JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Electrostatics

The three sheets carry surface charge densities $-\sigma$ (at $X=-a$), $-2\sigma$ (at $X=a$) and $+\sigma$ (at $X=3a$); point P lies between $X=a$ and $X=3a$.
Each infinite sheet gives a field $\frac{|\sigma |}{2{ϵ}_0}$; a positive sheet points the field away from itself and a negative sheet toward itself. Taking $+x$ as positive, at P:
from the $-\sigma$ sheet (P to its right): $-\frac{\sigma }{2{ϵ}_0}$;
from the $-2\sigma$ sheet (P to its right): $-\frac{2\sigma }{2{ϵ}_0}=-\frac{\sigma }{{ϵ}_0}$;
from the $+\sigma$ sheet (P to its left): $-\frac{\sigma }{2{ϵ}_0}$.

Magnitude $=\frac{2\sigma }{{ϵ}_0}$, so $x=2.$