The field lines of an isolated point charge spread out radially. A small area \(\Delta S\) is held normal to the lines at a distance \(r\) from the charge, and \(n\) field lines cross it. At a distance \(3r\) from the charge, the same \(n\) lines cross a normal area of
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(9\,\Delta S\)
A solid angle is the angle of a cone drawn out from the charge. The number of radial field lines inside a given solid angle \( \Delta\Omega \) stays the same at every distance. The area that this solid angle cuts out at a distance \( r \) is \( r^{2}\Delta\Omega \), so the area grows as the square of the distance.
At the distance \( r \), the solid angle of the small area is \( \Delta\Omega=\Delta S/r^{2} \). The same solid angle at the distance \( 3r \) cuts out the area \( \Delta S_2 \).
\( \begin{aligned} \Delta S_2 &= (3r)^{2}\,\Delta\Omega \\ &= 9r^{2}\times\frac{\Delta S}{r^{2}} \\ &= 9\,\Delta S \end{aligned} \)
Therefore, the same \( n \) lines are spread over nine times the area. The number of lines per unit area becomes one-ninth, which is how the field of a point charge falls as \( 1/r^{2} \).
NCERT (2026-27):
Electric Field Lines – “… Likewise, in three-dimensions the solid angle subtended by a small perpendicular plane area ΔS, at a distance r, can be written as ΔΩ = ΔS/r2. We know that in a given solid angle the number of radial field lines is the same. … for two points P1 and P2 at distances r1 and r2 from the charge, the element of area subtending the solid angle ΔΩ is r12 ΔΩ at P1 and an element of area r22 ΔΩ at P2, respectively. The number of lines (say n) cutting these area elements are the same.”