About Electrostatic Potential and Electric Field

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SUMMARY

The discussion focuses on determining the continuity of the electric field (E) and electrostatic potential (\Phi) at the surface of a spherical shell in electrostatics. It emphasizes the necessity for the potential functions, \varphi_1 and \varphi_2, derived from solving the Laplace equation, to satisfy the condition that the one-sided limits at the boundary (r=r_0) are equal. Specifically, continuity is confirmed if limr→r0+ \varphi_1(r) = limr→r0- \varphi_2(r). This principle is applicable in both simple and more complex scenarios involving multiple variables.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with the Laplace equation
  • Knowledge of spherical symmetry in potential functions
  • Basic calculus, specifically limits
NEXT STEPS
  • Study the continuity of multivariable functions in calculus
  • Explore the applications of the Laplace equation in electrostatics
  • Learn about boundary conditions in electrostatic problems
  • Investigate the implications of electric field discontinuities
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Students and professionals in physics, particularly those specializing in electromagnetism, as well as educators seeking to clarify concepts related to electric fields and potentials in electrostatics.

jhosamelly
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How do I know if E (electric field) and [itex]\Phi[/itex] (electrostatic potential) is continuous at the surface?

I'm asking this because I have a problem choosing which formula to use, if you know what I mean. There are a lot of formulas for E and [itex]\Phi[/itex] .
 
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Based on your question and you posting this in the physics section, I'm assuming you're not very interested in the precise mathematical definition, so I'm going to give you what hopefully is at least a bit more concrete example. If my assumption is wrong, say so and I'll be more precise.

Let's say you have two potential functions, [itex]\varphi_1[/itex] and [itex]\varphi_2[/itex], from, for example, solving the Laplace equation for a charge distribution in two distinct regions (say, for [itex]r>r_0[/itex] and [itex]r<r_0[/itex] in a sphere), and the potential has to be continuous on a surface in order to fill some boundary conditions.

In elementary electrostatics the solutions are often nice and symmetric, so let's assume further that the solutions have spherical symmetry [itex]\varphi_1=\varphi_1(r)[/itex] and [itex]\varphi_2=\varphi_2(r)[/itex] and we wish to check whether the potential is continuous on a spherical shell with [itex]r=r_0[/itex]. The only thing that needs to be done is to take the limit as [itex]r\rightarrow r_0[/itex] for both functions: The potential is continuous if (the one-sided limits exist and)

[tex]\lim_{r\rightarrow r_0+} \varphi_1(r) = \lim_{r\rightarrow r_0-}\varphi_2(r)[/tex]

Simple as that! Even more, if the functions are for example both defined at [itex]r_0[/itex], which could very well be the case in electrostatics, you naturally only need to check that [itex]\varphi_1(r_0)=\varphi_2(r_0)[/itex]!

In case you do want to know what happens in a more general case, with multiple variables or whatever, that's simple, as well. You basically just have to take take the (one-sided) limits of your function near the surface. I can elaborate on this (and you can easily find information on the continuity of a multivariable function yourself as well), I just don't want to be confuse you with unnecessary details.
 
DeIdeal said:
Let's say you have two potential functions, [itex]\varphi_1[/itex] and [itex]\varphi_2[/itex], from, for example, solving the Laplace equation for a charge distribution in two distinct regions (say, for [itex]r>r_0[/itex] and [itex]r<r_0[/itex] in a sphere), and the potential has to be continuous on a surface in order to fill some boundary conditions.

In elementary electrostatics the solutions are often nice and symmetric, so let's assume further that the solutions have spherical symmetry [itex]\varphi_1=\varphi_1(r)[/itex] and [itex]\varphi_2=\varphi_2(r)[/itex] and we wish to check whether the potential is continuous on a spherical shell with [itex]r=r_0[/itex]. The only thing that needs to be done is to take the limit as [itex]r\rightarrow r_0[/itex] for both functions: The potential is continuous if (the one-sided limits exist and)

[tex]\lim_{r\rightarrow r_0+} \varphi_1(r) = \lim_{r\rightarrow r_0-}\varphi_2(r)[/tex]

This answers my question, thank you! :)
 

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