A cylinder of radius R and length L is placed | vedclass

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(D) The cylinder has three surfaces: two circular end faces ($A$ and $B$) and one curved surface $(C)$.$1$. Flux through the left circular face $(A)$:

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(D) The cylinder has three surfaces: two circular end faces ($A$ and $B$) and one curved surface $(C)$.$1$. Flux through the left circular face $(A)$: The electric field lines enter the surface,so the angle between the area vector and the electric field is $180^{\circ}$. Thus,$\phi_{A} = E \cdot A \cdot \cos(180^{\circ}) = -E \cdot \pi R^2$.$2$. Flux through the right circular face $(B)$: The electric field lines exit the surface,so the angle between the area vector and the electric field is $0^{\circ}$. Thus,$\phi_{B} = E \cdot A \cdot \cos(0^{\circ}) = +E \cdot \pi R^2$.$3$. Flux through the curved surface $(C)$: At every point on the curved surface,the area vector $dS$ is perpendicular to the electric field $E$ (angle is $90^{\circ}$). Thus,$\phi_{C} = \int E \cdot dS \cdot \cos(90^{\circ}) = 0$.Total flux $\phi_{total} = \phi_{A} + \phi_{B} + \phi_{C} = -E \pi R^2 + E \pi R^2 + 0 = 0$.

$A$ cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by-

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