Two large plane parallel sheets shown in the figure have equal but opposite surface charge densities +σ and–σ. A point charge q placed at points P1, P2, and P3 experiences forces F1, F2, and F3 respectively. Then,

Choose the correct answer from the options given below.

The correct answer is:
${\overrightarrow{F}}_1=0,{\overrightarrow{F}}_2\ne 0,{\overrightarrow{F}}_3=0$

Step-by-Step Explanation:

1. Understanding the Setup from the Image:
The provided image shows two large, vertical, parallel plane sheets:
- The left sheet has a uniform positive surface charge density of $+\sigma$.
- The right sheet has a uniform negative surface charge density of $-\sigma$.
Three distinct points are marked:
- P₁: located in the region to the left of the positive sheet.
- P₂: located in the region between the two parallel sheets.
- P₃: located in the region to the right of the negative sheet.
At each of these points, a test point charge $q$ is placed, experiencing forces ${\overrightarrow{F}}_1$, ${\overrightarrow{F}}_2$, and ${\overrightarrow{F}}_3$ respectively.

2. Electric Field due to an Infinite Plane Sheet of Charge:
According to Gauss's Law, the magnitude of the electric field $E$ produced by an infinite, non-conducting plane sheet of uniform surface charge density $\sigma$ at any nearby point is constant (independent of the distance from the sheet) and is given by:
$E=\frac{\sigma }{2{\epsilon }_0}$
The direction of this electric field is:
- Directed perpendicularly away from the sheet if the charge is positive ($+\sigma$).
- Directed perpendicularly towards the sheet if the charge is negative ($-\sigma$).

3. Calculating the Net Electric Field and Force at Each Point:

At Point P₁ (Left Region):
- The electric field due to the positive sheet (${\overrightarrow{E}}_{+}$) points to the left (away from it).
- The electric field due to the negative sheet (${\overrightarrow{E}}_{-}$) points to the right (towards it).
Since the magnitudes are equal:
$E_1=E_{+}-E_{-}=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$
Therefore, the force experienced by the charge $q$ at P₁ is:
${\overrightarrow{F}}_1=q{\overrightarrow{E}}_1=0$

At Point P₂ (Middle Region):
- The electric field due to the positive sheet (${\overrightarrow{E}}_{+}$) points to the right (away from it).
- The electric field due to the negative sheet (${\overrightarrow{E}}_{-}$) also points to the right (towards it).
Since both fields point in the same direction, they reinforce each other:
$E_2=E_{+}+E_{-}=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}\ne 0$
Therefore, the force experienced by the charge $q$ at P₂ is non-zero:
${\overrightarrow{F}}_2=q{\overrightarrow{E}}_2=\frac{q\sigma }{{\epsilon }_0}\ne 0$

At Point P₃ (Right Region):
- The electric field due to the positive sheet (${\overrightarrow{E}}_{+}$) points to the right (away from it).
- The electric field due to the negative sheet (${\overrightarrow{E}}_{-}$) points to the left (towards it).
Since the magnitudes are equal and opposite:
$E_3=E_{+}-E_{-}=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$
Therefore, the force experienced by the charge $q$ at P₃ is:
${\overrightarrow{F}}_3=q{\overrightarrow{E}}_3=0$

4. Summary of Results:
- At P₁: ${\overrightarrow{F}}_1=0$
- At P₂: ${\overrightarrow{F}}_2\ne 0$
- At P₃: ${\overrightarrow{F}}_3=0$
This perfectly matches the correct option.