The electric field components in figure are Ex-b-x- Ey-Ez-0- Calculate the net flux through the cube-

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Gauss's Law

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Question

The electric field components in figure are $E_{x} = b \sqrt{x}, E_{y} = E_{z} = 0,$ Calculate the net flux through the cube.

(\sqrt{2} - 1) b a^{\frac{5}{2}}

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b a^{\frac{5}{2}}

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\sqrt{b} a^{\frac{5}{2}}

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(\sqrt{5} - 3) b a^{\frac{5}{2}}

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Solution

The correct option is A $(\sqrt{2} - 1) b a^{\frac{5}{2}}$
As $E_{y} = E_{z} = 0$ so only faces along x-axis will contribute the flux through the cube.
Entering flux through the face at $x = a$ is $ϕ_{i n} = E_{x} . a^{2} = b \sqrt{x} . a^{2} = b \sqrt{a} . a^{2} = b a^{5 / 2}$
Outgoing flux through the face at $x = 2 a$ is $ϕ_{o u t} = E_{x} . a^{2} = b \sqrt{2 a} . a^{2} = b \sqrt{2 a} . a^{2} = b a^{5 / 2} \sqrt{2}$
Thus , net flux through the cube is $ϕ = ϕ_{o u t} - ϕ_{i n} = b a^{5 / 2} \sqrt{2} - b a^{5 / 2} = b a^{5 / 2} (\sqrt{2} - 1)$

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