If \( +\sigma \) and \( -\sigma \) are the uniform surface charge densities of two parallel infinite thin plane sheets, then the net electric field at the midpoint of the two sheets will be given by: (Where \( \varepsilon_0 \) is the permittivity of free space, where the sheets are kept)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\sigma}{\varepsilon_0} \)
By Gauss's law, an infinite plane sheet with surface charge density \( \sigma \) produces a uniform field on both sides, directed away from a positive sheet and towards a negative sheet:
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} \)
Between the sheets, the field of the \( +\sigma \) sheet points away from it, towards the \( -\sigma \) sheet, and the field of the \( -\sigma \) sheet also points towards the \( -\sigma \) sheet. The two fields are in the same direction, so they add:
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \)
The field is the same at every point between the sheets, the midpoint included, and points from the positive sheet to the negative sheet. Outside the sheets the two fields are opposite and cancel, which is why zero is the field outside, not between them.
NCERT (2026-27):
The Parallel Plate Capacitor – “… In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A} \qquad (2.41) \)
The direction of electric field is from the positive to the negative plate.”