A uniform electric field E = 3 times 10^5 tex | vedclass

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(B) Given,electric field $\vec{E} = 3 	imes 10^5 \hat{j} 	ext{ NC}^{-1}$.Area of the rectangle $A = 10 	ext{ cm} 	imes 30 	ext{ cm} = 0.1 	ext{

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$9 	imes 10^3 	ext{ Vm}$

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(B) Given,electric field $\vec{E} = 3 	imes 10^5 \hat{j} 	ext{ NC}^{-1}$.Area of the rectangle $A = 10 	ext{ cm} 	imes 30 	ext{ cm} = 0.1 	ext{ m} 	imes 0.3 	ext{ m} = 3 	imes 10^{-2} 	ext{ m}^2$.Since the plane of the rectangle is parallel to the $ZX$-plane,its area vector $\vec{A}$ is directed along the $Y$-axis (normal to the $ZX$-plane).Therefore,$\vec{A} = 3 	imes 10^{-2} \hat{j} 	ext{ m}^2$.The electric flux $\phi$ is given by the dot product of the electric field and the area vector:$\phi = \vec{E} \cdot \vec{A} = (3 	imes 10^5 \hat{j}) \cdot (3 	imes 10^{-2} \hat{j})$$\phi = 9 	imes 10^3 	ext{ Vm}$.

$A$ uniform electric field $E = 3 \times 10^5 \text{ NC}^{-1}$ is acting along the positive $Y$-axis. The electric flux through a rectangle of area $10 \text{ cm} \times 30 \text{ cm}$ whose plane is parallel to the $ZX$-plane is

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