Let us today again discuss another application of gauss law of electrostatics that is to calculate Electric Field Due To Two Infinite Parallel Charged Sheets:-

Consider two parallel sheets of charge A and B with surface density of σ and –σ respectively .The magnitude of intensity of electric field on either side, near a plane sheet of charge having surface charge density σ is given by

E=σ/2ε_{0}

And it is directed normally away from the sheet of positive charge.

The resultant electric field intensity E at any point near the sheet,due to both the sheets A and B will be the vector sum due to the individual intensities set up by each sheet (try to make figure yourself).

We will now calculate the intensity of electric field at different points when the surface density of sheet B changes from +σ to –σ.The electric field E_{2} produced by it will be in opposite direction

At Point P:

At point P ,to the left of the sheets the intensities E_{1} and E_{2}due to both the sheets are in opposite directions.As they are equal in magnitude,the resultant intensity E would be zero,that is,

E=E_{1}-E_{2}=σ/2ε_{0} –σ/2ε_{0}=0

At point Q:

At a point Q ,mid way between the sheets,the intensities E_{1} and E_{2}due to individual sheets are directed normally away from the sheet A or towards the sheet B .therefore, the resultant intensity E at Q is given as

E=E_{1}+E_{2}= σ/2ε_{0} +σ/2ε_{0}

Or E=σ/ε_{0}

At Point R:

At a point R to the right of sheets,the intensities E_{1} and E_{2} are again in opposite directions.Since they are of equal magnitude ,the resultant intensity E would be zero,that is,

E=E_{1}-E_{2}= -σ/2ε_{0}+ σ/2ε_{0}=0

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.

the above are the results for Electric Field Due To Two Infinite Parallel Charged Sheets