Here is a surprising fact about electric flux. Put a point charge $q$ at the center of an imaginary sphere of radius $r$. The field on the sphere is $kq/r^2$, straight out through every patch, and the sphere's area is $4\pi r^2$. So the flux is $$\Phi_E = \frac{kq}{r^2}\cdot 4\pi r^2 = 4\pi k q = \frac{q}{\varepsilon_0}.$$ The radius dropped out. A bigger sphere has a weaker field over a larger area, and the two changes cancel exactly, because the field falls as $1/r^2$ and area grows as $r^2$.
Key idea
Why it works
Charges outside: a line from an outside charge that enters the surface must leave it again. It adds negative flux where it enters and the same positive flux where it leaves. The field on the surface changes, but the total does not.
Many charges: fields add, so fluxes add: $\Phi = (q_1 + q_2 + \dots)/\varepsilon_0$ over the charges inside.
Worked examples
Worked example: Counting what is inside
- Only the inside charges count: $q_{\text{enc}} = 4.0 - 1.0 = 3.0\ \text{nC}$.
- $\Phi_E = \dfrac{3.0 \times 10^{-9}}{8.85 \times 10^{-12}} = 339\ \text{N·m²/C}$.
- $q_3$ is close and strong, but its lines go in and come back out.
Answer: $\Phi_E \approx 339\ \text{N·m²/C}$
Worked example: One face of a cube
- The whole cube gets $q/\varepsilon_0$. The six faces are identical as seen from the center, so they share it equally.
- $\Phi_{\text{face}} = \dfrac{q}{6\varepsilon_0} = \dfrac{12 \times 10^{-9}}{6(8.85 \times 10^{-12})} = 226\ \text{N·m²/C}$.
Answer: $\Phi \approx 226\ \text{N·m²/C}$
Worked example: A charge at a corner
- Eight cubes meeting at that corner would surround the charge completely and share $q/\varepsilon_0$ equally.
- So one cube gets $\dfrac{q}{8\varepsilon_0} = 113\ \text{N·m²/C}$. (The three faces touching the corner get none: the field skims along them.)
Answer: $\Phi \approx 113\ \text{N·m²/C}$
Common mistake