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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?

1 Answer

4 votes

Answer:

The charge inside the cube is null.

Step-by-step 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}=(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=(q_0)/(\varepsilon_0) \longleftrightarrow q_0=0c

User Vishal Tiwari
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