A flat square of area \(A\) is placed in a uniform electric field \(E\). Match the position of the square in List-I with the electric flux through it in List-II.
| List-I | List-II |
|---|---|
| (A) Normal to the square at \(60^\circ\) to the field | (I) \(EA\) |
| (B) Plane of the square at \(60^\circ\) to the field | (II) Zero |
| (C) Plane of the square perpendicular to the field | (III) \(EA/2\) |
| (D) Plane of the square parallel to the field | (IV) \(\sqrt{3}\,EA/2\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The flux through a flat surface in a uniform field is \( \phi=EA\cos\theta \). The angle \( \theta \) is measured between the field and the area vector, which lies along the normal to the surface.
(A) Normal to the square at \( 60^\circ \) to the field — This matches (III) \( EA/2 \). Here \( \theta=60^\circ \) and \( \cos60^\circ=1/2 \).
(B) Plane of the square at \( 60^\circ \) to the field — This matches (IV) \( \sqrt{3}\,EA/2 \). The normal is at \( 90^\circ-60^\circ=30^\circ \) to the field, so \( \theta=30^\circ \) and \( \cos30^\circ=\sqrt{3}/2 \).
(C) Plane of the square perpendicular to the field — This matches (I) \( EA \). The normal is then along the field, so \( \theta=0^\circ \) and the flux is the largest.
(D) Plane of the square parallel to the field — This matches (II) Zero. The normal is at \( 90^\circ \) to the field, and the field lines run along the surface without crossing it.
NCERT (2026-27):
Electric Flux – “… Clearly, the normal to the plane specifies the orientation of the plane. Thus the direction of a planar area vector is along its normal.”
Electric Flux – “We now come to the definition of electric flux. Electric flux Δφ through an area element ΔS is defined by Δφ = E.ΔS = E ΔS cosθ …”
Electric Flux – “… When θ = 90°, field lines will be parallel to ΔS and will not cross it at all …”