Electric field intensity at a point due to an infinite sheet of charge having surface charge density `sigma` is `E`.If sheet were conducting electric intensity would be

Explanation

Consider two thin infinite plane charged sheet held parallel to each other.
Let `sigma_(A) and sigma_(B)` be charge densities on A and B
respectively. Let `sigma_(A) gt sigma_(B) gt0`.
in region I, we have `E_(1)=-E_(A)-E_(B)`
or `E_(1)=-(sigma_(A))/(2epsi_(0))-(sigma_(B))/(2epsi_(0))`
or `E_(1)=(-1)/(2epsi_(0))(sigma_(A)+sigma_(B))` . . (i)
In region II, we have `E_(II)=E_(A)-E_(B)`
or `E_(II)=(sigma_(A)-sigma_(B))/(2epsi_(0))` . . (i)

and in region III, we have `E_(III)=E_(A)+E_(B)=(sigma_(A)+sigma_(B))/(2epsi_(0))` . . (iii)
Special cases. (i) if `sigma_(A)=sigma and sigma_(B)=-sigma` i.e., charges on plates A and B are equal but opposite and held parallel (as in the case of capacitor), then ,br> `E_(I)=E_(III)=0 and E_(II)=(2sigma)/(2epsi_(0))=(sigma)/(epsi_(0))`
i.e. electric intensity between plates do not depend upon the distance between plates.
(ii) if `sigma_(A)=sigma_(B)=sigma`, i.e. both are charged equally and similarly, then
`E_(I)=(-sigma)/(epsi_(0)),E_(II)=0 and E_(III)=+(sigma)/(epsi_(0))`.