# Why field lines must begin or end on electric charges?

1. Jun 29, 2011

### pc2-brazil

1. The problem statement, all variables and given/known data
Use Gauss' law to explain why field lines must begin or end on electric charges.

2. Relevant equations
Gauss' law:
$$\epsilon_0 \phi_E=q$$

3. 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).