Density of flux lines

Question

In the given figure flux through surface $\,\rm S_1\,$ is $\,\Phi_1\,$ and through surface $\,\rm S_2\,$ is $\,\Phi_2$. Which option is correct?

(A) $\quad \Phi_1\boldsymbol{=}\Phi_2$
(B) $\quad \Phi_1\boldsymbol{>}\Phi_2$
(C) $\quad \Phi_1\boldsymbol{<}\Phi_2$

The answer to this question is option A. But according to inverse square law, the density of flux lines is inversely proportional to the square of the distance from the source. So according to this statement, shouldn't $\Phi_1>\Phi_2$ be the correct answer, since density of flux lines is greater in $S_1$?

See here https://math.meta.stackexchange.com/q/5020 . It seems that you want to use the symbol $\phi$, such as $\phi_1 = \phi_2$, etc. Edit your post and try it yourself. — The Pointer, May 03 '21 at 06:20
(a) is correct. Read the question again. Nowhere in the question does it actually define or ask about flux density. — DKNguyen, May 03 '21 at 06:33
Hint: Apply Gauss’ law to the closed surface of the cone that includes S1 and S2 — Bob D, May 03 '21 at 07:57
In general the flux through an oriented open or closed surface $,\rm S,$ due to a point charge $,Q,$ is \begin{equation} \Phi_{\mathrm{S}}=\dfrac{\Theta}{4\pi}\dfrac{Q}{\epsilon_{0}} \tag{F}\label{F} \end{equation} where $,\Theta,$ the solid angle by which the charge $,Q,$ $''$sees$''$ the surface, see my answer here Electric flux through an infinite plane due to point charge. — Frobenius, May 03 '21 at 09:17
...also related What is the electric field flux through the base of a cube from a point charge infinitesimally close to a vertex?... — Frobenius, May 03 '21 at 09:19
@Frobenius So, in this question, since the solid angle by which the charge sees the surface is same, that's why flux through S1 and flux through S2 are the same. Is my statement right? — Aashita, May 03 '21 at 09:34
@Aashita : Welcome to PSE. You said that $\Phi_1 =\Phi_2$...By the way, I vote to close your question as homework-and-exercises. — Frobenius, May 03 '21 at 09:35
Please take a minute to read our guidelines for homework and exercise questions as well as check-my-work questions. We intend our questions to be potentially useful to a broader set of users than just the one asking, and we prefer conceptual questions over those just asking for a specific computation. Your question looks on the on-topic side of the border to me, but you should read and be aware of the guidelines for future posts. — Emilio Pisanty, May 03 '21 at 09:56
Note also that we use MathJax to typeset mathematics; you can find a good tutorial here. — Emilio Pisanty, May 03 '21 at 09:56

score 1 · Accepted Answer · answered May 03 '21 at 09:55

according to inverse square law, the density of flux lines is inversely proportional to the square of the distance from the source. So according to this statement, shouldn't $\Phi_1>\Phi_2$ be the correct answer, since density of flux lines is greater in $S_1$?

The flux is defined as the product of the flux density (a.k.a. the density of flux lines) times the area of the surface. For your diagram, the flux density is higher in $S_1$, but the area is smaller.

The deep result of electrostatics is that these two changes completely cancel each other: the electric flux produced by a point charge does not depend on how far the surface is from the charge, but only on the solid angle that it subtends at the charge.

In your example, the two surfaces subtend exactly the same solid angle, and hence have identical electric flux through them.

Thanks a lot,this helped me to clarify my doubt. – Aashita May 03 '21 at 09:58 — Aashita, May 03 '21 at 09:58

score 1 · Answer 2 · answered May 03 '21 at 11:49

Gauss' law states that the net electric flux through a closed surface equals the enclosed net charge divided by the permittivity of the space. There is no charge within the closed surface formed by $S_1$, $S_2$ and the surface of the cone connecting them, thus the net flux is zero.

The electric flux is not flux density. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field. So it is the flux density times the area. While the density of the lines is greater through $S_1$ than $S_2$ the area of $S_1$ is less than $S_2$, such that the product of the density and area is the same.

Another way to look at it is if there is no net charge enclosed, the number of lines entering the volume enclosed by the surface equals the number of lines exiting the volume, for a net number of lines crossing the closed surface of zero.

Hope this helps.

Density of flux lines

2 Answers2