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Let's solve the problem step by step. The information provided includes:

1. The electric field \( \vec{E} \) is parallel to the \( x \)-axis, meaning \( E_y = E_z = 0 \) and only \( E_x \) is non-zero.
2. The value of \( E_x \) at \( x = 0 \) (on the \( yz \)-plane) is \( E_x = 125 \, \text{N/C} \).
3. There are two surfaces, labeled I and II, through which we need to calculate the electric flux.
4. A volume containing a charge of \( -24.0 \, \text{nC} \) is mentioned.
5. We need to determine the magnitude and direction of \( E \) on the opposite face of the block, as well as whether the electric field is produced by charges inside or outside the block.

Let's tackle each part of the question.

### Part (a): Electric Flux through Surface I
The electric flux \( \Phi \) through a surface is given by:

\[
\Phi = \int \vec{E} \cdot d\vec{A}
\]

For surface I, which is perpendicular to the \( x \)-axis, the area vector \( d\vec{A} \) points in the \( x \)-direction and has a magnitude of \( A = 3.0 \, \text{m} \times 2.0 \, \text{m} = 6.0 \, \text{m}^2 \).

Since \( E_x = 125 \, \text{N/C} \) at \( x = 0 \), the flux through surface I is:

\[
\Phi_I = E_x \times A = 125 \, \text{N/C} \times 6.0 \, \text{m}^2 = 750 \, \text{N} \cdot \text{m}^2/\text{C}
\]

### Part (b): Electric Flux through Surface II
For surface II, we need to determine \( E_x \) at \( x = 1.0 \, \text{m} \). This value could vary depending on the charge distribution within the block. 

If the electric field is constant (i.e., if \( E_x \) remains \( 125 \, \text{N/C} \) across the entire region), the flux through surface II would be:

\[
\Phi_{II} = -E_x \times A = -125 \, \text{N/C} \times 6.0 \, \text{m}^2 = -750 \, \text{N} \cdot \text{m}^2/\text{C}
\]

The negative sign indicates that the flux is in the opposite direction to the flux through surface I.

### Part (c): Electric Field on the Opposite Face
To find the electric field on the opposite face (at \( x = 1.0 \, \text{m} \)), we need to consider Gauss's law:

\[
\Phi_{\text{total}} = \frac{q_{\text{enc}}}{\epsilon_0}
\]

where \( q_{\text{enc}} = -24.0 \, \text{nC} = -24.0 \times 10^{-9} \, \text{C} \).

The total flux is the sum of the fluxes through surfaces I and II, which should equal \( \frac{q_{\text{enc}}}{\epsilon_0} \). Solving this will give us the exact value of \( E_x \) on the opposite face.

### Part (d): Determining the Source of the Electric Field
To determine whether the electric field is caused by charges inside or outside the block, we can use Gauss’s law and the fact that the net flux through the entire closed surface (all sides of the block) should match the enclosed charge divided by \( \epsilon_0 \). 

If there is no flux through other faces, it suggests that the electric field observed is due to the charge inside the block itself, as any external field would not contribute net flux across only two faces.

El campo eléctrico E en la figura es paralelo en todo lugar al eje x, por lo que las componentes Ey y Ez son iguales a cero. La componente x del campo Ex depende de x, pero no de y ni de z. En los puntos del plano yz (donde x = 0), Ex = 125 N/C. ¿Cuál es el flujo eléctrico a través de la superficie I en la figura? b) ¿Cuál es el flujo eléctrico a través de la superficie II? c) Si el volumen ilustrado tiene una carga total de -24.0 nC, ¿cuáles son la magnitud y dirección de E en la cara opuesta a la superficie I? d) ¿El campo eléctrico es producido por cargas dentro del bloque o también se debe a cargas fuera del bloque?

Solve an advanced physics problem involving electric flux through surfaces I and II in a rectangular block with an electric field along the x-axis. The problem includes applying Gauss's Law, calculating electric flux, and determining the electric field’s direction and magnitude based on a given charge of -24.0 nC inside the block. Ideal for high school students in Grades 11-12.

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