A uniformly charged infinite plane sheet of charge density $2 × 10^{-8}\, C\, m^{-2}$ is held in air. What will be the separation between two equipotential surfaces in the region of electric field produced by the sheet, if the potential difference between them is 10 V? (Take the value of the permittivity of free space = $8.85 × 10^{-12} C^2 N^{-1} m^{-2}$)

8.85 mm

The correct answer is Option (3) → 8.85 mm

Given:

Surface charge density: $\sigma = 2 \times 10^{-8}\ \text{C/m²}$

Potential difference: $V = 10\ \text{V}$

Permittivity of free space: $\epsilon_0 = 8.85 \times 10^{-12}\ \text{C²/N·m²}$

Electric field due to infinite plane sheet: $E = \frac{\sigma}{2 \epsilon_0}$

$E = \frac{2 \times 10^{-8}}{2 \cdot 8.85 \times 10^{-12}} = \frac{2 \times 10^{-8}}{1.77 \times 10^{-11}} \approx 1130\ \text{N/C}$

Separation between equipotential surfaces: $d = \frac{V}{E} = \frac{10}{1130} \approx 8.85 \times 10^{-3}\ \text{m} \approx 8.85\ \text{mm}$

∴ Separation between equipotential surfaces = 8.85 mm