Match the LIST-I with LIST-IILIST-ILIST-IIA.E is independent of rI.Fo…

The correct answer is Option (2) → A-IV, B-III, C-I, D-II

Each distance dependence follows from the standard field expression of the charge distribution.

(A) E is independent of r — this matches (IV) For a uniformly charged infinite plane sheet. Gauss's law gives a field that is the same at every distance from the sheet.

    \( \displaystyle E = \frac{\sigma}{2\varepsilon_0} \)

(B) E is inversely proportional to r — this matches (III) For a uniform line charge. A cylindrical Gaussian surface gives

    \( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0 r} \)

(C) E is inversely proportional to r² — this matches (I) For a point charge, by Coulomb's law:

    \( \displaystyle E = \frac{q}{4\pi\varepsilon_0 r^{2}} \)

(D) E is inversely proportional to r³ — this matches (II) For a short dipole. On the axis and on the equatorial line the fields of the two charges nearly cancel, leaving a faster fall-off:

    \( \displaystyle E_{\mathrm{axial}} = \frac{2p}{4\pi\varepsilon_0 r^{3}} \)

NCERT (2026-27):
Summary – “… The 1/r³ dependence of dipole electric fields should be noted in contrast to the 1/r² dependence of electric field due to a point charge.”

Summary – “… (i) Thin infinitely long straight wire of uniform linear charge density \( \lambda \)

    \( \displaystyle \mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 r}\,\hat{\mathbf{n}} \)

where r is the perpendicular distance of the point from the wire and \( \hat{\mathbf{n}} \) is the radial unit vector in the plane normal to the wire passing through the point.”

Summary – “… (ii) Infinite thin plane sheet of uniform surface charge density \( \sigma \)

    \( \displaystyle \mathbf{E} = \frac{\sigma}{2\varepsilon_0}\,\hat{\mathbf{n}} \)

where \( \hat{\mathbf{n}} \) is a unit vector normal to the plane, outward on either side.”