List-I gives a positively charged body and the Gaussian surface used with it. Match each with the flux through that surface in List-II.
| List-I | List-II |
|---|---|
| (A) Long straight wire; a cylinder with the wire along its axis | (I) Flux only through the curved surface |
| (B) Infinite plane sheet; a cylinder with its axis normal to the sheet, cut in the middle by the sheet | (II) Flux only through the two flat end faces |
| (C) Thin spherical shell; a concentric sphere smaller than the shell | (III) Zero flux through every part of the surface |
| (D) Thin spherical shell; a concentric sphere larger than the shell | (IV) Equal outward flux through equal areas all over the surface |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
The flux through a part of a surface is zero where the field runs along that part. It is largest where the field is normal to it.
(A) Long straight wire; a cylinder with the wire along its axis — This matches (I). The field of the wire is radial. It runs along the two flat ends, so they get no flux. It is normal to the curved surface at every point.
(B) Infinite plane sheet; a cylinder with its axis normal to the sheet, cut in the middle by the sheet — This matches (II). The field is normal to the sheet. It runs along the curved side and crosses both flat ends outward.
(C) Thin spherical shell; a concentric sphere smaller than the shell — This matches (III). This sphere encloses no charge, and the field is zero at every point inside the shell. So no part of the surface has any flux.
(D) Thin spherical shell; a concentric sphere larger than the shell — This matches (IV). Outside the shell the field is radial and has the same magnitude at every point of this sphere. So equal areas get equal outward flux.
NCERT (2026-27):
Field due to an infinitely long straight uniformly charged wire – “… Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero. At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r. …”
Field due to a uniformly charged infinite plane sheet – “… As seen from the figure, only the two faces 1 and 2 will contribute to the flux; electric field lines are parallel to the other faces and they, therefore, do not contribute to the total flux.”
Field due to a uniformly charged thin spherical shell – “… that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.”
Field due to a uniformly charged thin spherical shell – “… The electric field at each point of the Gaussian surface, therefore, has the same magnitude E and is along the radius vector at each point. Thus, E and ΔS at every point are parallel and the flux through each element is E ΔS. …”