The problem involves calculating the magnitude $E$ of the electric field given the electric flux $\Phi_E = 39.5 \, \text{N} \cdot \text{m}^2 / \text{C}$, the area of the square, and the angle $\theta = 145^\circ$ between the electric field and the normal vector $\hat{n}$.

The electric flux is related to the electric field by the equation:

$\Phi_E = E A \cos(\theta)$

where:

• $\Phi_E$ is the electric flux.
• $E$ is the magnitude of the electric field.
• $A$ is the area of the square.
• $\theta$ is the angle between the electric field and the normal to the surface.

Given:

• $\Phi_E = 39.5 \, \text{N} \cdot \text{m}^2 / \text{C}$
• $A = (0.100 \, \text{m})^2 = 0.0100 \, \text{m}^2$
• $\theta = 145^\circ$

Rearranging the equation to solve for $E$:

$E = \frac{\Phi_E}{A \cos(\theta)}$

Substituting the known values:

$E = \frac{39.5}{0.0100 \cdot \cos(145^\circ)}$

Now, calculating the value of $\cos(145^\circ)$:

$\cos(145^\circ) = -\cos(35^\circ) \approx -0.8192$

Thus:

$E = \frac{39.5}{0.0100 \cdot (-0.8192)}$

Let's calculate $E$. math

Given values

flux = 39.5 # N·m^2/C side_length = 0.100 # m theta_deg = 145 # degrees

Calculate the area of the square

area = side_length ** 2

Calculate the cosine of the angle

theta_rad = math.radians(theta_deg) cos_theta = math.cos(theta_rad)

Calculate the electric field magnitude E

E = flux / (area * cos_theta) EThe calculated magnitude of the electric field is approximately $4822 \, \text{N/C}$. The negative sign indicates that the direction of the electric field is opposite to the direction of the normal vector. Since we are looking for the magnitude, we take the absolute value:

$E = 4822 \, \text{N/C}$

Would you like further clarification or more questions on this topic? Here are some related questions you might find useful:

1. What is the concept of electric flux and how is it related to the electric field?
2. How does the angle $\theta$ affect the electric flux through a surface?
3. Can electric field magnitude be determined without knowing the angle?
4. How do different surface orientations affect the calculation of electric flux?
5. What other scenarios could you calculate electric flux for using different geometries?

Tip: Always double-check the angle's reference direction to ensure correct sign when calculating flux.