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: (a) (b) (c) (d)

Answer: (b). Each infinite sheet gives a uniform field (σ)/(2ε₀) normal to it, so in each quadrant the net field is uniform and the field lines are straight. Field of σ+ (horizontal sheet) points away from it vertically; field of σ- (vertical sheet) points towards it horizontally. Since |σ+| > |σ-|, the vertical component is larger, so lines are steeper than 45°, giving a tall diamond pattern going away from σ+ and towards σ-. So the answer is (b).

Two charged thin infinite plane sheets of uniform surface charge… | JEE Main 2020 (4 Sep, Shift 1)

Q38JEE Main 2020 · 4 SEP · SHIFT 1Medium

Electric Field & Field Lines

Two charged thin infinite plane sheets of

uniform surface charge density $\sigma_{+}$ and $\sigma_{-}$ ,

where | $\sigma_{+}$ | > | $\sigma_{-}$ |, intersect at right angle.

Which of the following best represents the

electric field lines for this system:

Show answer and solution

Answer: (b)

Each infinite sheet gives a uniform field $\dfrac{\sigma}{2\varepsilon_0}$ normal to it, so in each quadrant the net field is uniform and the field lines are straight.
Field of $\sigma_+$ (horizontal sheet) points away from it vertically; field of $\sigma_-$ (vertical sheet) points towards it horizontally.
Since $|\sigma_+| > |\sigma_-|$ , the vertical component is larger, so lines are steeper than $45^\circ$ , giving a tall diamond pattern going away from $\sigma_+$ and towards $\sigma_-$ .
So the answer is (b).

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Electric Field & Field Lines question from JEE Main 2020 (4 Sep, Shift 1)

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