Gauss' Law — Question 1

Question 1

Large exterior charges can change the field without changing enclosed flux

Use $\epsilon_0=8.854\times10^{-12}\,\mathrm{C^2/(N\,m^2)}$. A closed Gaussian surface contains charges $+5.00$ nC, $-8.00$ nC and $-2.00$ nC. A separate $+100$ nC charge is outside. None lies on the surface.

1. Find the net outward electric flux.

2. The exterior charge doubles while all three interior charges stay fixed. Find the change in total flux and explain whether the field at individual surface points must stay fixed.

Question 1 - Solution

Large exterior charges can change the field without changing enclosed flux

(a) Method and calculation

$$Q_{\mathrm{enc}}=(5-8-2)\,\mathrm{nC}=-5.00\times10^{-9}\,\mathrm C.$$$$\Phi=\frac{Q_{\mathrm{enc}}}{\epsilon_0}=\frac{-5.00\times10^{-9}}{8.854\times10^{-12}}\approx-564.716\,\mathrm{N\,m^2/C}.$$

(b) Interpretation and further calculation

$$\Delta Q_{\mathrm{enc}}=0\Rightarrow\Delta\Phi=0.$$$$\mathbf E_{\mathrm{total}}=\mathbf E_{\mathrm{inside}}+\mathbf E_{\mathrm{outside}},\quad\oint\Delta\mathbf E_{\mathrm{outside}}\cdot d\mathbf A=0.$$The external source generally changes the local field, while its net closed-surface contribution remains zero.

Physical check

Negative total flux means net inward field flux, not inward field at every point.