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

Question #124314
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?
1
Expert's answer
2020-06-29T14:04:41-0400

Since, flux through any closed surface is

ϕ=SEdS\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 EdS\overrightarrow{E}\bot d\overrightarrow{S} .

Now, electric flux through y-z plane is


ϕyz=ϕx=0+ϕx=2    ϕx=0=02025j(dydz)i=0ϕx=2=0202(6i+5j)(dydz)i=24    ϕyz=24\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

ϕzx=ϕy=0+ϕy=2    ϕy=0=0202(3xi+5j)(dxdz)j=20ϕy=2=0202(3xi+5j)(dxdz)j=20    ϕzx=0\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 24Vm24 \: Vm


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