1st Example Gauss' Law Problem

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Gauss' Law Example 1: Net charge on cube due to external source.

What we are shown is a cube in an external electric field. We are asked to prove that the net flux out of the cube is zero due to that external electric field. Easy right?

Remember that the geometry of the situation will lead us to the solution of the problem. Think about it! There are six faces on the cube and four of them eliminate themselves immediately! How? Well, remember that the formula asks for the divergence (or dot product) of the two variables E and dA. This means that the important contribution to the flux will be due to the field lines going in the same line of action as the area vector. There are two faces, dA1 and dA2. The dA2 area vector is in the same direction of the electric field, while the dA1 area vector is directly the opposite of dA2 BUT IN THE SAME LINE OF ACTION.

This fact is important because the dot product uses the cosine trigonometric function. The angle between both E and dA1 and E and dA2 is zero degrees. Thus, the product E * dA2 = EA2*cos (angle) is EA2 cos (0) = EA2 because cos (0) = 1. On the other hand the product E * dA1 * cos (180) = - EdA1. Since both the areas dA1 and dA2 are equal, then EdA2 - EdA1 = 0, which satisfies our proof. The net flux out of the cube is zero due to the electric field!

What about the other four faces? Well, in terms of trigonometry each of those faces is at RIGHT ANGLES, OR 90 DEGREES, WITH THE ELECTRIC FIELD. THE COS (90) = 0 SO , AS STATED, THEY DO NOT CONTRIBUTE TO THE NET FLUX.

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You can go to DE Ex2 here...D.E. Ex2.

You can go to DE Ex1 here ...D.E. Ex1.