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# Flux through a Cube (Eigure 1) A cube has one corner at the origin and the opposite corner at the point (L, L, L). The sides of the cube are parallel to the coordinate planes # Flux through a Cube (Eigure 1) A cube has one corner at the origin and the opposite corner at the point (L, L, L). The sides of the cube are parallel to the coordinate planes

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Other asked 2021-04-25
Flux through a Cube (Eigure 1) A cube has one corner at the origin and the opposite corner at the point (L, L, L). The sides of the cube are parallel to the coordinate planes  ## Answers (1) 2021-04-27
(A)
The electric flux through left surface is,
$$\displaystyle\phi_{{1}}=\vec{{{E}}}\cdot\vec{{{A}}}$$
$$\displaystyle={\left({\left({a}+{b}{x}\right)}\hat{{{i}}}+{c}\hat{{{j}}}\right)}\cdot{\left({L}^{{2}}{\left(-\hat{{{i}}}\right)}\right)}$$
$$\displaystyle=-{L}^{{2}}{\left({a}+{b}{x}\right)}$$
For left surface x=0.
$$\displaystyle\phi_{{1}}=-{a}{L}^{{2}}$$
Th electric flux through right surface is,
$$\displaystyle\phi_{{2}}=\vec{{{E}}}\cdot\vec{{{A}}}$$
$$\displaystyle={\left({\left({a}+{b}{x}\right)}\hat{{{i}}}+{c}\hat{{{j}}}\right)}\cdot{\left({L}^{{2}}{\left(\hat{{{i}}}\right)}\right)}$$
$$\displaystyle={L}^{{2}}{\left({a}+{b}{x}\right)}$$
For right surface x=L
$$\displaystyle\phi_{{2}}={L}^{{2}}{\left({a}+{b}{L}\right)}$$
$$\displaystyle={a}{L}^{{2}}+{b}{L}^{{3}}$$
The electric flux through top surface is,
$$\displaystyle\phi_{{3}}=\vec{{{E}}}\cdot\vec{{{A}}}$$
$$\displaystyle={\left({\left({a}+{b}{x}\right)}\hat{{{i}}}+{c}\hat{{{j}}}\right)}\cdot{\left({L}^{{2}}{\left(\hat{{{j}}}\right)}\right)}$$
$$\displaystyle={c}{L}^{{2}}$$
The electric flux through bottom surface is,
$$\displaystyle\phi_{{4}}=\vec{{{E}}}\cdot\vec{{{A}}}$$
$$\displaystyle={\left({\left({a}+{b}{x}\right)}\hat{{{i}}}+{c}\hat{{{j}}}\right)}\cdot{\left({L}^{{2}}{\left(-\hat{{{j}}}\right)}\right)}$$
$$\displaystyle=-{c}{L}^{{2}}$$
Since there is no field along z-direction, the elcetric flux through front and back surface of the the cube must be zero.
The total electric flux through surface of the cube is,
$$\displaystyle\phi_{{E}}=\phi_{{1}}+\phi_{{2}}+\phi_{{3}}+\phi_{{4}}$$
$$\displaystyle=-{a}{L}^{{2}}+{a}{L}^{{2}}+{b}{L}^{{3}}+{c}{L}^{{2}}-{c}{L}^{{2}}$$
$$\displaystyle={b}{L}^{{3}}$$
Thus, the electric flux through surface of the cube is $$\displaystyle{b}{L}^{{3}}$$
(C)
According to Gauss law,
$$\displaystyle\phi_{{E}}={\frac{{{q}}}{{\epsilon_{{0}}}}}$$
$$\displaystyle{q}=\epsilon_{{0}}\phi_{{E}}$$
$$\displaystyle{q}=\epsilon_{{0}}{b}{l}^{{3}}$$
Thus, the net charge inside the cube is $$\displaystyle\epsilon_{{0}}{b}{L}^{{3}}$$

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