The electric field components in figure are Ex=bx,Ey=Ez=0. Calculate the flux through the cube.

Use Gauss’s law to calculate flux.Ey=Ez=0 ⇒ ϕy=0; ϕz=0ϕx=a=-ba×a2; ϕx=2a=b2a×a2⇒ϕnet=(2-1)ba5/2

The electric field components in figure are Ex=bx,Ey=Ez=0. Calculate the flux through the cube.

Use Gauss’s law to calculate flux.Ey=Ez=0 ⇒ ϕy=0; ϕz=0ϕx=a=-ba×a2; ϕx=2a=b2a×a2⇒ϕnet=(2-1)ba5/2

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

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$(\sqrt{5} - 3) {ba}^{5 / 2}$

${ba}^{5 / 2}$

$\sqrt{2} {ba}^{5 / 2}$

$(\sqrt{2} - 1) {ba}^{5 / 2}$

answer is A.

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Use Gauss’s law to calculate flux.

$E_{y} = E_{z} = 0 \Rightarrow ϕ_{y} = 0; ϕ_{z} = 0$

$ϕ_{x = a} = - b \sqrt{a} \times a^{2}; ϕ_{x = 2 a} = b \sqrt{2 a} \times a^{2}$

$\Rightarrow ϕ_{net} = (\sqrt{2} - 1) {ba}^{5 / 2}$

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The electric field components in figure are Ex=bx,Ey=Ez=0. Calculate the flux through the cube.

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