Question

Two infinite sheets having charge densities `sigma_1` and `sigma_2` are placed in two perpendicular planes whose two-dimensional view is shown in figure. The charges are distributed uniformly on the sheets in electrostatic equilibrium condition. Four points are marked I, II, III and IV. The electric field intensities at these points are `vecE_1, vecE_2, vecE_3, and vecE_4`, respectively. The correct expression for the electric field intensities is

A. `|vecE_1|=|vecE_2|=(sqrt(sigma_1^2+sigma_2^2))/(2epsilon_0)!=|vecE_4|`
B. `|vecE_(2)|=|vecE_(4)|=(sqrt(sigma_1^2+sigma_2^2))/(2epsilon_0)`
C. `|vecE_(1)|=|vecE_(2)|=|vec(E)_(3)|=|vec(E )_(4)|(sqrt(sigma_1^2+sigma_2^2))/(2epsilon_0)`
D. None of the above

Answer 1

Correct Answer - c
Using principle of superposition `vec(E )` due to infinite plane sheet having surface density `sigma` on its one face is `(sigma)/(2 epsilon_(0))`.
So `vec(E _(1))=(sigma_(1))/(2 epsilon_(0))hat(i)+(sigma_(2))/(2epsilon_(0))hat(j)`
`vec(E _(2))= -(sigma_(1))/(2epsilon_(0))hat(i)+(sigma_(2))/(2 epsilon_(0))hat(j)`
`vec(E )_(3)=-(sigma_(1))/(2epsilon_(0))hat(i)-(sigma_(2))/(2epsilon_(0))hat(j)`
`vec(E )_(4)= (sigma_(1))/(2 epsilon_(0))hat(i)-(sigma_(2))/(2 epsilon_(0))hat(j)`

Comments

Thnx