A cube has sides of length L=10.0cm, the electric field is uniform, has a magnitude E=4.00×103N/C, and is parallel to the xy-plane at an angle of 36.9∘ measured from the +x−axis toward the +y−axis. What is the electric flux through the left-side cube face?

The electric flux (\(\Phi_E\)) through a surface is given by the equation:

\[
\Phi_E = \vec{E} \cdot \vec{A} = EA \cos \theta
\]

Where:
- \( \vec{E} \) is the electric field vector,
- \( \vec{A} \) is the area vector, which is perpendicular to the surface,
- \( A \) is the area of the surface,
- \( \theta \) is the angle between the electric field and the area vector.

### Step 1: Understanding the problem

- **Cube side length, \(L = 10.0 \, \text{cm} = 0.1 \, \text{m}**:**
  The cube has six faces. We are interested in the flux through the left-side face, which is the face perpendicular to the \(x\)-axis (where the area vector points in the negative \(x\)-direction).

- **Electric field \(E = 4.00 \times 10^3 \, \text{N/C}\)**:
  The electric field is in the \(xy\)-plane, at an angle of \(36.9^\circ\) from the \(+x\)-axis toward the \(+y\)-axis.

### Step 2: Area of the cube's left face

The area of the cube's face is:
\[
A = L^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2
\]

### Step 3: Direction of the electric field and the angle to the area vector

The electric field makes an angle of \(36.9^\circ\) with the \(+x\)-axis in the \(xy\)-plane. The left face of the cube has an area vector that points in the \(-x\)-direction, so the angle between the electric field and the normal to the left face is:
\[
\theta = 180^\circ - 36.9^\circ = 143.1^\circ
\]

### Step 4: Electric flux through the left-side cube face

Now, we can calculate the electric flux through the left face using the formula for flux:

\[
\Phi_E = EA \cos \theta
\]

Substitute the known values:

\[
\Phi_E = (4.00 \times 10^3 \, \text{N/C})(0.01 \, \text{m}^2) \cos(143.1^\circ)
\]

We need the cosine of \(143.1^\circ\):

\[
\cos(143.1^\circ) \approx -0.8
\]

Therefore:

\[
\Phi_E = (4.00 \times 10^3)(0.01)(-0.8)
\]

\[
\Phi_E = -32 \, \text{N} \cdot \text{m}^2/\text{C}
\]

### Final Answer:
The electric flux through the left-side cube face is \( \boxed{-32 \, \text{N} \cdot \text{m}^2/\text{C}} \).

Would you like more details or have any further questions?

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Here are 5 related questions for further exploration:

1. What would be the electric flux through the right-side face of the cube?
2. How does the electric flux change if the angle of the electric field changes?
3. What is the total electric flux through all faces of the cube?
4. How would the flux change if the cube's side length doubled?
5. How is the concept of electric flux used in Gauss's Law?

**Tip:** Always pay attention to the angle between the electric field and the normal to the surface when calculating flux—it can drastically affect the result.

A cube has sides of length L=10.0cm, the electric field is uniform, has a magnitude E=4.00×10^3N/C, and is parallel to the xy-plane at an angle of 36.9° measured from the +x−axis toward the +y−axis. What is the electric flux through the left-side cube face?

Learn how to calculate the electric flux through a cube face with a uniform electric field of magnitude E=4.00×10^3 N/C, parallel to the xy-plane at an angle of 36.9°. This solution covers key concepts such as vector dot products, the formula Φ_E = EAcosθ, and geometry principles. Suitable for high school physics or early university students.

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