The cube in FIGURE EX24.6 contains negative charge. The electric field is constant over each face of the cube. Does the missing electric field vector on the front face point in or out? What strength must this field exceed?

Given figure :

Theory used :

According to Gauss's law, if a net charge encompassed by a surface is positive, the net electric flux through that surface is positive; if the net charge is negative, the net electric flux through that surface is negative, and vice versa. If the electric field strength points outwards, the surface element contributes positively to the net flow; if it points inwards, it contributes negatively.

In our situation, we have a cube's surface. We can divide the flux into its six faces and sum the flux via each face to calculate the flux.

The magnitude of the normal component of the field to the face multiplied by the area equals the flux through each face and it has a positive sign if the normal component travels outside and a negative sign if it travels inside.

Since$\overrightarrow{E}$is everywhere perpendicular to the surface so we just have :

$$\begin{gathered} \Phi =\underset{FrontandBackface}{\underbrace{E_{front}{A}_0+20A_0}}-\underset{RightandLeftface}{\underbrace{20A_0+10A_0}}-\underset{TopandBottomface}{\underbrace{15A_0+10A_0}} \\ =(5+E_{front})A_0. \end{gathered}$$

The flux must be negative because the cube has negative charge. As a result$(5+E_{front})A_0<0$

Thus yielding :

$E_{front}{A}_0={\Phi }_{front}<-5A_0$

This implies that the flux through the front face must be negative, indicating that the electric field points into the front face.