- #1

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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] .

- Thread starter jhosamelly
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- #1

- 128

- 0

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] .

- #2

- 140

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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.

- #3

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This answers my question, thank you! :)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]

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