Electric Intensity at any point outside two parallel, plane, oppositely charged conductors. Suppose there are two parallel plane sheets A and B with uniform charge densities σ1 and σ2. Let sheet A be to the left of sheet B. Then electric field at a point can be found by the superposition law. There are three regions.
Region I is to the left of sheet A.
Region II between sheets A and B.
Region III is to the right of sheet B.
If the direction from left to right is taken +ve, then using the result, E = \(\frac{\sigma}{2ɛ_o}\).
For region I
E = (-E1 - E2) = \(\frac{\sigma}{2ɛ_o}\) (- σ1 - σ2) ......................(i)
For region II
E = (E1 - E2) = \(\frac{\sigma}{2ɛ_o}\) ( σ1 - σ2) ......................(ii)
For region III
E = (E1 + E2) = \(\frac{\sigma}{2ɛ_o}\) ( σ1 + σ2) ......................(iii)
A negative field means that it points from right to left and a + ve field means the opposite.
The case σ1 > σ2 > 0 is illustrated in Fig.
If the two sheets are oppositely charged such that σ1 = σ and σ2 = σ, then
From eq. (i), (ii) and (iii), we get
EI = 0, EII = \(\frac{1}{2ɛ_o}\) (2σ) = \(\frac{\sigma}{2ɛ_o}\) and EIII = 0
