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In a charge-free region, electric field lines can be taken to be continuous curves without any breaks

What is meant by charge-free region? I suppose it means absence of any external charge.

What if some other charge is present near the source of the field, will it make the field lines discontinuous?

Why does it state that "field lines can be taken to be continuous"? Are they not exactly continuous?

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  • $\begingroup$ Doesn't it indirectly state what Gauss' Law states? Considering a region in space where there is no charge, the Electric flux has to be $0$ (according to Gauss' Law) i.e, there is no sink or source within that region. Hence, field lines are continuous in that region. Am I right? $\endgroup$
    – Kaushik
    Nov 12, 2020 at 19:15

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enter image description here

By convention, electric field lines are said to start from a positive charge and end at a negative charge. As you can see in the above figure, the field lines come to an abrupt stop at the surface of the charge. When there isn't any charge, the electric field lines must be continuous. The only place where they can start or end is at a charge.


Bonus: Fields tend to be uniform at points far away from the charge source

enter image description here

In the above figure, consider that the big black circular patch is a charge. As you move away from the charge, the electric field lines not only tend to be continuous, they also have nearly the same strength.

For example, have a look at the region bounded by the red rectangle. The field lines appear to be parallel straight lines. Of course, they are continuous and as they appear to be parallel, their magnitude is nearly the same.

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The idea is that electric field lines can start on positive charges and finish on negative charges.
In such a case electric field lines would be discontinuous.

With no charges present the electric field lines must be continuous.

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  • $\begingroup$ If no charges are present, then the field doesn't have any magnitude and thus field lines wouldn't exist, right? $\endgroup$
    – Silica19
    Jul 9, 2022 at 4:00
  • $\begingroup$ That is correct but as I pointed out the lines just disappear. $\endgroup$
    – Farcher
    Jul 9, 2022 at 7:13
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Suppose that there exists a discontinuity in the field $E(x,y,z)$ at some $P(x',y',z')$. Then, consider an infinitesimally small gaussian sphere, which passes through the point $P(x',y',z)$. Clearly, $\nabla \cdot \vec E \ne 0$ inside this sphere. But from gauss law we have:

$\nabla \cdot \vec E = \frac{\rho}{\epsilon_0}$. This means that $\rho \ne 0$.

Contradiction, since the region is charge free.

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It only means that in a region far away from the charge which is creating the electric field, the electric field lines are continuous.it is stated that electric field lines are closed continuoudenter link description here

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