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 cylinder with 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 the ends is zero since the component of the electric field E through those surfaces is zero. 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 by that surface;

= 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 not an inverse-square field. The electric field from a line of charge decreases inversely as the distance from the line. It is a 1/r dependende, not a 1/r2 dependence.

Again, a direct integration would be far, far more difficult than this calculation using Gauss's Law.

Spherical Shell


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(c) Doug Davis, 2002; all rights reserved