Consider a uniform electric field vec{E} = 3 | vedclass

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(A) The electric field is given by $\vec{E} = 3 	imes 10^3 \hat{k} 	ext{ N C}^{-1}$.Since the plane of the square is parallel to the $yz$-plane,its

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(A) The electric field is given by $\vec{E} = 3 	imes 10^3 \hat{k} 	ext{ N C}^{-1}$.Since the plane of the square is parallel to the $yz$-plane,its area vector $\vec{A}$ must be perpendicular to the $yz$-plane,which means it is directed along the $x$-axis.Therefore,the area vector is $\vec{A} = A \hat{i} = (0.2 	ext{ m})^2 \hat{i} = 0.04 \hat{i} 	ext{ m}^2$.The electric flux $\phi$ is defined as the dot product of the electric field and the area vector: $\phi = \vec{E} \cdot \vec{A}$.Substituting the values: $\phi = (3 	imes 10^3 \hat{k}) \cdot (0.04 \hat{i})$.Since the dot product of orthogonal unit vectors $\hat{k} \cdot \hat{i} = 0$,the flux is $\phi = 0 	ext{ N m}^2 	ext{ C}^{-1}$.

Consider a uniform electric field $\vec{E} = 3 \times 10^3 \hat{k} \text{ N C}^{-1}$. The electric flux of this field through a square of $20 \text{ cm}$ on a side whose plane is parallel to the $yz$-plane is $....... \text{ N m}^2 \text{ C}^{-1}$.

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