A closed cylinder has two flat end faces, 1 and 2, each of area \(S\), and a curved surface 3. It is placed in a uniform electric field \(\vec{E}\) with its axis parallel to the field. The field lines enter through face 1 and leave through face 2. Using the outward normal for each surface, which of the following correctly gives the sign of the electric flux through each surface?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Negative through face 1, positive through face 2, zero through surface 3
The flux through a flat surface in a uniform field is \( ES\cos\theta \). For a closed surface, \( \theta \) is the angle between the field and the outward normal.
Face 1: the outward normal points against the field, so \( \theta=180^\circ \).
Face 2: the outward normal points along the field, so \( \theta=0^\circ \).
Curved surface 3 (area \( S_3 \)): the normal is perpendicular to the field everywhere, so \( \theta=90^\circ \).
\( \begin{aligned} \phi_1 &= ES\cos180^\circ \\ &= -ES \end{aligned} \)
\( \begin{aligned} \phi_2 &= ES\cos0^\circ \\ &= +ES \end{aligned} \)
\( \begin{aligned} \phi_3 &= ES_3\cos90^\circ \\ &= 0 \end{aligned} \)
From the above, the total flux is \( -ES+ES+0=0 \). Hence the result agrees with Gauss’s law, as the cylinder encloses no charge.
Option (1) Positive through face 1, negative through face 2, zero through surface 3 — Incorrect. This swaps the signs of faces 1 and 2. The field enters through face 1, so the outward normal of face 1 points against the field.
Option (2) Zero through all three surfaces — Incorrect. Only the total flux is zero. The fluxes \( -ES \) and \( +ES \) through the two end faces add up to zero.
Option (3) Positive through faces 1 and 2, zero through surface 3 — Incorrect. Positive flux through face 1 would need its outward normal along the field. That is true only for face 2.
Option (4) Negative through face 1, positive through face 2, zero through surface 3 — Correct. As worked out above, \( \phi_1=-ES \), \( \phi_2=+ES \) and \( \phi_3=0 \).
NCERT (2026-27):
Gauss’s Law – “… Now the normal to the surface 3 at every point is perpendicular to E, so by definition of flux, φ3 = 0. Further, the outward normal to 2 is along E while the outward normal to 1 is opposite to E. Therefore, φ1 = –E S1, φ2 = +E S2 …”