Re : Why can't elof be discontinuous :proof

[namanjain]

i had a question in my paper
why electrostatic field lines cannot be discontinuous in charge free region

i guessed a weird (but an innovative proof)
Tell me is it correct

So here it goes
"Let's assume that ELOF can be discontinuous

Then i draw a diagram of broken electric field
Now at one of the two free ends i assumed a small Gaussian volume(Only the free end)
Now using gauss law
ø:FLUX
ø = ∑Qenclosed/ε
ELOF ARE ENTERING BUT NOT ESCAPING SO ø≠0
BUT ∑qENCLOSED=0
SO OUR ASSUMPTION IS FALSE
H.P."

 

[UltrafastPED]

You can cast this into a more mathematical form by:

a. Noting that you are working with a vector field - there is a direction (and magnitude) at each point in space
b. The field lines are tangent to the vectors (parallel) at each point; the construction is done by tracing the line that "flows" from point to point. This construction is what guarantees the continuity.
c. The divergence theorem proves that there can be no field lines which do not terminate on sources/sinks, which are your charges.

The fundamental assumption is that you have a vector field; this comes from the vector nature of forces, and that the "field of forces" exists everywhere.

Your proof seems to be equivalent to this.

 

[sophiecentaur]

Electric Fields are fields. The 'lines of force' representation of a field is not rigorous and it is not a good idea to try to take such a simple model and fit it to every phenomenon.

 

[namanjain]

You can cast this into a more mathematical form by:

a. Noting that you are working with a vector field - there is a direction (and magnitude) at each point in space
b. The field lines are tangent to the vectors (parallel) at each point; the construction is done by tracing the line that "flows" from point to point. This construction is what guarantees the continuity.
c. The divergence theorem proves that there can be no field lines which do not terminate on sources/sinks, which are your charges.

The fundamental assumption is that you have a vector field; this comes from the vector nature of forces, and that the "field of forces" exists everywhere.

Your proof seems to be equivalent to this.

Electric Fields are fields. The 'lines of force' representation of a field is not rigorous and it is not a good idea to try to take such a simple model and fit it to every phenomenon.

Well thank your for your respective posts but I'm just in high school and concepts like divergence and curls have not been given to me
it was just a question in my paper for school exams preparation that came and i went thinking till this point
So just small question
Is there any blunder here(please a bit simpler way)

 

[sophiecentaur]

Well thank your for your respective posts but I'm just in high school and concepts like divergence and curls have not been given to me
it was just a question in my paper for school exams preparation that came and i went thinking till this point
So just small question
Is there any blunder here(please a bit simpler way)

All credit to you for trying to get this sorted out. However, it may be better (if you can stand the suspense lol) to wait until the Vector Mathematical treatment arrives on your course and you will find it all makes more sense. The vector operators are actually the 'simpler' way to describe this stuff - in the same way that the differential calculus (which I guess you will have dealt with) is a far simpler way to describe and analyse change than arm waving, sketches and loads of words, which is the alternative.

 

[Meir Achuz]

why electrostatic field lines cannot be discontinuous in charge free region

E must satisfy two boundary conditions across any surface:
1. E_normal is continuous across the surface. This follows from the divergence theorem applied to div E=0 (for a charge free region). This means E must be continuous along its vector direction.
2. E_tangential is continuous across the surface. This follows from Stokes' theorem applied to
curl E=-dB/dt. This means that E cannot have a discontinuous change in direction.