There exists an electric field of 1 N/C alon | vedclass

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(D) The electric field is given as $\vec{E} = 1 \hat{j} \ N/C$.Since the square is placed in the $XY$-plane,its area vector $\vec{A}$ is perpendicular

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(D) The electric field is given as $\vec{E} = 1 \hat{j} \ N/C$.Since the square is placed in the $XY$-plane,its area vector $\vec{A}$ is perpendicular to the $XY$-plane,which is along the $Z$-axis.Therefore,$\vec{A} = 1 \hat{k} \ m^2$.The electric flux $\phi$ is defined as the dot product of the electric field and the area vector:$\phi = \vec{E} \cdot \vec{A}$$\phi = (1 \hat{j}) \cdot (1 \hat{k})$Since the dot product of orthogonal unit vectors $\hat{j} \cdot \hat{k} = 0$,the flux is:$\phi = 0 \ Nm^2/C$.

There exists an electric field of $1 \ N/C$ along the $Y$-direction. The flux passing through a square of $1 \ m^2$ placed in the $XY$-plane inside the electric field is . . . . . . .

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