Flux of an Electric Field — Question 1

Question 1

A quadratic normal field makes a midpoint estimate systematically low

On the rectangle $0\le x\le L$, $0\le y\le W$, $z=0$, the electric field has normal component $E_z=E_0(1+x^2/L^2)$. Choose the positive-z normal; tangential components need not vanish.

1. Integrate the electric flux and calculate the mean normal field.

2. Compare with using the field at the rectangle center multiplied by its area; report the fractional error relative to the exact flux.

Question 1 - Solution

A quadratic normal field makes a midpoint estimate systematically low

(a) Method and calculation

$$\Phi=\int_0^W\!\int_0^L E_0(1+x^2/L^2)\,dx\,dy=\frac43E_0LW.$$$$\langle E_n\rangle=\Phi/(LW)=\frac43E_0.$$

(b) Interpretation and further calculation

$$\Phi_{\mathrm{mid}}=E_z(L/2)LW=\frac54E_0LW.$$$$\frac{\Phi-\Phi_{\mathrm{mid}}}\Phi=\frac{4/3-5/4}{4/3}=\frac1{16}=6.25\%.$$The midpoint rule underestimates this convex normal-field profile.

Physical check

Electric flux has units N m$^2$/C; tangential components do not contribute to this planar integral.