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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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In summary, the flux through a Gaussian surface does not change when the shape is changed, even though the net charge is still inside the surface.

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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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True, flux is integral of dot product of electric field and [itex]\vec{da}[/itex], 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.Idyia said:

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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So, basically as surface area changes, the electric field changes too, so the flux remains constant... Thank you so much!Avalanche_ said:True, flux is integral of dot product of electric field and [itex]\vec{da}[/itex], 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.

A Gaussian surface is an imaginary surface used in Gauss's law to calculate the electric flux through a closed surface. It is typically chosen to be a symmetrical shape, such as a sphere or a cube, to simplify calculations.

The shape of a Gaussian surface is determined by the symmetry of the distribution of charges. It is chosen to be a shape that has the same symmetry as the charge distribution, making it easier to calculate the electric flux.

The shape of a Gaussian surface is significant because it allows us to calculate the electric flux through a closed surface, which is a crucial concept in Gauss's law. It also simplifies calculations by using symmetrical shapes.

No, the shape of a Gaussian surface must have the same symmetry as the charge distribution. This means that it is not always possible to choose any shape as a Gaussian surface.

The shape of a Gaussian surface does not directly affect the electric field. However, it helps to simplify calculations and determine the electric field using Gauss's law by enclosing a symmetrical distribution of charges.

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