Match List-I with List-II (Symbols have their usual meaning)
| List-I | List-II | ||
|---|---|---|---|
| Shape of charged conductors | Expression for the electric fields | ||
| (A) | Uniformly charged infinitely long straight wire | (I) | \( \displaystyle\frac{\sigma}{\varepsilon_{0}} \) |
| (B) | Inside the uniformly charged thin spherical shell | (II) | \( \displaystyle\frac{\sigma}{2\varepsilon_{0}} \) |
| (C) | Uniformly charged infinite plane sheet | (III) | \( \displaystyle\frac{\lambda}{2\pi\varepsilon_{0}r} \) |
| (D) | On the surface of a uniformly charged thin spherical shell | (IV) | \( 0 \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Each field follows from Gauss’s law applied to a suitable Gaussian surface.
(A) Uniformly charged infinitely long straight wire — this matches (III) \( \frac{\lambda}{2\pi\varepsilon_{0}r} \). A coaxial cylinder of radius \( r \) and length \( l \) gives \( E(2\pi rl) = \lambda l/\varepsilon_{0} \).
(B) Inside the uniformly charged thin spherical shell — this matches (IV) 0. A Gaussian sphere inside the shell encloses no charge, so the field is zero everywhere inside.
(C) Uniformly charged infinite plane sheet — this matches (II) \( \frac{\sigma}{2\varepsilon_{0}} \). A pill-box across the sheet has flux through both faces: \( 2EA = \sigma A/\varepsilon_{0} \).
(D) On the surface of a uniformly charged thin spherical shell — this matches (I) \( \frac{\sigma}{\varepsilon_{0}} \). Just outside the shell, \( E = q/(4\pi\varepsilon_{0}R^{2}) \) with \( q = 4\pi R^{2}\sigma \), which gives \( \sigma/\varepsilon_{0} \); “on the surface” is taken, as usual, to mean just outside it.
NCERT (2026-27):
Field Due to an Infinitely Long Straight Uniformly Charged Wire – “… Gauss’s law then gives
\( \displaystyle E \times 2\pi r l = \lambda l/\varepsilon_0 \)
i.e.,
\( \displaystyle E = \frac{\lambda}{2\pi\varepsilon_0 r} \)”
Field Due to a Uniformly Charged Infinite Plane Sheet – “… Therefore by Gauss’s law,
\( \displaystyle 2EA = \sigma A/\varepsilon_0 \quad \text{or,} \quad E = \sigma/2\varepsilon_0 \)”
Field Due to a Uniformly Charged Thin Spherical Shell – “… By Gauss’s law
\( \displaystyle E \times 4\pi r^2 = \frac{\sigma}{\varepsilon_0} 4\pi R^2 \)
Or,
\( \displaystyle E = \frac{\sigma R^2}{\varepsilon_0 r^2} = \frac{q}{4\pi\varepsilon_0 r^2} \)”
Field Due to a Uniformly Charged Thin Spherical Shell – “… Gauss’s law then gives
\( \displaystyle E \times 4\pi r^2 = 0 \)
i.e., E = 0 (r < R ) (1.35) that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.”