Why is the flux no infinite around an isolated charge?

Question

I was wondering that the density of electric field lines determine the strength of the electric field .Now let's say you have an isolated charge ; you know the flux through a closed surface which I take to be a unit sphere suddenly limited by the Gauss law. If the flux is limited then it's bound to be the field lines should be also Limited as if its Infinite if then you know that $\int \int E.ds$ is infinite and the flux becomes infinite as well. Given that the electric field lines are not infinite and they indicate where a unit point charge goes if kept at that point now it create some empty spaces where there are no field lines; so does it suggest that if I put a point charge in those empty spaces for there are no field lines that it would not interact with the field or feel a force outward or inward depending on the charge? Please show that where am I misunderstanding?

Answer 1

I was wondering that the density of electric field lines determine the strength of the electric field

First we need to realize that electric field lines are used to visualize and analyze electric fields and therefore should be considered as a pictorial tool as opposed to a physical entity. For example, one cannot think of the space between the field lines in the picture, which gets larger and larger the further away we go from the point charge producing the field, to be “devoid” of an electric field.

Now let's say you have an isolated charge ; you know the flux through a closed surface which I take to be a unit sphere suddenly limited by the Gauss law.

Not quite sure what you mean here, but the total flux is the same for any spherical closed surface surrounding the charge, no mater how large the sphere. Since the flux is the product of field strength and area perpendicular to the field, as the surface area increases with increasing radius, the electric field strength decreases proportionally so that ExA is a constant.

If the flux is limited then it's bound to be the field lines should be also Limited as if its Infinite if then you know that the double integral of E.dS is infinite and the flux becomes infinite as well.

As said previously, pictorially the number of field lines is the same, but their density decreases with increasing area, so that the total flux bounded by any sized closed surface remains the same.

Given that the electric field lines are not infinite and they indicate where a unit point charge goes if kept at that point now it create some empty spaces where there are no field lines; so does it suggest that if I put a point charge in those empty spaces for there are no field lines that it would not interact with the field or feel a force outward or inward depending on the charge? Please show that where am I misunderstanding?

As previously said, field lines are a pictorial tool. You cannot think of the space between lines as having no field. A charge placed anywhere will experience the field, albeit weakly the further out you go. Instead of thinking of individual, discrete field lines, think of the electric field as a continuum or smear of field lines each line getting thinner and thinner to represent lessening strength. This would be much harder to show in a picture, thus we use the discrete field lines representation.

Hope this helps.

Answer 2

It is not true that if $E \rightarrow \infty$ then it should be the case that $\Phi_E \rightarrow \infty$.

Notice that in your example, the surface area of your sphere tends to zero $A \rightarrow 0$ as the sphere gets smaller. Since you're computing $\Phi_E = EA$, you're trying to multiply a quantity that tends to infinity by a quantity that vanishes. It turns out that $E$ grows exactly in a way to cancel out $A$ shrinking and the product $EA$ remains at a constant value of $q / \epsilon_0$.

If you're not convined that the product $EA$ could feasibly remain constant, consider the functions $f(x) = 1/x$ and $g(x) = x$. As $x \rightarrow 0$, we have $f \rightarrow \infty$ and $g \rightarrow 0$. However, if we consider their product $f(x) g(x)$, you'll probably agree that it remains at a constant value, even as $x \rightarrow 0$.

Answer 3

Gauss's law in integral form says that $\int d\vec A \cdot \vec E = q/\epsilon_0$, not infinite, so there is something not quite right in your reasoning.