A uniformly charged infinite plane sheet of charge density \( 2\times10^{-8}\,\mathrm{C\,m^{-2}} \) 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 = \( 8.85\times10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 8.85 mm
The field of an infinite plane sheet is uniform and perpendicular to the sheet, so the equipotential surfaces are planes parallel to it. In a uniform field the potential difference between two such planes a distance \( d \) apart is \( V = Ed \).
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} = \frac{2\times10^{-8}}{2 \times 8.85\times10^{-12}} = 1.13\times10^{3}\,\mathrm{V\,m^{-1}} \)
\( \displaystyle d = \frac{V}{E} = \frac{2\varepsilon_0 V}{\sigma} = \frac{2 \times 8.85\times10^{-12} \times 10}{2\times10^{-8}} \)
\( \displaystyle d = 8.85\times10^{-3}\,\mathrm{m} = 8.85\,\mathrm{mm} \)