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_1| = |vecE_4| = (sqrt(sigma_(1)^(2) + sigma_(2)^(2)))/(2epsilon_0) `
C. `|vecE_1| = |vecE_2| =|vecE_3| = |vecE_4| = (sqrt(sigma_(1)^(2) + sigma_(2)^(2)))/(2epsilon_0) `
D. none of these

Answer 1

Correct Answer - C
c. Using the principle of superposition `vecE` due to infinite plane
sheet having surface charge density `sigma` on its one face is `sigma//2epsilon_0`.
So
`vecE_1 = (sigma_1)/(2epsilon_0) I + (sigma_2)/(2epsilon_0)j`
`vecE_2 = -(sigma_1/(2epsilon_0))i + sigma_2/(2epsilon_0)j`
`vecE_3 = - sigma_1/(2epsilon_0)i - sigma_2/(2epsilon_0)j`
`vecE_4 = sigma_1/(2epsilon_0)i - sigma_2/(2epsilon_0)j`