An electric field overrightarrow{E} = (2xhat{ | vedclass

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(D) Given,electric field $\overrightarrow{E} = 2x\hat{i} 	ext{ NC}^{-1}$.The cube is placed between $x = 2 	ext{ m}$ and $x = 4 	ext{ m}$. The side

Vedclass · Accepted Answer

$16$

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(D) Given,electric field $\overrightarrow{E} = 2x\hat{i} 	ext{ NC}^{-1}$.The cube is placed between $x = 2 	ext{ m}$ and $x = 4 	ext{ m}$. The side length of the cube is $a = 2 	ext{ m}$,so the area of each face is $A = a^2 = (2)^2 = 4 	ext{ m}^2$.Electric flux is given by $\phi = \int \overrightarrow{E} \cdot d\overrightarrow{A}$.For the left face at $x = 2 	ext{ m}$,the area vector is $\overrightarrow{A}_1 = -4\hat{i} 	ext{ m}^2$ and $\overrightarrow{E}_1 = 2(2)\hat{i} = 4\hat{i} 	ext{ NC}^{-1}$.$\phi_{	ext{in}} = \overrightarrow{E}_1 \cdot \overrightarrow{A}_1 = (4\hat{i}) \cdot (-4\hat{i}) = -16 	ext{ Nm}^2 	ext{C}^{-1}$.For the right face at $x = 4 	ext{ m}$,the area vector is $\overrightarrow{A}_2 = 4\hat{i} 	ext{ m}^2$ and $\overrightarrow{E}_2 = 2(4)\hat{i} = 8\hat{i} 	ext{ NC}^{-1}$.$\phi_{	ext{out}} = \overrightarrow{E}_2 \cdot \overrightarrow{A}_2 = (8\hat{i}) \cdot (4\hat{i}) = 32 	ext{ Nm}^2 	ext{C}^{-1}$.The net electric flux through the cube is $\phi_{	ext{net}} = \phi_{	ext{in}} + \phi_{	ext{out}} = -16 + 32 = 16 	ext{ Nm}^2 	ext{C}^{-1}$.

An electric field $\overrightarrow{E} = (2x\hat{i}) \text{ NC}^{-1}$ exists in space. $A$ cube of side $2 \text{ m}$ is placed in the space as per the figure given below. The electric flux through the cube is .................. $\text{Nm}^2 \text{C}^{-1}$.

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