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A cubical Gaussian surface has side of length $a=10 \mathrm{~cm}$. Electric field lines are parallel to $X$-axis as shown in figure. The magnitudes of electric fields through surfaces $A B C D$ and $E F G H$ are $6 \mathrm{kNC}^{-1}$ and $9 \mathrm{kNC}^{-1}$ respectively. Then, the total charge enclosed by the cube is
[Take, $\varepsilon_0=9 \times 10^{-12} \mathrm{Fm}^{-1}$ ]
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[Take, $\varepsilon_0=9 \times 10^{-12} \mathrm{Fm}^{-1}$ ]
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Verified Answer
The correct answer is:
$0.27 \mathrm{nC}$
Total flux $\phi_E=E \cdot d s$
$\begin{aligned} & =9 \mathrm{k}-6 \mathrm{k} \times\left(10^{-2} \times 10\right)^2 \\ & =3 \times 10^3 \times 10^{-2}=30 \mathrm{NC}^{-1} \mathrm{~m}^2\end{aligned}$
According to Gauss' law, $\frac{q}{\varepsilon_0}=\phi_E$
$\begin{aligned} \Rightarrow q_{\text {endosed }} & =3 \times \varepsilon_0 \\ & =30 \times 9 \times 10^{-12} \mathrm{C}=270 \times 10^{-9} \times 10^{-3} \\ & =0.27 \times 10^{-9} \mathrm{C}=0.27 \mathrm{nC}\end{aligned}$
$\begin{aligned} & =9 \mathrm{k}-6 \mathrm{k} \times\left(10^{-2} \times 10\right)^2 \\ & =3 \times 10^3 \times 10^{-2}=30 \mathrm{NC}^{-1} \mathrm{~m}^2\end{aligned}$
According to Gauss' law, $\frac{q}{\varepsilon_0}=\phi_E$
$\begin{aligned} \Rightarrow q_{\text {endosed }} & =3 \times \varepsilon_0 \\ & =30 \times 9 \times 10^{-12} \mathrm{C}=270 \times 10^{-9} \times 10^{-3} \\ & =0.27 \times 10^{-9} \mathrm{C}=0.27 \mathrm{nC}\end{aligned}$
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