JEE Main 2020 — Electrostatics Question with Solution

Given, |$${\sigma }_{+}$$| > |$${\sigma }_{-}$$|

We know, $$\left|E_{+}\right|=\frac{\left|{\sigma }_{+}\right|}{{\epsilon }_0}$$

and $$\left|E_{-}\right|=\frac{\left|{\sigma }_{-}\right|}{{\epsilon }_0}$$

$$\therefore$$ $$\left|E_{+}\right|$$ > $$\left|E_{-}\right|$$

Thin infinite uniformly charged planes produces uniform electric field therefore option A and C are obviously wrong.

And as positive charge density is bigger in magnitude so its field along Y direction will be bigger than field of negative charge in X direction and this is evident in option B so it is correct.