Why field lines must begin or end on electric charges?

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SUMMARY

The discussion centers on the application of Gauss' law to explain why electric field lines must begin or end on electric charges. It establishes that when a Gaussian surface encloses a positive charge (+q), the electric flux (фE) is positive, indicating that field lines emanate outward. Conversely, if the surface encloses a negative charge (-Q), the net enclosed charge results in zero flux, necessitating that field lines entering the surface from the positive charge must also exit through the negative charge. The reasoning is reinforced by considering scenarios with multiple charges or continuous charge distributions, where any deviation would violate Gauss' law.

PREREQUISITES
  • Understanding of Gauss' law and its mathematical formulation: \(\epsilon_0 \phi_E=q\)
  • Familiarity with electric field lines and their representation
  • Knowledge of positive and negative charge behavior in electric fields
  • Concept of Gaussian surfaces in electrostatics
NEXT STEPS
  • Study the implications of Gauss' law in different charge configurations
  • Explore the concept of electric flux and its calculation in various scenarios
  • Learn about continuous charge distributions and their effect on electric fields
  • Investigate the relationship between electric field lines and potential energy
USEFUL FOR

Students of physics, educators teaching electrostatics, and anyone interested in understanding the principles governing electric fields and charges.

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Homework Statement


Use Gauss' law to explain why field lines must begin or end on electric charges.

Homework Equations


Gauss' law:
\epsilon_0 \phi_E=q

The Attempt at a Solution


I'm not sure whether this solution is incomplete or inconsistent.

First, I take a gaussian surface enclosing one positive charge +q. There, фE > 0. So, field lines are outwards from the gaussian surface. If I change the size of the surface, the flux must remain constant. If a field line finished abruptly, it would be possible to change the size of the gaussian surface for the line not to cut it; this would change the flux, violating Gauss' law.
If the field lines started from a point different from the positive charge, changing the size of the gaussian surface would also make it possible not to enclose some line, also changing the total flux. This reasoning also applies to the case where the charge inside the surface is negative.

Now, suppose that the surface encloses a negative charge -Q along with the positive charge +q:

If q = Q, as the charges will cancel each other out, the net enclosed charge will be zero, therefore Gauss' law states that the flux will be zero too. So, the number of field lines leaving a surface must be equal to the number of field lines entering the surface. And the field lines leaving the positive charge must be entering the negative charge. Otherwise, if a line finished in a point inside the surface which is not on any charge, it would be possible to modify the gaussian surface in order to change the flux even though it is still enclosing the two charges.

I'm not sure how to proceed if q is different from Q:
I think that, if q > Q, following Gauss' law, the flux will be positive but smaller than it was in the situation where there is only the positive charge +q. It means that some lines which come from the positive charge are not leaving the surface. These lines must be entering the negative charge, or else they would have to break in the middle of the way and changing the surface would change the flux. I think that this same reasoning can be applied if q < Q; the difference is that the flux will be negative (therefore, there will be more lines entering the surface than lines leaving it).

Thank you in advance.
 
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I think your solution is convincing, but not absolutely rigorous. What if there's 3 charges, or a continuous charge distribution? I would approach the problem by identifying the starting or ending point of a field line, and drawing a Gaussian sphere around the start/end point. If I shrink the sphere, the charge enclosed approaches 0 no matter what the charge distribution looks like, but the flux doesn't. This violates Gauss' law.
 

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