Flux of an Electric Field — Question 9

Question 9

A curved surface in a height-dependent field cannot use projection alone

A paraboloid cap is $z=h(1-r^2/R^2)$ above its circular base $z=0$, with $0\le r\le R$. Its normal points upward. The prescribed field is $\mathbf E=\lambda z\hat{\mathbf z}$ with $\lambda>0$.

1. Use $n_zdA=dA_{xy}$ to integrate flux through the cap.

2. Include the base and compare the closed flux with $\lambda$ times the enclosed volume. Explain why multiplying projected area by the base field fails.

Question 9 - Solution

A curved surface in a height-dependent field cannot use projection alone

(a) Method and calculation

$$\Phi_{\mathrm{cap}}=\int_0^R\lambda h\left(1-\frac{r^2}{R^2}\right)2\pi r\,dr=\frac{\lambda\pi R^2h}{2}.$$$$E_z(0)=0\Rightarrow\Phi_{\mathrm{base}}=0.$$

(b) Interpretation and further calculation

$$V=\int_0^Rh(1-r^2/R^2)2\pi r\,dr=\frac{\pi R^2h}{2}.$$$$\Phi_{\mathrm{closed}}=\lambda V.$$Projected area handles the orientation factor, but the field must still be evaluated at the actual height of each cap element.

Physical check

The flux vanishes as $h\to0$, when the cap collapses into the zero-field base plane.