Question

What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side $20\,\text{cm}$ oriented so that its faces are parallel to the coordinate planes?

Hint:

For a uniform field, flux into one face equals flux out of the opposite face. Side faces (normals perpendicular to E) carry zero flux. Net flux through a closed surface with no enclosed charge is zero by Gauss's law.

Answer 1

The same uniform field \(\vec E = 3\times10^{3}\,\hat\imath\,\text{N/C}\) passes through a cube of side \(20\,\text{cm}\) whose faces are parallel to the coordinate planes.

Step 1: Concept. The net flux through a closed surface is the sum of the flux through every face, counting outward normals as positive.

\[\phi_{net} = \oint \vec E \cdot d\vec A.\]

Step 2: Look at each pair of faces. The field points only along the x-axis. The four faces parallel to the x-axis (the top, bottom, front and back faces, with normals along \(\pm\hat\jmath\) or \(\pm\hat k\)) have normals perpendicular to \(\vec E\), so \(\vec E \cdot d\vec A = 0\) for them. They contribute zero flux.

Step 3: The two faces perpendicular to the x-axis. Let each have area \(A = (0.20)^{2} = 4\times10^{-2}\,\text{m}^{2}\). For the face on the +x side the outward normal is \(+\hat\imath\):

\[\phi_{+x} = +E\,A = (3\times10^{3})(4\times10^{-2}) = +120\,\text{N m}^{2}/\text{C}.\]

For the face on the -x side the outward normal is \(-\hat\imath\):

\[\phi_{-x} = -E\,A = -120\,\text{N m}^{2}/\text{C}.\]

Step 4: Add up.

\[\phi_{net} = \phi_{+x} + \phi_{-x} + 0 = 120 - 120 = 0.\]

Whatever flux enters one face leaves the opposite face, so the net flux is zero. This agrees with Gauss's law since there is no charge inside the cube.

\[\boxed{\phi_{net} = 0}\]

Answer 2

Gauss's-law (enclosed-charge) argument:

Step 1: Gauss's law states that the net flux through any closed surface depends only on the charge enclosed:

\[\phi_{net} = \frac{q_{enclosed}}{\varepsilon_0}.\]

Step 2: A uniform field is the field of charges that lie far outside the cube; there is no source or sink of field lines inside the cube. Hence \(q_{enclosed} = 0\).

Step 3: Substituting,

\[\phi_{net} = \frac{0}{\varepsilon_0} = 0.\]

Step 4 (physical picture): Field lines of a uniform field are straight and parallel. Every line that pierces the cube going in through one face comes straight out through the opposite face, so the inward and outward fluxes cancel exactly. The result does not depend on the cube's size (here 20 cm) or orientation as long as no charge sits inside.

\[\boxed{\phi_{net} = 0}\]