The electric field vec{E} = E_0 y hat{j} acts | vedclass

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Question

(C) According to Gauss's Law, the total electric flux $\phi$ through a closed surface is given by $\phi = \oint \vec{E} \cdot d\vec{s} = \frac{Q_{in}}

Vedclass · Accepted Answer

$E_0 \varepsilon_0 \pi r^2 l$

Vedclass · Answer

(C) According to Gauss's Law, the total electric flux $\phi$ through a closed surface is given by $\phi = \oint \vec{E} \cdot d\vec{s} = \frac{Q_{in}}{\varepsilon_0}$.For a cylinder with its axis along the $y$-axis, the electric field $\vec{E} = E_0 y \hat{j}$ is parallel to the curved surface and perpendicular to the circular end faces.However, the flux through the curved surface is zero because $\vec{E} \cdot d\vec{s} = 0$ (as $d\vec{s}$ is radial).At the bottom face $(y = 0)$, $\vec{E} = 0$, so flux is $0$.At the top face $(y = l)$, $\vec{E} = E_0 l \hat{j}$ and the area vector $d\vec{s} = dA \hat{j}$.The flux through the top face is $\phi = \int E_0 l dA = E_0 l \int dA = E_0 l (\pi r^2)$.Thus, the total flux $\phi = E_0 l \pi r^2$.Using Gauss's Law, $Q_{in} = \varepsilon_0 \phi = \varepsilon_0 E_0 \pi r^2 l$.

The electric field $\vec{E} = E_0 y \hat{j}$ acts in a space where a cylinder of radius $r$ and length $l$ is placed with its axis parallel to the $y$-axis. The charge inside the volume of the cylinder is:

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