Gauss's law

Description

You 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.

Question

From 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.

Instructor Resources

Reference: See Giancoli-PA: Appendix D, Giancoli-SE: 22-3.

Answer: 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 Gauss’s law can be used to determine the electric field on the surface of the two detectors. In the intermediate situation, Gauss’s 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 vectors.

In addition to discussing electric field calculations, this problem can be used to discuss the meaning of Gauss’s 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. Gauss’s 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 Gauss’s law be used to easily calculate the electric field? No. The symmetry is gone even though the flux calculation remains the same.

Script Author: Mario Belloni