Finding volume charge density of nonconducting spherical shell

[DTownStudent]

The figure below shows a closed Gaussian surface in the shape of a cube of edge length 2.20 m. It lies in a region where the electric field is given by = [ (3.00x + 4.00) + 6.00 + 7.00 ] N/C, where x is in meters. What is the net charge contained by the cube?

So i used the x value of the electric field and took that times the the cube length squared or area divided by the permittivity of space and then times two because there are two sides. This probably isn't right but that's what I thought you had to do.

3 * 2.2^2 *2 / 8.85E-12

 

[anathema]

* 2 = 2.97E-8 C

I would like to provide a more accurate and detailed explanation for finding the net charge contained by the cube.

First, we need to understand the concept of a Gaussian surface. A Gaussian surface is an imaginary surface that is used to simplify the calculation of electric fields. It is a closed surface that encloses a certain area and is chosen in such a way that the electric field is constant and perpendicular to the surface. In this case, the Gaussian surface is a cube with an edge length of 2.20 m.

Next, we need to understand the given electric field equation: E = (3.00x + 4.00) + 6.00 + 7.00 N/C. This equation shows that the electric field is varying with distance (x) and has three components - x, y, and z. For simplicity, we can consider only the x-component, which is (3.00x + 4.00) N/C. This means that the electric field is changing with x and has a constant value of 4.00 N/C.

To find the net charge contained by the cube, we can use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of space (ε0). Mathematically, it can be represented as Φ = Q/ε0.

In this case, the electric flux through the Gaussian surface is equal to the electric field (4.00 N/C) multiplied by the area of the surface (2.20 m * 2.20 m) and multiplied by the number of sides (6). This can be written as Φ = 4.00 * 2.20 * 2.20 * 6 = 105.6 Nm^2/C.

Now, to find the net charge (Q) contained by the cube, we can rearrange the equation to Q = Φ * ε0. Plugging in the values, we get Q = 105.6 * 8.85E-12 = 9.34E-10 C.

Therefore, the net charge contained by the cube is 9.34E-10 C. It is important to note that this calculation is based on the assumption that the electric field is constant and perpendicular to the surface, which may not be the