Given the electric field in the region E = 2x î, find the net electric flux through the cube and the charge enclosed by it.

To find the net electric flux through a closed surface and the charge enclosed by it, you can use Gauss's Law. Gauss's Law relates the electric flux (Φ) through a closed surface to the charge enclosed (Q) by that surface. The formula for Gauss's Law is:

Φ = ∮ E ⋅ dA = Q / ε₀

Where:

Φ is the electric flux through a closed surface.
E is the electric field.
dA is an infinitesimal area vector element on the closed surface.
Q is the charge enclosed by the closed surface.
ε₀ is the permittivity of free space (a constant).
In your case, you're given the electric field E = 2x î (where î is the unit vector in the x-direction). To find the net electric flux through a cube, let's assume the cube has six faces, and each face has an area A.

Now, consider a cube centered at the origin with sides of length 'a.' Since the electric field E varies with position, you'll need to integrate over each face of the cube separately to find the total electric flux. The electric field is constant on each face of the cube, so you can find the flux through each face and then sum them up.

Flux through the front face (x = a/2):
Φ1 = ∮ E ⋅ dA = ∫(2x î) ⋅ dA
The normal vector (dA) for the front face is î, and the limits of integration for x are from -a/2 to a/2. So the integral becomes:

Φ1 = ∫(2x î) ⋅ î dA
Φ1 = ∫2x dA from x = -a/2 to x = a/2

Calculate Φ1 using these limits.

Flux through the back face (x = -a/2):
Φ2 = ∮ E ⋅ dA = ∫(2x î) ⋅ (-î) dA
The normal vector (dA) for the back face is -î, and the limits of integration for x are from -a/2 to a/2. So the integral becomes:

Φ2 = ∫(2x î) ⋅ (-î) dA
Φ2 = -∫2x dA from x = -a/2 to x = a/2

Calculate Φ2 using these limits.

Flux through the remaining four faces (left, right, top, and bottom):
Since the electric field is parallel to these faces, the dot product E ⋅ dA is zero, and there is no contribution to the flux from these faces.
Now, calculate the total electric flux (Φ_total) by summing up Φ1 and Φ2.

Φ_total = Φ1 + Φ2

Once you have the total electric flux, you can use Gauss's Law to find the enclosed charge Q:

Φ_total = Q / ε₀

Solve for Q:

Q = Φ_total * ε₀

Now you have the net electric flux through the cube and the charge enclosed by it.