Question

The cube in FIGURE EX24.8 contains no net 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 is the field strength?

Short answer

The flux through the front face must have a strength of$5$, indicating that theelectric field points into the front face.

Step-by-step solution

Given information and Theory used

Given figure :

Theory used :

According toGauss'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.

Determining if the missing electric field vector on the front face point in or out and what strength must this field exceed 

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 :

role="math" localid="1649702478098" $$\begin{gathered} \Phi =\underset{FrontandBackface}{\underbrace{E_{front}{A}_0+15A_0}}-\underset{RightandLeftface}{\underbrace{20A_0+10A_0}}-\underset{TopandBottomface}{\underbrace{15A_0+15A_0}} \\ =(5+E_{front})A_0. \end{gathered}$$

The flux must be zero because the cube has no charge. As a result

$(5+E_{front})A_0=0$

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

This implies that the flux through the front face must have a strength of $5$, indicating that theelectric field points into the front face.