How would a radial component of induced electric field makes flux ≠ 0

In summary, a radial component of an induced electric field contributes to a non-zero electric flux by creating a circulation of electric field lines that intersect a given surface. According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electric field, which can have components in different directions. When the radial component is present, it affects the overall electric field distribution, leading to a net flux through the surface, even if the surface is oriented perpendicular to the magnetic field. Thus, the induced electric field can generate a measurable flux due to its radial nature.
  • #1
KnightTheConqueror
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I don't understand this paragraph of Resnick halliday Krane where it says that if a radial component of induced electric field exists, it would mean net electric flux is not zero. I guess I am not exactly able to visualise the gaussian surface and how the flux is not zero. Please help
 

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  • #2
You appear to have a uniform magnetic field pointing out of the page. The Gaussian surface being considered is any arbitrary cylindrical surface with its axis pointing out of the page. The symmetry of the situation under rotation about the axis of our arbitrary cylinder means that if the electric field points outwards at any point on the surface it points outwards everywhere on the surface, so ##\vec E\cdot d\vec S## has the same value everywhere and hence the integral is necessarily non-zero.

Other symmetry arguments are available under these circumstances, but that seems to be the one they picked, if I understand right.
 
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  • #3
Thank you. After reading your answer I read the paragraph in the book again and now everything is clear
 
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