A cube of side 0.20 m is placed in a non‑uniform electric field that varies linearly with the vertical coordinate: $E_z=5000,z\text{N/C}$, where $z$ is measured in metres from the cube’s bottom face. The cube’s faces are perpendicular to the coordinate axes. What is the net electric flux through the cube?

Step-by-step solution

1. Electric flux through a surface is $\Phi =\int \vec{E}\cdot d\vec{A}$.
2. The cube’s bottom face lies at $z=0$, so $E_z=0$ there; its area is $A=0.20^2=0.04{\text{m}}^2$. The flux through the bottom face is $0\times 0.04=0$.
3. The top face is at $z=0.20\text{m}$, giving $E_z=5000\times 0.20=1000\text{N/C}$. Flux through the top face is $1000\times 0.04=40{\text{N m}}^2/\text{C}$.
4. Side faces are perpendicular to $x$ or $y$ and the field has only a $z$ component; hence their fluxes are zero.
5. Add contributions: ${\Phi }_{\text{net}}=0+40=40{\text{N m}}^2/\text{C}$.

Concept tested: Flux through a surface in a varying field

Published 4 October 2026 · How this solution was prepared and checked