Question

A point charge of $2.0\,\mu\text{C}$ is at the centre of a cubic Gaussian surface $9.0\,\text{cm}$ on edge. What is the net electric flux through the surface?

Hint:

Use Gauss's law: net flux = q/epsilon_0. The flux is independent of the size of the cube, so the edge length is not needed.

Answer 1

Step 1: State Gauss's law. The net electric flux through any closed surface equals the total charge enclosed divided by \(\epsilon_0\):
\[\Phi = \frac{q}{\epsilon_0}\]
Step 2: Note that the flux depends only on the enclosed charge, not on the size or shape of the Gaussian surface. So the edge length of \(9.0\,\text{cm}\) is not needed.
Step 3: Substitute \(q = 2.0\,\mu\text{C} = 2.0\times10^{-6}\,\text{C}\) and \(\epsilon_0 = 8.85\times10^{-12}\,\text{C}^2\,\text{N}^{-1}\text{m}^{-2}\):
\[\Phi = \frac{2.0\times10^{-6}}{8.85\times10^{-12}}\]
Step 4: Do the arithmetic:
\[\Phi = 2.26\times10^{5}\,\text{N m}^2\,\text{C}^{-1}\]
\[\boxed{\Phi = 2.26\times10^{5}\,\text{N m}^2\,\text{C}^{-1}}\]

Answer 2

Flux-scaling argument.
Step 1: For a point charge at the centre of a cube, the field spreads equally through all 6 faces by symmetry. The flux through one face is \(\Phi_1 = \dfrac{q}{6\epsilon_0}\).
Step 2: The total flux through the whole cube is just 6 times the per-face flux:
\[\Phi = 6\,\Phi_1 = 6\times\frac{q}{6\epsilon_0} = \frac{q}{\epsilon_0}\]
This confirms the size of the cube is irrelevant; the \(6\) cancels.
Step 3: Substitute \(q = 2.0\times10^{-6}\,\text{C}\):
\[\Phi = \frac{2.0\times10^{-6}}{8.85\times10^{-12}} = 2.26\times10^{5}\,\text{N m}^2\,\text{C}^{-1}\]
\[\boxed{\Phi = 2.26\times10^{5}\,\text{N m}^2\,\text{C}^{-1}}\]