Final Answer:
The electric flux through the shaded surface is Φ = 2π * (2 + 3i) * A, where A is the area of the shaded surface.
Step-by-step explanation:
To find the electric flux through the shaded surface, we first need to find the enclosed charge. Since the charge distribution is a rectangular box with a height of 2 units, a width of 3 units, and a depth of 1 unit, the enclosed charge is given by:
Q = ∫∫∫(ρ(x,y,z) d^3r
where ρ(x,y,z) is the charge density at each point in space. Since the charge density is uniform and given by ρ(x,y,z) = q/V, where V is the volume of the box, we have:
Q = ∫∫∫(q/V d^3r
Since the volume of the box is V = 2 * 3 * 1 = 6, we have:
Q = ∫∫∫(q/6 d^3r
Now, we need to integrate over the entire volume of the box. The integration limits are from 0 to 2 in the x and y directions, and from 0 to 1 in the z direction. Therefore, we have:
Q = ∫0^2 ∫0^2 ∫0^1 (q/6) d^3r
Using the appropriate integration rules, we can simplify the integral to:
Q = 2 * 3 * 1 * q = 6 * q
Next, we need to find the electric flux through the shaded surface. The electric flux is given by the equation:
Φ = ∫∫∫(E * dA)
where E is the electric field at each point in space, and dA is the area element of the surface. Since the electric field is given by:
E = -∇(k)
where k is the charge density, we have:
E = -∇(q/6)
Now, we need to integrate the electric field over the entire surface of the box. The surface of the box has an area of A = 2 * 3 * 1 = 6. Therefore, we have:
Φ = ∫∫∫(E * dA)
= ∫∫∫(-∇(q/6) * dA)
= -∫∫∫(q/6 * dA)
Since the area element dA is given by:
dA = dx * dy * dz
we have:
Φ = -∫∫∫(q/6 * dA)
= -∫0^2 ∫0^2 ∫0^1 (q/6) dx * dy * dz
Using the same integration rules as before, we can simplify the integral to:
Φ = -2 * 3 * 1 * q = -6 * q
Therefore, the electric flux through the shaded surface is Φ = -6 * q.