Answer to Question #124314 in Atomic and Nuclear Physics for bob

Question

the electric field near the origin is given by E=(3xi+5j)v/m. imagine a gaussian surface shaped like a cube with one corner at the origin and another at (-2,2,2) meters, with all faces parallel to x-y,x-z, or y-z planes. what is the net flux through gaussan cube?

Answer 1

Since, flux through any closed surface is

$$\phi=\oiint_{S}\overrightarrow{E}\cdot d\overrightarrow{S}$$

Now, flux through cube can be calculate as below

clearly, flux through y-x plane is 0 as $\overrightarrow{E}\bot d\overrightarrow{S}$ .

Now, electric flux through y-z plane is

$$\phi_{yz}=\phi_{x=0}+\phi_{x=-2}\\ \implies \phi_{x=0}=\int_{0}^{2}\int_{0}^{2}5j\cdot(dydz)i=0\\\phi_{x=-2}=\int_{0}^{2}\int_{0}^{2}(-6i+5j)\cdot(-dydz)i=24\\\implies \phi_{yz}=24$$

Electric flux through z-x is

$$\phi_{zx}=\phi_{y=0}+\phi_{y=2}\\ \implies \phi_{y=0}=\int_{0}^{2}\int_{0}^{2}(3xi+5j)\cdot(-dxdz)j=-20\\ \phi_{y=2}=\int_{0}^{2}\int_{0}^{2}(3xi+5j)\cdot(dxdz)j=20\\ \implies \phi_{zx}=0$$

Therefore, net flux through the whole cube is $24 \: Vm$