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(D) According to Gauss's Law, the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enc}}{\varepsilon_0}$.The wire passes

Vedclass · Accepted Answer

$2.16$

Vedclass · Answer

(D) According to Gauss's Law, the net electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enc}}{\varepsilon_0}$.The wire passes through two maximally displaced corners of the cube, which means it passes along the body diagonal of the cube.The length of the body diagonal of a cube with side length $a$ is $L = \sqrt{3}a$.Given $a = \sqrt{3} \ cm = \sqrt{3} 	imes 10^{-2} \ m$, the length of the wire inside the cube is $L = \sqrt{3} 	imes (\sqrt{3} 	imes 10^{-2} \ m) = 3 	imes 10^{-2} \ m$.The charge enclosed by the cube is $q_{enc} = \lambda \cdot L = (2 	imes 10^{-9} \ C/m) 	imes (3 	imes 10^{-2} \ m) = 6 	imes 10^{-11} \ C$.Using $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9$, we have $\frac{1}{\varepsilon_0} = 36 \pi 	imes 10^9$.Therefore, the flux $\phi = q_{enc} \cdot \frac{1}{\varepsilon_0} = (6 	imes 10^{-11}) 	imes (36 \pi 	imes 10^9) = 216 \pi 	imes 10^{-2} = 2.16 \pi \ Nm^2 C^{-1}$.Comparing this with $x\pi$, we get $x = 2.16$.

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