are asked to calculate the electric field due to a charge filament as shown in
the animation. You are given three different detectors and three different
viewpoints: intermediate distance, very close, and very far from the filament.
The bar graph (not shown) displays the electric flux passing through the cubical
and spherical Gaussian surfaces. The option of showing the electric
field vectors is also given.
what viewpoint(s) can you safely calculate the electric field using the flux
detector's reading and Gauss's law? You may drag around each detector.
Reference: See Giancoli-PA: Appendix D, Giancoli-SE: 22-3.
The near and far
views have a symmetry that can be exploited. In the classroom, there are several
ways to have students come to this conclusion. Primarily, is there symmetry? If
so, what is it? In the near view, the charge filament looks long and
approximates an infinite line charge. For the far view, the charge looks like a
point charge. In both these cases, there is an appropriate symmetry and
Gaussís law can be used to determine the electric field on the surface of the
two detectors. In the intermediate situation, Gaussís law cannot be used to
determine the electric field, as the symmetry is not exact. These conclusions
can be further emphasizes by turning on the option of showing the electric field
In addition to
discussing electric field calculations, this problem can be used to discuss the
meaning of Gaussís law and electric flux. For example, in the far view, as
long as the charge filament is enclosed, the electric flux does not change as
the detector is moved around. Gaussís law states that this should be the case,
and it always is. Now, if the spherical Gaussian surface is not centered on the
charge filament can Gaussís law be used to easily calculate the electric
field? No. The symmetry is gone even though the flux calculation remains the
Script Author: Mario Belloni