Gauss' Law — Question 8

Question 8

Deforming a Gaussian surface changes flux only by swept charge

A closed surface is expanded into a second surface without losing any previously enclosed region. The added region has volume $\Delta V$ and uniform volume charge density $\rho$. Arbitrary fixed charges may also lie inside the original surface and outside the final one.

1. Derive the change in outward flux and evaluate its sign for negative $\rho$.

2. Now form the boundary of just the added region, including the original inner surface. Explain the orientation of that inner boundary and recover the same flux difference.

Question 8 - Solution

Deforming a Gaussian surface changes flux only by swept charge

(a) Method and calculation

$$Q_2=Q_1+\rho\Delta V,\quad\Phi_i=Q_i/\epsilon_0.$$$$\Delta\Phi=\Phi_2-\Phi_1=\rho\Delta V/\epsilon_0<0\quad(\rho<0).$$

(b) Interpretation and further calculation

$$\hat{\mathbf n}_{\mathrm{inner,layer}}=-\hat{\mathbf n}_{1},\quad\Phi_{\mathrm{layer}}=\Phi_2-\Phi_1.$$$$\Phi_{\mathrm{layer}}=Q_{\mathrm{layer}}/\epsilon_0=\rho\Delta V/\epsilon_0.$$The inner boundary normal points into the removed central volume, outward from the layer itself.

Physical check

Surface shape is irrelevant to the difference once the added charge is specified; no field symmetry is required.