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Three infinite long plane sheets carrying charge densities $\sigma _1 =+3 \sigma , \sigma _2 =+2 \sigma $ and $\sigma _3 =- \sigma $ are placed parallel to $x-z$ plane at $y=4a,y=3a$ and $y=a $ respectively. Electric field at point P is :

$(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
answered Jan 15, 2014 by meena.p

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