© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Recall Gauss’s Law
Gauss’s law states that the total electric flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space:
\Phi_{\mathrm{total}} \;=\; \frac{q_{\mathrm{enc}}}{\varepsilon_0}.
Here, q_{\mathrm{enc}} is the net charge enclosed by the surface, and \varepsilon_0 is the permittivity of free space.
Step 2: Identify the Electric Field and Cuboid Dimensions
The given electric field is
\vec{E} \;=\; 2\,x^{2}\,\hat{i} \;-\; 4\,y\,\hat{j} \;+\; 6\,\hat{k} \quad \text{(in N/C)}.
The dimensions of the cuboid are 1 \times 2 \times 3\,\text{m}^3. Let us assume:
x -dimension: 0 \rightarrow 1\,\text{m}
y -dimension: 0 \rightarrow 2\,\text{m}
z -dimension: 0 \rightarrow 3\,\text{m}
Step 3: Calculate the Flux Through Each Pair of Opposite Faces
(a) Flux through the faces parallel to the y\!-\!z plane
These faces are at x = 0 and x = 1. The outward normals to these faces point along \pm \hat{i}.
At x = 0 : The electric field’s x -component is E_x = 2\,(0)^2 = 0. Thus, flux = 0 \times \text{(area)} = 0.
At x = 1 : The electric field’s x -component is E_x = 2\,(1)^2 = 2. The area of each face parallel to y\!-\!z is 2 \times 3 = 6\,\text{m}^2. Because the outward normal is \hat{i}, the flux is
\Phi_{x=1} \;=\; (2)\times 6 \;=\; 12 \,\text{N}\cdot\text{m}^2/\text{C}.
Hence, the total flux through these two faces is 12 - 0 = 12.
(b) Flux through the faces parallel to the x\!-\!z plane
These faces are at y = 0 and y = 2. The outward normals to these faces point along \pm \hat{j}.
At y = 0 : The electric field’s y -component is E_y = -4\,(0) = 0. So the flux at y=0 is 0.
At y = 2 : The electric field’s y -component is E_y = -4\,(2) = -8. The area of each face parallel to x\!-\!z is 1 \times 3 = 3\,\text{m}^2. The outward normal is +\hat{j}, so
\Phi_{y=2} \;=\; (-8)\times 3 \;=\; -24\,\text{N}\cdot\text{m}^2/\text{C}.
Hence, the total flux through these two faces is -24 - 0 = -24.
(c) Flux through the faces parallel to the x\!-\!y plane
These faces are at z = 0 and z = 3. The outward normals to these faces point along \pm \hat{k}.
At z = 0 : E_z = 6. The area is 1 \times 2 = 2\,\text{m}^2. The outward normal for the z=0 face is -\hat{k} , giving flux (6)\times 2 \times \cos(180^\circ) = -12.
At z = 3 : E_z = 6. The area is also 2\,\text{m}^2, and the outward normal is +\hat{k}, giving flux (6)\times 2 = 12.
Hence, the net flux through these two faces is 12 + (-12) = 0.
Step 4: Sum of the Total Electric Flux
Adding up the contributions from all three pairs of opposite faces:
\Phi_{\text{total}} \;=\; 12 + (-24) + 0 \;=\; -12\,\text{N}\cdot\text{m}^2/\text{C}.
Step 5: Relate Total Flux to Enclosed Charge
According to Gauss’s law, we have
\Phi_{\text{total}} \;=\; \frac{q_{\mathrm{enc}}}{\varepsilon_0}.
Thus,
-12 \;=\; \frac{q_{\mathrm{enc}}}{\varepsilon_0} \quad \Longrightarrow \quad q_{\mathrm{enc}} \;=\; -12\,\varepsilon_0.
The question asks for the magnitude of the enclosed charge in the form n\,\varepsilon_0, so
\bigl|q_{\mathrm{enc}}\bigr| \;=\; 12\,\varepsilon_0.
Hence, n = 12.
Final Answer: n = 12
