About Electrostatic Potential and Electric Field

AI Thread Summary
To determine if the electric field (E) and electrostatic potential (Φ) are continuous at a surface, one must check the limits of the potential functions from both sides of the boundary. For two potential functions, Φ1 and Φ2, defined in different regions, continuity is confirmed if the one-sided limits as the radius approaches the boundary are equal. Specifically, for a spherical shell at r = r0, continuity holds if the limits from both sides match, or if both functions are defined at r0 and equal. This method applies to both simple and more complex scenarios involving multiple variables. Understanding these continuity conditions is crucial for solving electrostatic problems effectively.
jhosamelly
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How do I know if E (electric field) and \Phi (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 \Phi .
 
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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, \varphi_1 and \varphi_2, from, for example, solving the Laplace equation for a charge distribution in two distinct regions (say, for r>r_0 and r<r_0 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 \varphi_1=\varphi_1(r) and \varphi_2=\varphi_2(r) and we wish to check whether the potential is continuous on a spherical shell with r=r_0. The only thing that needs to be done is to take the limit as r\rightarrow r_0 for both functions: The potential is continuous if (the one-sided limits exist and)

\lim_{r\rightarrow r_0+} \varphi_1(r) = \lim_{r\rightarrow r_0-}\varphi_2(r)

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

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, \varphi_1 and \varphi_2, from, for example, solving the Laplace equation for a charge distribution in two distinct regions (say, for r>r_0 and r<r_0 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 \varphi_1=\varphi_1(r) and \varphi_2=\varphi_2(r) and we wish to check whether the potential is continuous on a spherical shell with r=r_0. The only thing that needs to be done is to take the limit as r\rightarrow r_0 for both functions: The potential is continuous if (the one-sided limits exist and)

\lim_{r\rightarrow r_0+} \varphi_1(r) = \lim_{r\rightarrow r_0-}\varphi_2(r)

This answers my question, thank you! :)
 
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