Field At The Surface Of A Conductor

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Two students A and B were taught that electric field near a uniformly charged large surface is normal to the surface and is equal to (sigma)/(2 in_(0)) . They were also told that field near the surface of a conductor is (sigma)/(in_(0)) normal to the conductor where s is charge density on the conductor surface. Now both of them were asked to write field between the plates of an ideal parallel plate capacitor having charge density sigma and -sigma on its plates. Student A said that field can be seen as superposition of field due to two large charged surfaces. He wrote the answer as E=(sigma)/(2in_(0))+(sigma)/(2in_(0))=(sigma)/(in_(0)) Student B thought that a capacitor has conducting plates and therefore field due to each plate Must be (sigma)/(in_(0)) . He wrote his answer as E=(sigma)/(in_(0))+(sigma)/(in_(0))=(2sigma)/(in_(0)) Who is wrong and where is the flaw in thinking ?

Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?

Statement-1 Electrostatic field lines are discontinuous at the surface of a conductor Statement-2 Magnetic field lines produced by permanent magnets are never discontinuous

If q is the charge per unit area on the surface of a conductor, then the electric field intensity at a point on the surface is

If S is the surface area of charged conductor on which the surface density of charge is and K is the dielectric constant of the medium around it, then outward force acting on the surface of the conductor is

Show that the electric field at the surface of a charged conductor is proportional to the surface charge density.

The electric field intensity on the surface of a charged conductor is