Question

A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with uniform magnitude E0 over each of those two faces. No flux crosses the other four faces. How much charge is inside the cube?

Answer 1

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}$

If we consider than the direction of the electric field is $\vec{E}=E_0\hat{x}$, we can solve the problem differentiating  the integral for each face of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6$

E₀ is a constant and each surface is equal to each other, so: $S_1=S_2=S_i=S$

Therefore:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c$