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(D) The electric field due to an infinite plane sheet of charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\varepsilon_0} \hat{n}$, where

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$-\frac{5 \sigma}{2 \varepsilon_{0}} \hat{j}$

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(D) The electric field due to an infinite plane sheet of charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\varepsilon_0} \hat{n}$, where $\hat{n}$ is the unit vector normal to the sheet pointing away from it.At the point $(0, 2a, 0)$, the position is $Y=2a$.$1$. For the sheet at $Y=a$ with charge density $-\sigma$: The point is to the right of the sheet. The field points towards the sheet (negative $Y$-direction). $\vec{E}_1 = -\frac{-\sigma}{2\varepsilon_0} \hat{j} = \frac{\sigma}{2\varepsilon_0} \hat{j}$.$2$. For the sheet at $Y=3a$ with charge density $2\sigma$: The point is to the left of the sheet. The field points away from the sheet (negative $Y$-direction). $\vec{E}_2 = -\frac{2\sigma}{2\varepsilon_0} \hat{j} = -\frac{\sigma}{\varepsilon_0} \hat{j}$.$3$. For the sheet at $Y=4a$ with charge density $4\sigma$: The point is to the left of the sheet. The field points away from the sheet (negative $Y$-direction). $\vec{E}_3 = -\frac{4\sigma}{2\varepsilon_0} \hat{j} = -\frac{2\sigma}{\varepsilon_0} \hat{j}$.The net electric field is $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 = \left( \frac{\sigma}{2\varepsilon_0} - \frac{2\sigma}{2\varepsilon_0} - \frac{4\sigma}{2\varepsilon_0} \right) \hat{j} = \frac{\sigma - 2\sigma - 4\sigma}{2\varepsilon_0} \hat{j} = -\frac{5\sigma}{2\varepsilon_0} \hat{j}$.

Three infinite plane sheets carrying uniform charge densities $-\sigma, 2 \sigma, 4 \sigma$ are placed parallel to the $XZ$ plane at $Y=a, 3a, 4a$ respectively. The electric field at the point $(0, 2a, 0)$ is

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