Two charged thin infinite plane sheets of
uniform surface charge density σ + and σ - ,
where |σ + | > |σ - |, intersect at right angle.
Which of the following best represents the
electric field lines for this system :

Solution

Physics sir · 4 steps

Each sheet gives a uniform field, so add the two quadrant by quadrant: straight lines, steeper than 45°.

1. Field of each sheet. An infinite sheet gives a uniform field of magnitude |σ|/2ε₀ at right angles to it. For σ₊ (the horizontal sheet) the field points away from it, up above and down below. For σ₋ (the vertical sheet) the field points towards it from both sides.
2. Add in one quadrant. In the upper-right quadrant the net field is E⃗ = 1/2ε₀(-|σ₋| î + |σ₊| ĵ), which points up and to the left. It is the same at every point of that quadrant, so the field lines there are parallel straight lines. They bend only where they cross a sheet.

   |E_y/Eₓ| = |σ₊|/|σ₋| > 1

3. Shape of the lines. Because the slope is greater than 1, the lines make more than 45° with the σ₊ sheet. Joined across the four quadrants, they form a diamond stretched along the σ₋ sheet. The arrows run away from the σ₊ sheet and towards the σ₋ sheet.

   Option	Field lines drawn	Verdict
   (a)	closed circles	electrostatic field lines never close on themselves
   (b)	straight lines, diamond stretched along σ₋, away from σ₊ and towards σ₋	matches
   (c)	closed ellipses	electrostatic field lines never close on themselves
   (d)	straight lines, diamond stretched along σ₊	slope less than 1, as if |σ₋| > |σ₊|

4. So. Only option (b) shows straight lines steeper than 45° that start at the σ₊ sheet and end on the σ₋ sheet. The answer is (b).

The idea to keep. For crossed infinite sheets, the field in each region is uniform, so the field lines are straight. Their slope is the ratio of the two charge densities, and the stronger sheet decides which way the lines lean.

Answer from the NTA final answer key. Solution by PrepWiser.