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Hijaz Aslam

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Gauss's Law states that if a Gaussian Surface encloses a charge ##q_{enc}## then the electric flux through the Gaussian Surface is given by ##\phi=q_{enc}/\varepsilon_{o}## .

It also states that any external field does not contribute to the Electric Flux through the Gaussian Surface.

I am bit confused over there. If we have Gaussian Surface which is a cube placed in a non-uniform linear electric charge (by an infinite sheet for instance, and the Electric Field is parallel (anti-parallel) to the area vector of one of the faces (and the opposite face) ) the Flux through the two opposite faces of the cube does not cancel out. Is this true? What does the Gauss's Law actually states (along with conditions)?

It also states that any external field does not contribute to the Electric Flux through the Gaussian Surface.

I am bit confused over there. If we have Gaussian Surface which is a cube placed in a non-uniform linear electric charge (by an infinite sheet for instance, and the Electric Field is parallel (anti-parallel) to the area vector of one of the faces (and the opposite face) ) the Flux through the two opposite faces of the cube does not cancel out. Is this true? What does the Gauss's Law actually states (along with conditions)?

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