## Gauss' Law, Flux and Fields

It is found experimentally that the electric field in a certain region of the Earth's atmosphere is directed vertically downward. At an altitude of 300 m, E = 60 N/C, and at 200 m, E = 100 N/C. Find the net charge contained in a Gaussian cube 100 m on edge, with horizontal faces at altitudes of 200 m and 300 m. Neglect the curvature of the Earth.To solve this problem we will use Gauss' law. The net flux through the cube is proportional to the charge contained within the cube.

As the field is vertical, there is only flux through the horizontal faces. The upper face has a field of 60N/C across an area of 10 000 m^{2} – so the total flux through this surface is 600 000 Nm^{2}/C (which we’ll take as negative because it goes into the cubical volume). Similarly, the flux through the lower surface 1 000 000 Nm^{2}/C.

The net flux through the surface then is 400 000 Nm^{2}/C

By Gauss' Law, we know that the net flux through the surface is related to the internal charge by the fomula:

- Flux

_{net}= Q

_{tot}/ epsilon

_{0}

Q_{tot} = Flux_{net} * epsilon_{0} = 400 000 Nm^{2}/C* 8.8542 x 10^{-12} C^{2}/(Nm^{2})

Q_{tot} = 3.54 x 10^{-6} C

## 1 comment:

Please solve it step by step..

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