Consider a wire with linear charge density. Assuming that for symmetry reasons the field is directed at every point orthogonally to the wire axis and depends only on the distance from the wire, we can conclude that the flux of the electric field through the coaxial cylind er (Gauss surface) is
Consider a wire with linear charge density. Assuming that for symmetry reasons the field is directed at every point orthogonally to the wire axis and depends only on the distance from the wire, we can conclude that the flux of the electric field through the coaxial cylind er (Gauss surface) is
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Transcribed Image Text:Consider a wire with linear charge density A. Assuming that for symmetry reasons the field is directed at every point orthogonally to the
wire axis and depends only on the distance from the wire, we can conclude that the flux of the electric field through the coaxial cylind
er (Gauss surface) is
OO(E) = 2 r h E(r)
O(E) = 2 r h E(r) + 2π r² E(r)
=
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