Since, flux through any closed surface is
ϕ = ∯ S E → ⋅ d S → \phi=\oiint_{S}\overrightarrow{E}\cdot d\overrightarrow{S} ϕ = ∬ S E ⋅ d S Now, flux through cube can be calculate as below
clearly, flux through y-x plane is 0 as E → ⊥ d S → \overrightarrow{E}\bot d\overrightarrow{S} E ⊥ d S .
Now, electric flux through y-z plane is
ϕ y z = ϕ x = 0 + ϕ x = − 2 ⟹ ϕ x = 0 = ∫ 0 2 ∫ 0 2 5 j ⋅ ( d y d z ) i = 0 ϕ x = − 2 = ∫ 0 2 ∫ 0 2 ( − 6 i + 5 j ) ⋅ ( − d y d z ) i = 24 ⟹ ϕ y z = 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 ϕ yz = ϕ x = 0 + ϕ x = − 2 ⟹ ϕ x = 0 = ∫ 0 2 ∫ 0 2 5 j ⋅ ( d y d z ) i = 0 ϕ x = − 2 = ∫ 0 2 ∫ 0 2 ( − 6 i + 5 j ) ⋅ ( − d y d z ) i = 24 ⟹ ϕ yz = 24 Electric flux through z-x is
ϕ z x = ϕ y = 0 + ϕ y = 2 ⟹ ϕ y = 0 = ∫ 0 2 ∫ 0 2 ( 3 x i + 5 j ) ⋅ ( − d x d z ) j = − 20 ϕ y = 2 = ∫ 0 2 ∫ 0 2 ( 3 x i + 5 j ) ⋅ ( d x d z ) j = 20 ⟹ ϕ z x = 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 ϕ z x = ϕ y = 0 + ϕ y = 2 ⟹ ϕ y = 0 = ∫ 0 2 ∫ 0 2 ( 3 x i + 5 j ) ⋅ ( − d x d z ) j = − 20 ϕ y = 2 = ∫ 0 2 ∫ 0 2 ( 3 x i + 5 j ) ⋅ ( d x d z ) j = 20 ⟹ ϕ z x = 0 Therefore, net flux through the whole cube is 24 V m 24 \: Vm 24 Vm
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