A cube of side $0.50,\text{m}$ is placed in a uniform electric field of magnitude $2000,\text{N/C}$ directed along the +$z$ axis. What is the net electric flux through the cube?

Step-by-step solution

1. Step 1: Identify the faces perpendicular to the field: the top (normal +$z$) and bottom (normal –$z$).
2. Step 2: Compute the flux through the top face: ${\Phi }_{\text{top}}=E,A=2000\times (0.50\times 0.50)=2000\times 0.25=500,{\text{N m}}^2\text{/C}$.
3. Step 3: Compute the flux through the bottom face: ${\Phi }_{\text{bottom}}=-E,A=-500,{\text{N m}}^2\text{/C}$ (negative because the normal is opposite the field).
4. Step 4: Add contributions from all six faces. The side faces contribute zero because the field is parallel to them.
5. Step 5: Net flux: $500-500=0,{\text{N m}}^2\text{/C}$.

Concept tested: Flux through cube in uniform field

Published 7 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked