A long, thin straight wire carries a uniform charge along its length. The electric field at a distance of 4 cm from the wire is \(9\times10^{4}\ \mathrm{N\,C^{-1}}\). The electric field at a distance of 12 cm from the wire is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3\times10^{4}\ \mathrm{N\,C^{-1}}\)
For a long straight wire with linear charge density \( \lambda \), Gauss’s law with a cylinder around the wire gives \( E=\lambda/2\pi\varepsilon_0 r \). The field falls as \( 1/r \), so tripling the distance makes the field one-third.
\( \begin{aligned} \frac{E_2}{E_1} &= \frac{r_1}{r_2} \\ E_2 &= 9\times10^{4}\times\frac{4}{12} \\ &= 3\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
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”
\( E\times2\pi rl=\lambda l/\varepsilon_0 \)
i.e., \( \displaystyle E=\frac{\lambda}{2\pi\varepsilon_0 r} \)