Gaussian Surface and its Properties

An imaginary, arbitrary closed three-dimensional surface chosen to calculate the electric flux due to a charge distribution using Gauss's law.

Valid Gaussian surfaces: sphere, cylinder, cube (all closed surfaces).
Invalid Gaussian surfaces: disc, square, open hemisphere (not closed).

• Applies to any closed surface, regardless of shape.
• Total flux = algebraic sum of all enclosed charges divided by ε₀​.
• Independent of the position of charge within the surface (only total enclosed charge matters).
• Especially useful for high-symmetry charge distributions (spherical, cylindrical, planar).
• Zero flux does not imply zero electric field — external charges can still produce a field at the surface.
• Directly derivable from and consistent with Coulomb's Law.

Given: Point charge q = 5 × 10⁻⁶ C at the center of a sphere in vacuum.
Find: Electric flux through the sphere; flux if placed in a dielectric medium with K = 5.

Solution:
In vacuum: \[\frac{q}{\varepsilon_0}=\frac{5\times10^{-6}}{8.85\times10^{-12}}\approx5.65\times10^5\mathrm{N}\cdot\mathrm{m}^2/\mathrm{C}\]

In dielectric: Φ_K = \[\frac {Φ}{K}\] ≈ 1.13 × 105 N ⋅ m²/C

Answer: Flux reduces by a factor of K in a dielectric medium.

Given: Point charge q placed at one corner of a cube.
Find: Flux through one shaded face of the cube.

Solution: A charge at a corner is shared symmetrically among 8 cubes (imagining the corner as the center of a larger symmetric arrangement); each cube has 3 visible faces at that corner, giving flux per face:

\[\Phi_{face}=\frac{q}{24\varepsilon_0}\]

Answer: \[\Phi_{face}=\frac{q}{24\varepsilon_0}\]

Given: Charge q = 8.85 μC enclosed in a cube.
Find: Total flux; effect of moving the charge inside the cube.

Solution: Φ = \[\frac {q}{ε_0}\] = \[\frac {8.85×10^{−6}{8.85×10^{−12}\] = 1 × 10⁶ N ⋅ m²/C

Answer: Total flux remains constant regardless of the charge's position inside; only the distribution of flux across the six faces changes.