Match the LIST-I with LIST-II
| LIST-I (Charge distribution) | LIST-II (Variation of Electric field E with distance r) |
| A. Point charge | I. E ∝ r−2 |
| B. Infinite line charge | II. E ∝ r−1 |
| C. Charged infinite plane sheet | III. E independent of r |
| D. Short electric dipole | IV. E ∝ r−3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-I, B-II, C-III, D-IV
The way a field falls off with distance follows from Gauss’s law: the flux spreads over a surface whose area grows as \( r^{2} \) (sphere), as r (cylinder) or not at all (plane).
(A) Point charge — this matches (I) E ∝ r−2. The flux spreads over a sphere of area \( 4\pi r^{2} \), giving Coulomb’s law.
\( \displaystyle E = \frac{q}{4\pi\varepsilon_0 r^{2}} \)
(B) Infinite line charge — this matches (II) E ∝ r−1. The flux passes through a cylinder whose curved area \( 2\pi rl \) grows only as r.
\( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0 r} \)
(C) Charged infinite plane sheet — this matches (III) E independent of r. The field lines stay parallel, so the field is uniform on each side of the sheet.
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} \)
(D) Short electric dipole — this matches (IV) E ∝ r−3. The fields of the two opposite charges nearly cancel far away, so the field falls faster than that of a single charge.
\( \displaystyle E_{\mathrm{axial}} = \frac{2p}{4\pi\varepsilon_0 r^{3}} \)