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1. Electric Charges and Fields
hard
A cube is placed inside an electric field, $\overrightarrow{{E}}=150\, {y}^{2}\, \hat{{j}}$. The side of the cube is $0.5 \,{m}$ and is placed in the field as shown in the given figure. The charge inside the cube is $.....\times 10^{-11} {C}$
A
$3.8$
B
$8.3$
C
$0.38$
D
$830$
(JEE MAIN-2021)
Solution
As electric field is in $y-$direction so electric flux is only due to top and bottom surface
Bottom surface ${y}=0$
$\Rightarrow {E}=0 \Rightarrow \phi=0$
Top surface ${y}=0.5\, {m}$
$\Rightarrow {E}=150(0.5)^{2}=\frac{150}{4}$
Now flux $\phi={EA}=\frac{150}{4}(.5)^{2}=\frac{150}{16}$
By Gauss's law $\phi=\frac{Q_{\text {in }}}{\epsilon_{0}}$
$\frac{150}{16}=\frac{{Q}_{\text {in }}}{\epsilon_{0}}$
${Q}_{\text {in }}=\frac{150}{16} \times 8.85 \times 10^{-12}=8.3 \times 10^{-11} \,{C}$
Standard 12
Physics
