Field of a charged cube

Problem

Challenge: field of a charged cube. A charge $Q$ is uniformly distributed over the outer surface of a cube of side $L$. Without using closed-form formulas, reason about the symmetry of the electric field flux through the faces of the cube. What fraction of the total flux passes through each face? Why? Is the field uniform on a single face?

Solution

Total flux. Enclose the cube in a Gaussian surface (the cube itself). Gauss’s theorem fixes the total outward flux, regardless of shape: $\Phi_\text{tot} = \frac{Q}{\varepsilon_0}.$

Fraction per face. A cube has six geometrically equivalent faces: none is privileged over the others (the cube’s symmetry maps each face to every other one). Since the charge is uniformly distributed, the flux must split equally among the six faces: $\ev{\Phi_\text{face} = \frac{\Phi_\text{tot}}{6} = \frac{Q}{6\varepsilon_0}}.$ This is a pure symmetry argument: there is no need to know the detailed shape of the field, only that the six faces are interchangeable.

Is the field uniform on the face? No. Even though the flux through each face is the same, the field is not constant point by point: near the edges and vertices of the cube the field lines bunch up (point effect), while at the centre of each face the field is weaker. The distribution of charge on a cubic conductor is not uniform, nor is the field just outside it. Symmetry guarantees equal total flux per face, not local uniformity — this is why there is no simple closed-form formula like that for the sphere.

Connections

Topics: Electric field and potential Concepts: Gauss’s theorem · Electric field Skills: Applying Gauss’s theorem Methods: Gauss’s theorem Objects: Infinite charged plane