**Electric field due to two charged parallel sheets:**

Consider a thin plane infinite sheet having positive charge density σ. In order to find the electric field intensity at a point p, which is at a perpendicular distance r from the plane shell, we choose a closed cylinder of length 2r, whose ends have an area as the Gaussian surface.

Let M and N be two parallel charged conductors [Figure]. The sheet M is positively charged and the sheet N is negatively charged and are charged with the same density. The charge density of both of them is σ. Electric field E^{→} at a point between the two sheets is to be found out.

Now, the magnitude of the electric field at point P due to the sheet M will be E_{1} = σ / 2ε_{0}; and its direction will be along MN. Again for the sheet N magnitude of the electric field at point P will be E_{2} = σ / 2ε_{0}; and since the sheet N is charged negatively, so the direction of E_{2} will be along MN.

Hence magnitude of the total electric field at point P will be,

E = E_{1} + E_{2}

= (σ / 2ε_{0}) + (σ / 2ε_{0})

So, E = σ / ε_{0} … … (1)

The direction of the electric field will be from sheet M to N i.e., from positive charge to negative charge.

At a point Q outside the two sheets, the magnitude of the electric field due to the sheets M and N will be E_{1} = σ / 2ε_{0} and E_{2} = σ / 2ε_{0} respectively. But as they act opposite to each other so the net field will be zero. That means, there will not be electric field outside the two sheets.

In Conclusion, E due to two oppositely charged infinite plates is σ/ε_{0} at any point between the plates and is zero for all external points.

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