same, so if something changes, it changes on that part of your surface that you ''touched''. You moved part of your surface away from charges so area of that part is now bigger than it was, BUT also the electric field doesn't have the same value that it had on ''old'' surface. Since that part of surface is further from charges, electric field has a smaller value but their product is the same. (surface area raises, field decreases). Especially for one point charge and a spherical gaussian surface, field decreases as 1/r^2 , and surface area increases as r^2, so their product is constant.
 
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Avalanche_ said:
True, flux is integral of dot product of electric field and \vec{da}, that's an infinitesimal element of a surface. But if you change shape of your surface, let's say you move some part of surface away from charges, the flux through the part of your surface that you didn't touch stays the same, so if something changes, it changes on that part of your surface that you ''touched''. You moved part of your surface away from charges so area of that part is now bigger than it was, BUT also the electric field doesn't have the same value that it had on ''old'' surface. Since that part of surface is further from charges, electric field has a smaller value but their product is the same. (surface area raises, field decreases). Especially for one point charge and a spherical gaussian surface, field decreases as 1/r^2 , and surface area increases as r^2, so their product is constant.
So, basically as surface area changes, the electric field changes too, so the flux remains constant... Thank you so much!
 
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