Consider a cylinder of length L and radius R | vedclass

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(D) Let the cylinder be placed with its axis along the $x$-axis in a uniform electric field $\vec{E} = E\hat{i}$.$1$. Flux through the left circular f

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(D) Let the cylinder be placed with its axis along the $x$-axis in a uniform electric field $\vec{E} = E\hat{i}$.$1$. Flux through the left circular face (Area $A = \pi R^2$): The area vector points outward,so $\vec{A}_1 = -\pi R^2\hat{i}$. The flux is $\phi_1 = \vec{E} \cdot \vec{A}_1 = (E\hat{i}) \cdot (-\pi R^2\hat{i}) = -E\pi R^2$.$2$. Flux through the right circular face: The area vector points outward,so $\vec{A}_2 = \pi R^2\hat{i}$. The flux is $\phi_2 = \vec{E} \cdot \vec{A}_2 = (E\hat{i}) \cdot (\pi R^2\hat{i}) = +E\pi R^2$.$3$. Flux through the curved surface: At every point on the curved surface,the area vector $d\vec{A}$ is perpendicular to the electric field $\vec{E}$ (i.e.,$	heta = 90^{\circ}$). Therefore,$d\phi = \vec{E} \cdot d\vec{A} = E dA \cos 90^{\circ} = 0$. Thus,the total flux through the curved surface is $0$.$4$. Total flux $\phi_{total} = \phi_1 + \phi_2 + \phi_{curved} = -E\pi R^2 + E\pi R^2 + 0 = 0$.

Consider a cylinder of length $L$ and radius $R$ placed in a uniform electric field $E$ such that its axis is parallel to the electric field. The total electric flux associated with the cylinder is .......

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