In the figure given below, plate 1 and plate 2 are separated by a small distance d. If the air is the medium between the plates, then
A. The direction of the electric field is from plate 1 to plate 2.
B. The direction of the electric field is from plate 2 to plate 1.
C. The electric field in region I (region above plate 1) is zero.
D. The electric field in the region between plates is zero.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A and C only
Each large charged plate produces a uniform field \( \sigma/2\varepsilon_0 \) on each side, directed away from a positive plate and towards a negative plate (\( \sigma = q/A \)); the field at any point is the sum of the two.
\( \displaystyle \text{Between the plates: } E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{q}{A\varepsilon_0} \)
\( \displaystyle \text{Outside the plates: } E = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0 \)
(A) The direction of the electric field is from plate 1 to plate 2. — Correct. Between the plates the field of the positive plate 1 (away from it) and that of the negative plate 2 (towards it) both point from plate 1 to plate 2.
(B) The direction of the electric field is from plate 2 to plate 1. — Incorrect. Field lines run from the positive plate to the negative plate, not the other way.
(C) The electric field in region I (region above plate 1) is zero. — Correct. Above plate 1 the field of plate 1 points upwards and that of plate 2 points downwards, with equal magnitudes, so they cancel.
(D) The electric field in the region between plates is zero. — Incorrect. Between the plates the two fields add, giving \( E = \sigma/\varepsilon_0 \).
NCERT (2026-27):
The Parallel Plate Capacitor – “… Plate 1 has surface charge density \( \sigma = Q/A \) and plate 2 has a surface charge density \( -\sigma \). Using Eq. (1.33), the electric field in different regions is:
Outer region I (region above the plate 1),
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0 \qquad (2.39) \)
Outer region II (region below the plate 2),
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0 \qquad (2.40) \)
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.”