Question:
Two charged thin infinite plane sheets of uniform surface charge density \(\sigma_+\) and \(\sigma_-\), where \(|\sigma_+| \gt |\sigma_-|\), intersect at right angle. Which of the following best represents the electric field lines for this system :
Two charged thin infinite plane sheets of uniform surface charge density \(\sigma_+\) and \(\sigma_-\), where \(|\sigma_+| \gt |\sigma_-|\), intersect at right angle. Which of the following best represents the electric field lines for this system :
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Each sheet gives a uniform field \(\sigma/2\varepsilon_0\) perpendicular to itself (away from the positive sheet, towards the negative sheet); add the two as vectors in each quadrant.
Updated On: Nov 5, 2026
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The Correct Option is A
Solution and Explanation
Seen edge-on, the sheet \(\sigma_+\) is the horizontal line and \(\sigma_-\) is the vertical line in the figures.
Field of an infinite sheet: \(E = \sigma/2\varepsilon_0\), normal to the sheet. The \(\sigma_+\) sheet gives a vertical field of magnitude \(\sigma_+/2\varepsilon_0\), pointing up above the sheet and down below it. The \(\sigma_-\) sheet gives a horizontal field of magnitude \(|\sigma_-|/2\varepsilon_0\) pointing towards that sheet, i.e. leftwards on its right side and rightwards on its left side.
In each of the four quadrants the net field is uniform, so the field lines are straight, and the vertical component is larger because \(|\sigma_+| \gt |\sigma_-|\). Thus: upper-right region: up and left; upper-left: up and right; lower-right: down and left; lower-left: down and right; every line makes a steep angle (more than \(45^\circ\)) with the horizontal.
Option 1 (tall diamond of steep straight lines, arrows away from the \(\sigma_+\) line and towards the \(\sigma_-\) line) matches. Option 2 has the same directions but flatter lines (horizontal component larger), which would correspond to \(|\sigma_-| \gt |\sigma_+|\). Options 3 and 4 are closed loops, which are impossible for an electrostatic field.
Answer: (1) steep straight-line field pattern (tall diamond)
Field of an infinite sheet: \(E = \sigma/2\varepsilon_0\), normal to the sheet. The \(\sigma_+\) sheet gives a vertical field of magnitude \(\sigma_+/2\varepsilon_0\), pointing up above the sheet and down below it. The \(\sigma_-\) sheet gives a horizontal field of magnitude \(|\sigma_-|/2\varepsilon_0\) pointing towards that sheet, i.e. leftwards on its right side and rightwards on its left side.
In each of the four quadrants the net field is uniform, so the field lines are straight, and the vertical component is larger because \(|\sigma_+| \gt |\sigma_-|\). Thus: upper-right region: up and left; upper-left: up and right; lower-right: down and left; lower-left: down and right; every line makes a steep angle (more than \(45^\circ\)) with the horizontal.
Option 1 (tall diamond of steep straight lines, arrows away from the \(\sigma_+\) line and towards the \(\sigma_-\) line) matches. Option 2 has the same directions but flatter lines (horizontal component larger), which would correspond to \(|\sigma_-| \gt |\sigma_+|\). Options 3 and 4 are closed loops, which are impossible for an electrostatic field.
Answer: (1) steep straight-line field pattern (tall diamond)
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