An infinite plane sheet of charge density 10−8 Cm−2 is held in air. In this situation how far apart are two equipotential surfaces, whose potential difference is 5 V?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8.85 mm
The field of an infinite plane sheet of charge is uniform and perpendicular to the sheet, so the equipotential surfaces are planes parallel to the sheet. In a uniform field the distance between two equipotentials is the potential difference divided by the field.
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} \)
\( \displaystyle E = \frac{10^{-8}\,\mathrm{C\,m^{-2}}}{2 \times 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}} \)
\( \displaystyle E \approx 565\,\mathrm{V\,m^{-1}} \)
\( \displaystyle d = \frac{\Delta V}{E} = \frac{5\,\mathrm{V}}{565\,\mathrm{V\,m^{-1}}} \)
\( \displaystyle d \approx 8.85\times10^{-3}\,\mathrm{m} = 8.85\,\mathrm{mm} \)