Gauss's Law

Example: Cylindrical Charge Consider a line of charge with charge density of

By symmetry, the electric field is everywhere radially outward, perpendicular to the line of charge.

For our "Gaussian surface" -- our surface of integration, use a

cylinderwith its axis of symmetry on the line of charge (as shown in the sketches here). Consider the radius r and the length L. The flux through theendsiszerosince the component of the electric field E through those surfaces iszero. There is flux only through the cylindrical surface. There the electric field is perpendicular to the surface so the flux is just the electric field E multiplied bythatsurface;= E ( 2 r L ) and we know, from Gauss's Law, that

= Q / Q = L

E ( 2 r L ) = L /

E = L / [ ( 2 r L )]

E = / [ (2 r)]

E =[ /( 2 )] /r

k = 1/4

2 k = 1/ 2

E =2 k /r

Notice that this is

notan inverse-squarefield. The electric field from a line of charge decreases inversely as the distance from the line. It is a 1/r dependende,nota 1/r^{2}dependence.Again, a direct integration would be far, far more difficult than this calculation using

Gauss's Law.

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(c) Doug Davis, 2002; all rights reserved