Question 23.4: Flux Through a Cube Consider a uniform electric field E orie...
Flux Through a Cube
Consider a uniform electric field \overrightarrow{E} oriented in the x direction in empty space. A cube of edge length \ell is placed in the field, oriented as shown in Figure 23.9. Find the net electric flux through the surface of the cube.
Conceptualize Examine Figure 23.9 carefully. Notice that the electric field lines pass through two faces perpendicularly and are parallel to four other faces of the cube.
Categorize We evaluate the flux from its definition, so we categorize this example as a substitution problem.
The flux through four of the faces (➂, ➃, and the unnumbered faces) is zero because \overrightarrow{E} s parallel to the four faces and therefore perpendicular to d \overrightarrow{A} on these faces.
Write the integrals for the net flux through faces ➀ and ➁:
\Phi_E=\int_1 \overrightarrow{E} \cdot d \overrightarrow{A}+\int_2 \overrightarrow{E} \cdot d \overrightarrow{A}For face ➀, \overrightarrow{E} is constant and directed inward but d \overrightarrow{A}_1 is directed outward \left(\theta=180^{\circ}\right). Find the flux through this face:
\int_1 \overrightarrow{E} \cdot d\overrightarrow{A}=\int_1 E\left(\cos 180^{\circ}\right) d A=-E \int_1 d A=-E A=-E \ell^2For face ➁, \overrightarrow{E} is constant and outward and in the same direction as d \overrightarrow{A}_2\left(\theta=0^{\circ}\right). Find the flux through this face:
\int_2 \overrightarrow{E} \cdot d\overrightarrow{A}=\int_2 E\left(\cos 0^{\circ}\right) d A=E \int_2 d A=+E A=E \ell^2Find the net flux by adding the flux over all six faces:
\Phi_E=-E \ell^2+E \ell^2+0+0+0+0=0In the next section, we generate a fundamental principle that explains this zero value.