The problem describes an electric field $\mathbf{E} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ (in units of $\text{N/C}$) and asks for the flux through the three surfaces shown, each forming a square of side length $L = 20 \text{ cm}$.

To calculate the electric flux $\Phi$ through each surface, we use the formula:

$\Phi = \mathbf{E} \cdot \mathbf{A}$

Where:

• $\mathbf{E}$ is the electric field vector.
• $\mathbf{A}$ is the area vector of the surface, pointing outward and perpendicular to the surface. Its magnitude is $A = L^2$ and its direction corresponds to the surface's orientation.

Step 1: Calculate the Area

Given $L = 20 \text{ cm} = 0.2 \text{ m}$:

$A = L^2 = (0.2)^2 = 0.04 \text{ m}^2$

Step 2: Find the Flux Through Each Surface

There are three surfaces perpendicular to the $x$, $y$, and $z$ axes. Let's find the flux for each:

1. Surface Perpendicular to the $x$-axis:

   • The normal vector $\mathbf{A}_x = -\mathbf{i} \cdot A$ (points inward).
   • $\mathbf{E} \cdot \mathbf{A}_x = \mathbf{E} \cdot (-\mathbf{i})A = -(2) \cdot 0.04 = -0.08 \text{ Nm}^2/\text{C}$

2. Surface Perpendicular to the $y$-axis:

   • The normal vector $\mathbf{A}_y = -\mathbf{j} \cdot A$ (points inward).
   • $\mathbf{E} \cdot \mathbf{A}_y = \mathbf{E} \cdot (-\mathbf{j})A = -(-1) \cdot 0.04 = 0.04 \text{ Nm}^2/\text{C}$

3. Surface Perpendicular to the $z$-axis:

   • The normal vector $\mathbf{A}_z = -\mathbf{k} \cdot A$ (points inward).
   • $\mathbf{E} \cdot \mathbf{A}_z = \mathbf{E} \cdot (-\mathbf{k})A = -(3) \cdot 0.04 = -0.12 \text{ Nm}^2/\text{C}$

Step 3: Sum the Flux Values

$\Phi_{\text{total}} = \Phi_x + \Phi_y + \Phi_z = -0.08 + 0.04 - 0.12 = -0.16 \text{ Nm}^2/\text{C}$

The total electric flux through the three surfaces is $\boxed{-0.16 \text{ Nm}^2/\text{C}}$.

Would you like further details on any part of this solution?

Related Questions:

1. What is the physical significance of electric flux in this context?
2. How would the result change if the electric field was different?
3. What would happen if the surfaces were oriented differently relative to the electric field?
4. How does the orientation of the area vector affect the sign of the flux?
5. How is the concept of flux applied in Gauss's law?

Tip: Remember that the direction of the area vector is crucial in determining the sign of the electric flux. Always ensure it points outward from the surface.