Three large plates having uniform surface charge densities are shown in figure. Electric field at point 'P' will be :

Options

A. `(2sigma)/epsilon_0 hatk`
B. `-(2sigma)/epsilon_0 hatk`
C. `(4sigma)/epsilon hatk`
D. `-(4sigma)/epsilon_0 hatk`

Answer

B

Explanation

Electric field intensity due to surface charge distribution is given by `E=sigma/(2epsilon_0)` , here `sigma` is charge per unit area of the plate.
Electric field intensity due to positive charge is directed away from it and due to negative charge it is directed towards it.
Electric field due to `-sigma` at point P will be directed opposite to Z-axis. `vecE_1=-sigma/(2epsilon_0)hatk`
Electric field due to `-2sigma` at point P will be directed opposite to Z-axis.`vecE_2=-(2sigma)/(2epsilon_0) hatk=-sigma/epsilon_0 hatk`
Electric field due to `sigma` at point P will be directed opposite to Z-axis. `vecE_3=-sigma/(2epsilon_0)hatk`
Net electric field at point P can be written as follows :
`vecE=vecE_1+vecE_2+vecE_3=-sigma/(2epsilon_0) hatk -sigma/epsilon_0 hatk -sigma/(2epsilon_0) hatk =- (2sigma)/epsilon_0 hatk`