There are two thin parallel charged sheets having surface charge density \( \sigma \) placed at a small distance apart. Which of the following statements are correct?
(A) If both the sheets are positively charged, then the electric field between them is \( \sigma/\varepsilon_0 \)
(B) If both the sheets are negatively charged, then the electric field between them is \( \sigma/\varepsilon_0 \)
(C) If the left sheet is positively charged and the right sheet is negatively charged the field between them is \( \sigma/\varepsilon_0 \)
(D) If the right sheet is positively charged and the left sheet is negatively charged then the field between them is \( \sigma/\varepsilon_0 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
By Gauss's law, each thin sheet produces a uniform field of magnitude
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} \)
on both of its sides, directed away from a positive sheet and towards a negative sheet. The field between the sheets is the vector sum of the two sheet fields (the superposition principle).
(A) If both the sheets are positively charged, then the electric field between them is \( \sigma/\varepsilon_0 \) — Incorrect. The two fields point away from their own sheets, so between the sheets they cancel: the field is zero.
(B) If both the sheets are negatively charged, then the electric field between them is \( \sigma/\varepsilon_0 \) — Incorrect. Both fields now point towards their own sheets; again they cancel and the field is zero.
(C) If the left sheet is positively charged and the right sheet is negatively charged the field between them is \( \sigma/\varepsilon_0 \) — Correct. Between the sheets both fields point from the positive sheet towards the negative sheet, so they add.
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} \)
(D) If the right sheet is positively charged and the left sheet is negatively charged then the field between them is \( \sigma/\varepsilon_0 \) — Correct. This is (C) seen from the other side; the field is again \( \sigma/\varepsilon_0 \), now from right to left.
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. Thus, the electric field is localised between the two plates and is uniform throughout.”