Two parallel large thin metal sheets have equal charge densities of $5.31 × 10^{-11}\, C m^{-2}$ of opposite signs. If the permittivity of free space is equal to $8.85 × 10^{-12}\, C^2\, N^{-1}m^{-2}$, the electric field between these sheets will be

$6\, N C^{-1}$

The correct answer is Option (3) → $6\, N C^{-1}$

Given:

Surface charge density, $\sigma = 5.31 \times 10^{-11} \ \text{C/m}^2$

Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \ \text{C}^2/\text{N·m}^2$

Electric field due to a single sheet: $E_\text{sheet} = \frac{\sigma}{2 \epsilon_0}$

For two parallel sheets with opposite charges, the fields add up between the sheets:

$E = 2 \cdot \frac{\sigma}{2 \epsilon_0} = \frac{\sigma}{\epsilon_0}$

Substitute values:

$E = \frac{5.31 \times 10^{-11}}{8.85 \times 10^{-12}} \approx 6.0 \ \text{N/C}$

Electric field between the sheets: $E \approx 6.0 \ \text{N/C}$