Two infinite sheets having charge densities σ1 and σ2 are placed in two perpendicular planes whose two-dimensional view is shown in Fig. 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 E→1,E→2,E→3 and E→4 respectively. The correct expression for the electric field intensities is
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a
|E→1| = |E→2|=σ12+σ222ε0≠|E→4|
b
|E→2| = |E→4|=σ12+σ222ε0
c
|E→1| = |E→2| = |E→3|= |E→4| =σ12+σ222ε0
d
none of these
answer is C.
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Detailed Solution
Using the principle of superposition E due to infinite plane sheet having surface charge density con its one face is σ2ε0.E→1=σ12ε0i+σ22ε0j, E→2=−σ12ε0i+σ22ε0j, E→3=−σ12ε0i−σ22ε0j, E→4=σ12ε0i−σ22ε0j