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(D) The side length of the cube is $a = 10 \,cm = 0.1 \,m$. The area of each face is $A = a^2 = (0.1 \,m)^2 = 0.01 \,m^2$.Electric field lines enter t

Vedclass · Accepted Answer

$0.27$

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(D) The side length of the cube is $a = 10 \,cm = 0.1 \,m$. The area of each face is $A = a^2 = (0.1 \,m)^2 = 0.01 \,m^2$.Electric field lines enter through surface $ABCD$ and exit through surface $EFGH$.The flux through surface $ABCD$ is $\phi_1 = -E_1 A = -(6 	imes 10^3 \,NC^{-1}) 	imes (0.01 \,m^2) = -60 \,Nm^2C^{-1}$.The flux through surface $EFGH$ is $\phi_2 = E_2 A = (9 	imes 10^3 \,NC^{-1}) 	imes (0.01 \,m^2) = 90 \,Nm^2C^{-1}$.The flux through all other four faces is zero because the electric field is parallel to these surfaces.The net flux through the cube is $\phi_{net} = \phi_1 + \phi_2 = -60 + 90 = 30 \,Nm^2C^{-1}$.According to Gauss's law, $\phi_{net} = \frac{q_{enclosed}}{\varepsilon_0}$.Therefore, $q_{enclosed} = \phi_{net} 	imes \varepsilon_0 = 30 \,Nm^2C^{-1} 	imes 9 	imes 10^{-12} \,Fm^{-1} = 270 	imes 10^{-12} \,C = 0.27 	imes 10^{-9} \,C = 0.27 \,nC$.

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