In each case of List-I, the electric field at a point has magnitude \(E\) at first. The charge (or charge density) and the distance of the point are then changed as stated.
Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) Infinitely long straight wire, with \(\lambda\) doubled and distance from the wire doubled | (I) \(E/4\) |
| (B) Infinite plane sheet, with \(\sigma\) doubled and distance from the sheet doubled | (II) \(E/2\) |
| (C) Thin spherical shell, point outside, with charge doubled and distance from the centre doubled | (III) \(E\) |
| (D) Infinitely long straight wire, with \(\lambda\) halved and distance from the wire doubled | (IV) \(2E\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Gauss’s law gives the field of each charge distribution. For a long wire \( E=\lambda/2\pi\varepsilon_0r \), for a plane sheet \( E=\sigma/2\varepsilon_0 \), and outside a thin shell \( E=q/4\pi\varepsilon_0r^2 \). The new field follows from the change in charge and in distance.
(A) Infinitely long straight wire, with \( \lambda \) doubled and distance from the wire doubled — This matches (III) \( E \). Since \( E\propto\lambda/r \), the field changes by a factor \( 2/2=1 \).
(B) Infinite plane sheet, with \( \sigma \) doubled and distance from the sheet doubled — This matches (IV) \( 2E \). The field of a sheet depends on \( \sigma \) but not on the distance, so it doubles.
(C) Thin spherical shell, point outside, with charge doubled and distance from the centre doubled — This matches (II) \( E/2 \). Since \( E\propto q/r^2 \), the factor is \( 2/4=1/2 \).
(D) Infinitely long straight wire, with \( \lambda \) halved and distance from the wire doubled — This matches (I) \( E/4 \). Since \( E\propto\lambda/r \), the factor is \( (1/2)/2=1/4 \).
NCERT (2026-27):
Field due to an infinitely long straight uniformly charged wire – “The surface includes charge equal to λ l. Gauss’s law then gives E × 2πrl = λl/ε0 …”
\( \displaystyle E=\frac{\lambda}{2\pi\varepsilon_0 r} \)
Field due to a uniformly charged infinite plane sheet – “2 EA = σA/ε0 or, E = σ/2ε0 …”
“… Note that the above application of the Gauss’ law has brought out an additional fact: E is independent of x also.”
Field due to a uniformly charged thin spherical shell – “… Thus for points outside the shell, the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre.”