Use Gauss s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities `sigma and -sigma` respectively .

Explanation

Electric field due to two infinite parallel sheets of a charge : Suppose two infinite , plane non - conducting positively charged sheets 1 and 2 are placed parallel to each other in vacuum or air (as shown in fig.). Let `sigma_(1)andsigma_(2)` be the surface densities of charge on sheets 1 and 2 respectively . As the magnitude of electric intensity `vecE` on either side close to a plane sheet of charge of density `sigma` is :
`E=(sigma)/(2epsilon_(0))`
`vecE` acts perpendicular to the sheet , directed away from the sheets (if charge is positive) or towards the sheet (if charge is negative).
Let `vecE_(1)andvecE_(2)` be the electric intensities at any point due to sheets 1 and 2 respectively . Then , for points outside the sheets , like P. , we have
`E_(1)=(sigma_(1))/(2epsilon_(0))`
and `E_(2)=(sigma_(2))/(2epsilon_(0))`

Since `E_(1) and E_(2)` are in the same direction the magnitude of the resultant intesity at point P . is given by
`E=E_(1)+E_(2)`
`=(sigma_(1))/(2epsilon_(0))+(sigma_(2))/(2epsilon_(0))`
`=(1)/(2epsilon_(0))(sigma_(1)+sigma_(2))`,
At a point in between the sheets , like P we have
`E_(1)=(sigma_(1))/(2epsilon_(0))` (away from sheet 1)
and `E_(2)=(sigma_(2))/(2epsilon_(0))` (away from sheet 2)
Now `E_(1)andE_(2)` are oppositely - directed , so
`E=E_(1)-E_(2)`
`=(sigma_(1))/(2epsilon_(0))-(sigma_(2))/(2epsilon_(0))`
`=(1)/(2epsilon_(0))xx(sigma_(1)-sigma_(2))`.