An electric field in free space is \(\mathbf{E}=\left(5 z^{2} / \epsilon_{0}\right) \hat{\mathbf{a}}_{z} \mathrm{~V} / \mathrm{m}\). Find the total charge contained within a cube, centered at the origin, of \(4-\mathrm{m}\) side length, in which all sides are parallel to coordinate axes (and therefore each side intersects an axis at \(\pm 2\) ).

The given electric field has only the z-component and can be expressed as: \(\mathbf{E} = (0, 0, 5z^2/\epsilon_0)\)

To find the total electric flux, we'll calculate the electric flux through each of the six faces of the cube. The electric field is only along the z-direction, so for the top and bottom faces, the normal components of the electric field are equal to the z-component. Surfaces parallel to the xy-plane (Top and bottom faces): Flux through each face = \(\int \mathbf{E} \cdot \hat{\mathbf{n}}_{top/bottom} dS = \frac{5z^2}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\) As for the other four faces, the normal components have no contribution from the electric field, so the electric flux through those faces is zero. Now let's compute the electric flux through top face (z=2): Flux through top face (z=2) = \(\frac{5(2)^2}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy= \frac{20}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\) Similarly for bottom face (z=-2): Flux through bottom face (z=-2) = \(\frac{5(-2)^2}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy= \frac{20}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\)

Now let's calculate the total electric flux from the 6 faces of the cube. The total electric flux is the sum of electric fluxes of all individual faces: Total Electric flux = Flux through top face + Flux through bottom face + Flux through 4 side faces Since the electric flux through the side faces is zero, we only need to add the contributions from the top and bottom faces: Total Electric Flux = \(2 \cdot \frac{20}{\epsilon_0} \int_{-2}^{2}\int_{-2}^{2} dxdy\)

According to Gauss's law, the total enclosed charge Q is given by: \(\oint \mathbf{E}\cdot \hat{\mathbf{n}} dS = \frac{Q}{\epsilon_0}\) Applying this to our scenario: Total Electric Flux = \(\frac{Q}{\epsilon_0}\) Therefore, \(Q = Total Electric Flux \cdot \epsilon_0 = 2 \cdot 20 \int_{-2}^{2}\int_{-2}^{2} dxdy \cdot \epsilon_0 = 80 \int_{-2}^{2}\int_{-2}^{2} dxdy \cdot \epsilon_0\) Calculating the integral: \(Q = 80 \left[ (2+2)\cdot(2+2)\right] \cdot \epsilon_0 = 80\cdot 16 \cdot \epsilon_0 = 1280\epsilon_0\)

The total charge contained within the cube is \(Q = 1280\epsilon_0\)