An infinite plane sheet of charge density $10^{-8} C m^{-2}$ is held in air. In this situation, the distance between two equipotential surfaces, whose potential difference is 10 V, will be

17.7 mm

The correct answer is Option (1) → 17.7 mm

Given:

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

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 an infinite plane sheet: $E = \frac{\sigma}{2 \epsilon_0} = \frac{10^{-8}}{2 \cdot 8.85 \times 10^{-12}} \approx 564.97\ \text{N/C}$

Separation between equipotential surfaces: $d = \frac{V}{E} = \frac{10}{564.97} \approx 0.0177\ \text{m} \approx 1.77\ \text{cm}$

∴ Distance between the equipotential surfaces = 1.77 cm