What is the net flux of the uniform electric | vedclass

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$(A)$ The electric field is uniform and given by $E = 3 	imes 10^{3} \hat{i} \; N/C$.Since the cube is oriented with its faces parallel to the coordi

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$0 \; N \cdot m^{2}/C$

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$(A)$ The electric field is uniform and given by $E = 3 	imes 10^{3} \hat{i} \; N/C$.Since the cube is oriented with its faces parallel to the coordinate planes, the electric field lines enter through one face and exit through the opposite face.According to Gauss's Law, the net electric flux $\phi$ through a closed surface is given by $\phi = \oint E \cdot dA$.For a uniform electric field passing through a closed surface (like a cube) where there is no enclosed charge, the flux entering the surface is equal to the flux leaving the surface.Mathematically, the flux through the face at $x = 0$ is $\phi_{in} = -E \cdot A = -(3 	imes 10^{3}) 	imes (0.2)^{2} = -120 \; N \cdot m^{2}/C$.The flux through the face at $x = 0.2 \; m$ is $\phi_{out} = +E \cdot A = +(3 	imes 10^{3}) 	imes (0.2)^{2} = +120 \; N \cdot m^{2}/C$.The net flux is $\phi_{net} = \phi_{in} + \phi_{out} = -120 + 120 = 0 \; N \cdot m^{2}/C$.

What is the net flux of the uniform electric field of $E = 3 \times 10^{3} \hat{i} \; N/C$ through a cube of side $20 \; cm$ oriented so that its faces are parallel to the coordinate planes?

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