# Why is the electric potential continuous

Staff Emeritus
Gold Member

## Main Question or Discussion Point

When solving E&M boundary problems, we usually use the condition that the electric potential should be continuous across the boundary. Why is that?

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jtbell
Mentor
Hint: Suppose the electric potential is discontinuous at a certain point. What would the electric force be, on a test charge located at that point?

Claude Bile
Discontinuity implies an infinite electric field at the point of the discontinuity. Such fields would thus require an infinite amount of energy to establish and as far as we know thw energy in our universe is finite.

Claude.

Gza
Excellent answers; I was actually pondering the same question a few days back.

what about point charges? There is discontinuity in potential at origin and although they take an infinite amount of energy they are still used in describing phenomena

well one could answer since the gradient of potential has to be the field there otherwise a discontinuity implies so and so field but remember the gradient condtion itself assumes an differenitiable potential hence continious .
the best answer probably is there never has been a case except at the point charge where the potential has been known to blow up or is discontinious.
but if u meant surface charge then it can be very easily proven just take a short line segment across the boundary of the surface charge and calculate the line integral then u can know that as the line get's smaller so does the integrand and hence the potential difference.
note that the integral does exist in the above case though we 'pass' through a 'charge'.remeber we can always integrate function which are discontinious or blow up at finite points the best example would be the greatest integer or mod valued function....
but all these reasons are secondary the most important being experimental verification and nothing else

although i am wondering whether there is a mathematical proof to show that the potential is continuous everywhere except for point charges.

i am wondering whether there is a mathematical proof to show that the potential is continuous everywhere except for point charges.
it's the same as that for surface charges as in my last post take a small line and calculate the line integral

just one last question, will the potential also be discontinuous on a line charge?

ZapperZ
Staff Emeritus
2018 Award
I'm a bit puzzled by not the original question, but by the responses.

First of all, the original question is:

ahrkron said:
When solving E&M boundary problems, we usually use the condition that the electric potential should be continuous across the boundary. Why is that?
I'm surprised that no one actually pointed out clearly that this is actually not true all the time. While the parallel component of the E-field to the surface is continuous across the boundary, the normal component of the E-field need not be continuous all the time. The existence of any surface charge on that boundary will cause a discontinuity in this normal component of the E-field.

Check out here for the detailed derivation using nothing more than Gauss's law.

So no, E-field need not be continuous in all cases.

Zz.

But the question was about the potential, not the electric field; in fact it was because of this discontinuity of the field that i got confused about the continuity of potential

ZapperZ
Staff Emeritus
2018 Award
Oy! You may smack me now.

That's why no one found it puzzling! I need my eyes checked!

Zz.

Note on point charges

About the discontinuity at point charges.

Point charges are really just an approximation. Any real charge distribution, even an electron, takes up a finite amount of space. So the potential in any real situation is continuous.

Ben Niehoff
Gold Member
The potential can actually be discontinuous across a dipole layer (in the same way that the electric field is discontinuous across a surface charge). A dipole layer is a surface that has no net charge, but has a dipole moment per area (oriented perpendicular to the surface). Dipole layers can come up in certain situations as a means of approximating two oppositely-charged surfaces, very close together.

However, a dipole layer imposes a very specific condition at the boundary, and so the potential problem is still well-posed.

Andy Resnick