Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): A flat surface is placed in a uniform electric field so that field lines cross it. If the direction of its area vector is reversed, the electric flux through the surface changes sign.
Reason (R): The electric flux through a flat surface is zero when the electric field is parallel to the surface.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Both (A) and (R) are correct but (R) is NOT the correct explanation of (A).
Assertion (A) A flat surface is placed in a uniform electric field so that field lines cross it. If the direction of its area vector is reversed, the electric flux through the surface changes sign. — True. A flat surface has two normals, pointing in opposite directions. Reversing the area vector changes its angle with the field from \( \theta \) to \( 180^\circ-\theta \). Since \( \cos(180^\circ-\theta)=-\cos\theta \), the flux \( E\,\Delta S\cos\theta \) keeps its size but changes its sign.
Reason (R) The electric flux through a flat surface is zero when the electric field is parallel to the surface. — True. When the field is parallel to the surface, it is at \( 90^\circ \) to the area vector. Since \( \cos90^\circ=0 \), no field lines cross the surface and the flux is zero.
Why Reason (R) does not explain Assertion (A)?
R is about one special position of the surface, where the flux is zero. In (A) the field lines do cross the surface, so the flux is not zero.
The sign changes only because the normal is chosen the other way, and this enters through \( \cos\theta \). R says nothing about the choice of the normal.
NCERT (2026-27):
Electric Flux – “Notice one ambiguity here. The direction of an area element is along its normal. But a normal can point in two directions. …”
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 …”