Question

the cube in (figure 1) has sides of length l = 10.0 cm. the electric field is uniform, has a magnitude e = 4.00 × 10³ n/c, and is parallel to the xy-plane at an angle of 36.9° measured from the +x - axis toward the +y - axis.

part e
what is the electric flux through the cube face s₅?
express your answer in newton times meters squared per coulomb.
φ₅ =
n·m²/c

part f
content partially visible, likely a similar flux question for another face

part g
what is the electric flux through the cube face s₆?
express your answer in newton times meters squared per coulomb.
φ₆ =
n·m²/c

Explanation:

Let's solve for the electric flux through face \( S_6 \) (back face) of the cube.

Step 1: Recall the formula for electric flux

The electric flux \( \Phi \) through a surface is given by \( \Phi = \vec{E} \cdot \vec{A} = EA\cos\theta \), where \( E \) is the magnitude of the electric field, \( A \) is the area of the surface, and \( \theta \) is the angle between the electric field vector and the normal to the surface.

Step 2: Determine the area of the cube face

The length of each side of the cube is \( L = 10.0 \, \text{cm} = 0.100 \, \text{m} \). The area of one face of the cube is \( A = L^2 \).
\[
A = (0.100 \, \text{m})^2 = 0.0100 \, \text{m}^2
\]

Step 3: Determine the angle \( \theta \) for face \( S_6 \)

The electric field is parallel to the \( xy \)-plane and makes an angle of \( 36.9^\circ \) with the \( +x \)-axis. For the back face \( S_6 \), the normal to the surface is along the \( -z \)-direction (assuming the cube is oriented with \( S_5 \) as the front face along \( +z \) and \( S_6 \) as the back face along \( -z \)). Since the electric field is in the \( xy \)-plane, the angle between the electric field (in \( xy \)-plane) and the normal to \( S_6 \) (along \( -z \)) is \( 90^\circ \) (because the electric field has no \( z \)-component, so it's perpendicular to the normal of \( S_6 \)).

Step 4: Calculate the electric flux through \( S_6 \)

Using the flux formula \( \Phi = EA\cos\theta \), with \( \theta = 90^\circ \) (so \( \cos 90^\circ = 0 \)):
\[
\Phi_6 = E \cdot A \cdot \cos(90^\circ)
\]
We know \( E = 4.00 \times 10^3 \, \text{N/C} \) and \( A = 0.0100 \, \text{m}^2 \), and \( \cos(90^\circ) = 0 \).
\[
\Phi_6 = (4.00 \times 10^3 \, \text{N/C}) \times (0.0100 \, \text{m}^2) \times 0 = 0 \, \text{N·m}^2/\text{C}
\]

Answer:

\( \boxed{0} \)