
#1
Dec1212, 08:25 PM

P: 3,844

I read in the book regarding a point charge at the origin where [itex]Q(t)= \rho_{(t)}Δv'\;[/itex]. The wave eq is.
[tex]\nabla^2V\mu\epsilon\frac{\partial^2 V}{\partial t^2}= \frac {\rho_v}{\epsilon}[/tex] For point charge at origin, spherical coordinates are used where: [tex] \nabla^2V=\frac 1 {R^2}\frac {\partial}{\partial R}\left( R^2 \frac {\partial V}{\partial R}\right)[/tex] This is because point charge at origin, [itex]\frac {\partial}{\partial \theta} \hbox{ and }\; \frac {\partial}{\partial \phi}[/itex] are all zero. My question is this: The book then said EXCEPT AT THE ORIGIN, V satisfies the following homogeneous equation: [tex]\frac 1 {R^2}\frac {\partial}{\partial R}\left( R^2 \frac {\partial V}{\partial R}\right)\mu\epsilon \frac {\partial^2 V}{\partial t^2}=0[/tex] The only reason I can think of why this equation has to exclude origin is because R=0 and origin and this won't work. Am I correct or there's another reason? Thanks Alan 



#3
Dec1212, 10:57 PM

P: 3,844

Thanks
Alan 



#4
Dec1312, 12:38 AM

P: 3,844

Verify about the solution of wave equation of potential.
I have another question follow up with the original wave equation:
[tex]\frac 1 {R^2}\frac {\partial}{\partial R}\left( R^2 \frac {\partial V}{\partial R}\right)\mu\epsilon \frac {\partial^2 V}{\partial t^2}=0[/tex] According to the book, to simplify this equation, Let: [tex]V_{(R,t)}=\frac 1 R U_{(R,t)}[/tex] This will reduce the wave equation to: [tex]\frac {\partial^2U_{(R,t)}}{\partial R^2}\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}=0[/tex] One of the solution is [itex]U_{(R,t)}=f\left(t\frac R {\mu\epsilon}\right)[/itex] My question is I want to find the solution if U(R,t) is time harmonic which is a continuous sine wave. The book does not show the anything. So I am using the regular solution of time harmonic wave equation and using direction to be [itex]\hat R[/itex]. So the formula become: [tex]\frac {\partial^2U_{(R,t)}}{\partial R^2}\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}= \frac {\partial^2U_{(R,t)}}{\partial R^2}+\omega^2\mu\epsilon U= 0.[/tex] [tex] δ^2=\omega^2\mu\epsilon\;\Rightarrow\; \frac {\partial^2U_{(R,t)}}{\partial R^2}δ^2 U= 0.[/tex] This is a 2nd order ODE with constant coef and the solution is: [tex]U_{(R,t)}= V_0^+ e^{δ R}+V_0^ e^{δ R}[/tex] For potential that reach to infinity space, there will be no reflection so the second term disappeared, whereby: [tex]U_{(R,t)}= V_0^+ e^{δ R}[/tex] Do I get this right? This might look very obvious, but I am working on retarded potential and I am interpreting what the book don't say, I have to be careful to make sure I get everything right. Thanks Alan 



#5
Dec1312, 12:43 AM

P: 985

[itex]\frac {\partial^2U_{(R,t)}}{\partial R^2}\mu\epsilon\frac {\partial^2U_{(R,t)}}{\partial t^2}= \frac {\partial^2U_{(R,t)}}{\partial R^2}+\omega^2\mu\epsilon U= 0.[/itex]
just from the last in it you can write the periodic solution.Don't substitute any δ after. 



#6
Dec1312, 12:47 AM

P: 3,844

[tex]U_{(R,t)}= V_0^+ e^{\alpha R} e^{j\beta R}\; \hbox { where } δ=\alpha + j\beta[/tex] 



#7
Dec1312, 12:53 AM

P: 985

a plus sign before ω^{2} in eqn is necessary for periodic solution.First write solution in terms of ω and then introduce other definitions.




#8
Dec1312, 02:33 AM

P: 3,844

[tex]\frac {\partial ^2 U}{\partial R^2} δ^2U=0 \hbox { is }\; U_{(R,t)}= U_0^+ e^{δR}\;=\; U_0^+ e^{\alpha R} e^{j\beta R}[/tex] This is a decay periodic function. For lossless, we use [itex] k=\omega\sqrt{\mu\epsilon}\hbox { where ε is real.}[/itex] Then [tex]\frac {\partial ^2 U}{\partial R^2} +k^2U=0 \hbox { is }\; U_{(R,t)}= U_0^+ e^{jkR}[/tex] Actually because you brought this up, I dig deep into this and answer the question why the books use δ for lossy media and k for lossless media. I never quite get this all these years until now. 



#9
Dec1312, 07:11 AM

P: 985

that is why it is better to go with + sign.but it is ok,now.In case,ε is complex it is still possible to define things after.But you already got it.




#10
Dec1312, 02:52 PM

P: 3,844

Thanks.



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