Re : Why can't elof be discontinuous :proof

  • Thread starter namanjain
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In summary, the electrostatic field lines cannot be discontinuous in a charge-free region because they must satisfy two boundary conditions: continuity along both the normal and tangential directions. This is proven through the divergence and Stokes' theorems.
  • #1
namanjain
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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.:tongue:"
 
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  • #2
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.
 
  • #3
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.
 
  • #4
UltrafastPED said:
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 said:
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:tongue:)
 
  • #5
namanjain said:
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:tongue:)

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.
 
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  • #6
namanjain said:
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.
 
Last edited:

1. Why is it important to prove that elof cannot be discontinuous?

It is important to prove that elof, or the limit of a function, cannot be discontinuous because it is a fundamental concept in mathematics. It helps us understand the behavior of a function and its values as it approaches a certain point. Additionally, the continuity of a function is a crucial aspect in many mathematical and scientific applications.

2. What does it mean for a function to be discontinuous?

A function is discontinuous at a point if its limit at that point does not exist, or if it is not equal to its function value at that point. This means that there is a break or gap in the function's graph at that specific point.

3. Can you provide an example of a function that is discontinuous?

One example of a discontinuous function is the step function, defined by f(x) = 0 for x < 0 and f(x) = 1 for x ≥ 0. This function has a clear break at x = 0, where the limit from the left is 0 and the limit from the right is 1.

4. How can we prove that a function is discontinuous?

To prove that a function is discontinuous, we can use the definition of continuity. If the limit of the function at a point does not exist or is not equal to the function value at that point, then the function is discontinuous at that point. We can also use other tools such as the epsilon-delta definition of continuity or the intermediate value theorem to prove discontinuity.

5. Is it possible for a function to be discontinuous at every point?

Yes, it is possible for a function to be discontinuous at every point. An example of such a function is the Dirichlet function, defined by f(x) = 1 for irrational values of x and f(x) = 0 for rational values of x. This function is discontinuous at every point, as there is a break in the graph between every rational and irrational number.

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