A very large, thin, flat plate of aluminum of area A has a total charge Q unifor

Question

A very large, thin, flat plate of aluminum of area A has a total charge Q uniformly distributed over its surfaces. Assuming the same charge is spread uniformly over the upper surface of an otherwise identical glass plate, compare the electric fields just above the center of the upper surface of each plate. (Use any variable or symbol stated above along with the following as necessary: ?0.)
(a) electric field above aluminum plate
=
--Direction-- vertically upward vertically downward parallel to the plate .

(b) electric field above glass plate
=
--Direction-- vertically upward vertically downward parallel to the plate

Explanation / Answer

For Al E= rac{sigma}{epsilon_o} for Glass E= rac{sigma}{2epsilon_o} Someone tell me if I am wrong, please What about Q and A? The charge Q spreads out over both surfaces of the aluminum and there is 0 charge inside the metal. So sigma = Q/2A The field of the Al sheet is (E_{top} + E_{bot}) = rac{Q/2A}{epsilon_0} So: [tex]E_{top} = rac{Q/2A}{2epsilon_0}[/itex] For the glass plate, Q is distributed over area A (not 2A), so sigma = Q/A [tex]E_{top} = rac{Q/A}{2epsilon_0}[/itex]

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