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(B) The electric field due to an infinite sheet with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\varepsilon_0} \hat{n}$,whe

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$-\frac{2\sigma}{\varepsilon_0} \hat{k}$

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(B) The electric field due to an infinite sheet with surface 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.Point $P$ is located between $Z = a$ and $Z = 3a$.$1$. For the sheet at $Z = 3a$ with charge density $\sigma$: Point $P$ is below it,so the field is directed downwards: $\vec{E}_1 = -\frac{\sigma}{2\varepsilon_0} \hat{k}$.$2$. For the sheet at $Z = a$ with charge density $-2\sigma$: Point $P$ is above it,so the field is directed towards the sheet (downwards): $\vec{E}_2 = \frac{-2\sigma}{2\varepsilon_0} \hat{k} = -\frac{\sigma}{\varepsilon_0} \hat{k}$.$3$. For the sheet at $Z = -a$ with charge density $-\sigma$: Point $P$ is above it,so the field is directed towards the sheet (downwards): $\vec{E}_3 = \frac{-\sigma}{2\varepsilon_0} \hat{k}$.The net electric field at $P$ is $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 = -\frac{\sigma}{2\varepsilon_0} \hat{k} - \frac{\sigma}{\varepsilon_0} \hat{k} - \frac{\sigma}{2\varepsilon_0} \hat{k} = -\frac{2\sigma}{\varepsilon_0} \hat{k}$.

Three infinite sheets are placed as shown in the figure. Find the electric field at point $P$.

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