Flux of an Electric Field — Question 7

Question 7

Closing a tetrahedron determines the flux through its oblique face

A tetrahedron occupies $x,y,z\ge0$ and $x/a+y/b+z/c\le1$. A uniform field is $\mathbf E=(E_x,E_y,E_z)$. All face normals point outward. The vector areas of a closed surface sum to zero.

1. Use the three coordinate-plane faces to derive the vector area of the oblique face.

2. Calculate its flux for $a=2$ m, $b=3$ m, $c=4$ m and $\mathbf E=(5,-2,1)$ N/C. Verify the full closed flux.

Question 7 - Solution

Closing a tetrahedron determines the flux through its oblique face

(a) Method and calculation

$$\mathbf A_x=-\tfrac12bc\hat{\mathbf x},\quad\mathbf A_y=-\tfrac12ac\hat{\mathbf y},\quad\mathbf A_z=-\tfrac12ab\hat{\mathbf z}.$$$$\mathbf A_{\mathrm{oblique}}=\tfrac12(bc,ac,ab).$$

(b) Interpretation and further calculation

$$\Phi_{\mathrm{oblique}}=\tfrac12(E_xbc+E_yac+E_zab)=30-8+3=25\,\mathrm{N\,m^2/C}.$$$$\Phi_{\mathrm{coordinate}}=-25\,\mathrm{N\,m^2/C},\quad\Phi_{\mathrm{closed}}=0.$$

Physical check

The oblique face area vector already contains the projection factors; multiplying by an extra cosine would count them twice.