Let $σ$ be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different region E_I, E_II and E_III are:

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${\vec{E}}_{I} = \frac{2 σ}{ϵ_{0}} \hat{n}, {\vec{E}}_{II} = 0, {\vec{E}}_{III} = \frac{2 σ}{ϵ_{0}} \hat{n}$

${\vec{E}}_{I} = - \frac{σ}{ϵ_{0}} \hat{n}, {\vec{E}}_{II} = 0, {\vec{E}}_{III} = \frac{σ}{ϵ_{0}} \hat{n}$

${\vec{E}}_{I} = \frac{σ}{2 ϵ_{0}} \hat{n}, {\vec{E}}_{II} = 0, {\vec{E}}_{III} = \frac{σ}{2 ϵ_{0}} \hat{n}$

${\vec{E}}_{I} = 0, {\vec{E}}_{II} = \frac{σ}{ϵ_{0}} \hat{n}, E_{III} = 0$

answer is C.

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Detailed Solution

$E_{I} = \frac{σ}{2 ϵ_{0}} + \frac{σ}{2 ϵ_{0}} = \frac{σ}{ϵ_{0}} = E_{III}$
E_I, E_III are in opposite directions.

$E_{II} = \frac{σ}{2 ϵ_{0}} - \frac{σ}{2 ϵ_{0}} = 0$

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