Section 2.8: Flux via Surface Integrals
Electric flux measures the "amount" of electric field passing through a surface. It is calculated using surface integrals of the electric field vector over the surface.
\[ \Phi_E = \iint_S \vec{E} \cdot d\vec{A} \] where \( \vec{E} \) is the electric field and \( d\vec{A} \) is the infinitesimal area vector on the surface \( S \).
Flux depends on the orientation of the surface relative to the field. If the field is perpendicular to the surface, flux is maximized; if parallel, flux is zero.
Example 1
Compute the flux through a flat square surface of area \( A \) perpendicular to a uniform field \( \vec{E} \).
\[ \Phi_E = \vec{E} \cdot \vec{A} = EA \]
Since the surface is perpendicular to the field, the dot product simplifies to the product of magnitudes.
Practice Problems
- Define electric flux in words and explain its physical meaning.
- Calculate the flux through a square surface tilted at 45° to a uniform electric field \( E = 10 \, \text{N/C} \) with area \( A = 2 \, \text{m²} \).
- For a cube of side \( L \) in a uniform field \( \vec{E} \), find the total flux through the cube.
- Explain the difference between flux through open vs closed surfaces.
- Show that flux through a closed surface enclosing zero charge is zero.