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Ruslan
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Why doesis electric flux through a cube isthe same as that of electric flux through a spherical shell?

If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\varepsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimallyinfinitesimal area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$cosine of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electric field vector changes, i.e., they are no more parallel, still the flux remains the same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

Why does electric flux through a cube is same as that of electric flux through a spherical shell?

If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\varepsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimally area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$ of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electric field vector changes, i.e., they are no more parallel, still the flux remains same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

Why is electric flux through a cube the same as electric flux through a spherical shell?

If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\varepsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimal area of the shell is parallel to the electric field vector, arising from the point charge, which makes the cosine of the dot product unity, which is understandable. But for the cube, the electric field vector is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electric field vector changes, i.e., they are no more parallel, still the flux remains the same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

Converted \epsilon to more appropriate \varepsilon; fixed a typo
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Vishnu
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If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\epsilon_0$$q/\varepsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimally area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$ of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electicelectric field vector changes, i.e., they are no more parallel, still the flux remains same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\epsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimally area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$ of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electic field vector changes, i.e., they are no more parallel, still the flux remains same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\varepsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimally area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$ of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electric field vector changes, i.e., they are no more parallel, still the flux remains same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

added 4 characters in body; edited tags
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Qmechanic ♦
Qmechanic ♦
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If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\epsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimally area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$ of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electic field vector changes, i.e., they are no more parallel, still the flux remains same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\epsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimally area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$ of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electic field vector changes, i.e., they are no more parallel, still the flux remains same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

If a point charge $q$ is placed inside a cube (at the center), the electric flux comes out to be $q/\epsilon_0$, which is same as that if the charge $q$ was placed at the center of a spherical shell.

The area vector for each infinitesimally area of the shell is parallel to the electric field vector, arising from the point charge, which makes the $cosine$ of the dot product unity, which is understandable. But for the cube, the electric field vector, is parallel to the area vector (of one face) at one point only, i.e., as we move away from centre of the face, the angle between area vector and electic field vector changes, i.e., they are no more parallel, still the flux remains same?

To be precise, I guess, I am having some doubt about the angles between the electric field vector and the area vector for the cube.

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kamer_kane
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