The figure at the right shows a closed Gaussian surface in the shape of a cube w

Question

The figure at the right shows a closed Gaussian surface in the shape of a cube with edge length 4.20 m. It lies in a region where the non-uniform electric

field is given by E! = (3.00y+4.00)ˆi+6.00ˆj+(4.00z+3.00)kˆ, N/C, with y and z in meters. Determine the net charge contained by the cube.

These are my teachers requirements if you could please follow them I would really appreciate it thank you :)

In addition to being neat and clear, and actually answering the question, you must:

1) show the original principle (often in equation form)

2) substitute variables as needed

3) solve it first (before substituting numbers)

4) show every substitution (with units and correct sig figs)

5) present a boxed answer (with units and correct sig figs)

In a Force problem in more than one dimension, you MUST start with a Free Body Diagram.

Explanation / Answer

I'm not sure if you've learned this but there is also a differential form of Gauss' Law.
Although Randy's explanation is correct, the integral form of Gauss' Law ,he is explaining, is typically used to find an unknown E-field , given the charge distribution.
The differential form is typically easier to use to find the charge, given the E-field, as in your problem.
The differential form is;
dEx/dx + dEy/dy + dEz/dz = D/eo (where D is the charge density at the location of the derivatives)

In your problem only the first and last terms are non-zero and immediately gives;
3 + 4 = D/eo

Since the left side is constant then the density is constant. So the charge in the cube is just;
q = (7)eoa^3

where a^3 is the volume of the cube and "a" is the given length of a side.

q = 7 x 8.854 x 10^-12 x (4.20)^3

q = 4.591 x 10^-9 C

Is this work for you ?

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