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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 . |
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Answer» Solution :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 ofcharge on sheets 1 and 2 respectively . As the magnitude of electric intensity `vecE` on either side CLOSE to a plane sheet ofcharge 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))`. |
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