Match List-I with List-II
| List-I | List-II |
| Electric Field | Expression |
| (A) Field due to an infinitely long straight uniformly charged wire | (I) \( \sigma\,(2\varepsilon_0)^{-1} \) |
| (B) Field due to a uniformly charged thin spherical shell (inside the shell) | (II) \( q\,(4\pi\varepsilon_0 r^{2})^{-1} \) |
| (C) Field due to a uniformly charged infinite plane sheet | (III) \( \lambda\,(2\pi\varepsilon_0 r)^{-1} \) |
| (D) Field due to a uniformly charged thin spherical shell (outside the shell) | (IV) zero |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Each field follows from Gauss's law, using a Gaussian surface that matches the symmetry of the charge.
(A) Field due to an infinitely long straight uniformly charged wire — this matches (III) \( \lambda\,(2\pi\varepsilon_0 r)^{-1} \). A cylindrical Gaussian surface of radius \( r \) and length \( l \) encloses charge \( \lambda l \), and the flux goes out only through the curved surface.
\( \displaystyle E\,(2\pi r l) = \frac{\lambda l}{\varepsilon_0} \;\Rightarrow\; E = \frac{\lambda}{2\pi\varepsilon_0 r} \)
(B) Field due to a uniformly charged thin spherical shell (inside the shell) — this matches (IV) zero. A Gaussian sphere drawn inside the shell encloses no charge, so the field there is zero.
(C) Field due to a uniformly charged infinite plane sheet — this matches (I) \( \sigma\,(2\varepsilon_0)^{-1} \). A cylinder through the sheet with end faces of area \( A \) encloses charge \( \sigma A \), and the flux leaves through both end faces.
\( \displaystyle E\,(2A) = \frac{\sigma A}{\varepsilon_0} \;\Rightarrow\; E = \frac{\sigma}{2\varepsilon_0} \)
(D) Field due to a uniformly charged thin spherical shell (outside the shell) — this matches (II) \( q\,(4\pi\varepsilon_0 r^{2})^{-1} \). Outside, the shell acts as if its whole charge \( q \) were at the centre.
\( \displaystyle E\,(4\pi r^{2}) = \frac{q}{\varepsilon_0} \;\Rightarrow\; E = \frac{q}{4\pi\varepsilon_0 r^{2}} \)
NCERT (2026-27):
Summary – “… 15. Gauss’s law: The flux of electric field through any closed surface S is 1/ε0 times the total charge enclosed by S. The law is especially useful in determining electric field E, when the source distribution has simple symmetry:
(i) Thin infinitely long straight wire of uniform linear charge density λ
\( \displaystyle \mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 r}\,\hat{\mathbf{n}} \)
… (ii) Infinite thin plane sheet of uniform surface charge density σ
\( \displaystyle \mathbf{E} = \frac{\sigma}{2\varepsilon_0}\,\hat{\mathbf{n}} \)
… (iii) Thin spherical shell of uniform surface charge density σ
\( \displaystyle \mathbf{E} = \frac{q}{4\pi\varepsilon_0 r^{2}}\,\hat{\mathbf{r}} \qquad (r \geq R) \)
\( \displaystyle \mathbf{E} = 0 \qquad (r \lt R) \)
… The electric field outside the shell is as though the total charge is concentrated at the centre. … The field is zero at all points inside the shell.”