A cube of side 0.15 m is placed in a uniform electric field of magnitude 400 N/C directed along the +z–axis. What is the net electric flux through the cube?

Step-by-step solution

1. Only the faces perpendicular to the field (the top and bottom) contribute to the net flux; the side faces have field parallel to them, giving zero contribution.
2. The area of each face is $A=a^2=0.15^2=0.0225{\text{m}}^2$.
3. The flux through one such face is $\Phi =E,A=400\times 0.0225=9.0{\text{N⋅m}}^2/\text{C}$.
4. The total net flux through the cube is therefore $9.0{\text{N⋅m}}^2/\text{C}$, as the contributions from top and bottom add and the side faces cancel out.

Concept tested: Electric flux through a closed surface

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