Flux of an Electric Field — Question 2

Question 2

A triangular aperture samples a linear field at its centroid

A triangular surface lies in $z=0$ with vertices $(0,0)$, $(a,0)$ and $(0,b)$. Its normal is positive z. The field on it is $E_z=E_0+\alpha x+\beta y$.

1. Integrate the flux over the triangle in terms of the given parameters.

2. Show that the same result equals area times field at the centroid, and explain why that shortcut is not generally exact for a quadratic field.

Question 2 - Solution

A triangular aperture samples a linear field at its centroid

(a) Method and calculation

$$\Phi=\int_0^a\!\int_0^{b(1-x/a)}(E_0+\alpha x+\beta y)\,dy\,dx.$$$$\Phi=\frac{abE_0}{2}+\frac{\alpha a^2b}{6}+\frac{\beta ab^2}{6}.$$

(b) Interpretation and further calculation

$$A=ab/2,\quad(x_c,y_c)=(a/3,b/3).$$$$\Phi=A\left(E_0+\frac{\alpha a}{3}+\frac{\beta b}{3}\right)=AE_z(x_c,y_c).$$A linear field depends only on the first area moments; a quadratic profile also depends on second moments.

Physical check

If both gradients vanish, the expression reduces to $E_0ab/2$.