The flux of the electric field overrightarrow | vedclass

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Question

(A) The electric field is given by $\overrightarrow{E} = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} 	ext{ NC}^{-1}$.Since the area lies on the $yz$ plane,i

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$480 	ext{ Nm}^2 	ext{C}^{-1}$

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(A) The electric field is given by $\overrightarrow{E} = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} 	ext{ NC}^{-1}$.Since the area lies on the $yz$ plane,its area vector $\vec{A}$ is directed along the $x$-axis.Thus,$\vec{A} = 20 \hat{i} 	ext{ m}^2$.The electric flux $\Phi_E$ is defined as the dot product of the electric field and the area vector: $\Phi_E = \overrightarrow{E} \cdot \vec{A}$.Substituting the values: $\Phi_E = (24 \hat{i} + 30 \hat{j} + 28 \hat{k}) \cdot (20 \hat{i})$.Using the dot product property $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \hat{j} = \hat{i} \cdot \hat{k} = 0$,we get:$\Phi_E = 24 	imes 20 = 480 	ext{ Nm}^2 	ext{C}^{-1}$.

The flux of the electric field $\overrightarrow{E} = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} \text{ NC}^{-1}$ through an area of $20 \text{ m}^2$ on the $yz$ plane is

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