$(A)\;zero \\ (B)\;\frac{-3 \sigma}{\in _0} \hat j \\ (C)\; \frac{-2 \sigma}{\in _0} \hat j \\ (D)\;\frac{3 \sigma }{\in _0} \hat j $

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Due to infinitely long plane carrying surface charge density $\sigma$, electric field at any point is given by

$E= \large\frac{\sigma}{2 \in _0}$

Due to sheet 1,

$E_1 =\large\frac{3 \sigma}{ 2 \in _0} $$(- \hat j)$

Due to sheet 2,

$E_2 =\large\frac{2 \sigma}{ 2 \in _0} $$(- \hat j)$

Due to sheet 3,

$E_3 =\large\frac{- \sigma}{ 2 \in _0} $$(- \hat j)$

$E_1+E_2+E_3=\large\frac{-6 \sigma}{2 \in _0} \hat j $

=>$ \large\frac{-3 \sigma}{\in _0} $$\hat j$

Hence B is the correct answer

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