The electric field ( overrightarrow{E} in N C | vedclass

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(C) The electric field is given by $\overrightarrow{E} = 3 \hat{i} + 5 \hat{j} \ N C^{-1}$.Since the square area is parallel to the $y-z$ plane,its ar

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$12 \ N C^{-1} \ m^2$

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(C) The electric field is given by $\overrightarrow{E} = 3 \hat{i} + 5 \hat{j} \ N C^{-1}$.Since the square area is parallel to the $y-z$ plane,its area vector $\overrightarrow{A}$ will be directed along the $x$-axis.The area of the square is $A = (2 \ m) 	imes (2 \ m) = 4 \ m^2$.Therefore,the area vector is $\overrightarrow{A} = 4 \hat{i} \ m^2$.The electric flux $\phi$ is given by the dot product of the electric field and the area vector:$\phi = \overrightarrow{E} \cdot \overrightarrow{A} = (3 \hat{i} + 5 \hat{j}) \cdot (4 \hat{i})$$\phi = (3 	imes 4) \hat{i} \cdot \hat{i} + (5 	imes 4) \hat{j} \cdot \hat{i}$Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = 0$,we get:$\phi = 12 	imes 1 + 20 	imes 0 = 12 \ N C^{-1} \ m^2$.

The electric field ( $\overrightarrow{E}$ in $N C^{-1}$ ) in a region is given by $\overrightarrow{E} = 3 \hat{i} + 5 \hat{j}$. The net electric flux through a square area of side $2 \ m$ parallel to the $y-z$ plane is:

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