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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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 image
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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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