An electric field,overrightarrow{E} = frac{2 | vedclass

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(A) The electric flux $\phi$ is given by the dot product of the electric field vector $\overrightarrow{E}$ and the area vector $\overrightarrow{A}$.Th

Vedclass · Accepted Answer

$12$

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(A) The electric flux $\phi$ is given by the dot product of the electric field vector $\overrightarrow{E}$ and the area vector $\overrightarrow{A}$.The area vector is defined as $\overrightarrow{A} = A \hat{n} = 4 \left( \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} ight) \ m^2$.The electric flux is calculated as:$\phi = \overrightarrow{E} \cdot \overrightarrow{A}$$\phi = \left( \frac{2 \hat{i} + 6 \hat{j} + 8 \hat{k}}{\sqrt{6}} ight) \cdot \left( 4 \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} ight)$$\phi = \frac{4}{6} 	imes (2 	imes 2 + 6 	imes 1 + 8 	imes 1)$$\phi = \frac{4}{6} 	imes (4 + 6 + 8)$$\phi = \frac{4}{6} 	imes 18$$\phi = 4 	imes 3 = 12 \ Vm$.

An electric field,$\overrightarrow{E} = \frac{2 \hat{i} + 6 \hat{j} + 8 \hat{k}}{\sqrt{6}} \ V/m$,passes through a surface of $4 \ m^2$ area having a unit normal vector $\hat{n} = \left( \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$. The electric flux through that surface is:

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