Find the expression of the electric field due to an arrangement of two closely separated
Answers
Consider first a single infinite conducting plate. In order to apply Gauss's law with one end of a cylinder inside of the conductor, you must assume that the conductor has some finite thickness. In doing this, the surface charge density σ must be spread over both sides (think of this as a finite plate with a small thickness and then stretch it out to infinity. Using Gauss's law with this plate (either putting one end of the cylinder in the conductor or one end on both sides) gives a result of E=σϵ0=Q2Aϵ0.
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Now imagine bringing the second plate, with opposite charge density −σ in from infinity. Because these plates are conductors, charges in each plate will move around to cancel the field from the opposite plate inside of the conductor (remember E=0 inside of a conductor). Because the electric field produced by each plate is constant, this can.