How can I solve Poisson's Equation using an identity from Griffiths?

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SUMMARY

This discussion focuses on solving Poisson's Equation, specifically problem 2.29 from David J. Griffiths' textbook on electromagnetism. The equation is given as \(\nabla^2 V = -\frac{\rho}{\epsilon}\), and the identity \(\nabla^2\left(\frac{1}{r}\right) = -4\pi \delta^3(r)\) is utilized. The user expresses confusion regarding the application of the Dirac delta function in the context of the potential \(V(r) = -\frac{1}{\epsilon}\int p(r')\delta^3(r)dr'\). Key points include the importance of precise notation and the distinction between the symbols \(p\) and \(\rho\).

PREREQUISITES
  • Understanding of Poisson's Equation in electromagnetism
  • Familiarity with Dirac delta functions and their properties
  • Knowledge of vector calculus, specifically gradient and Laplacian operators
  • Proficiency in Griffiths' textbook notation and concepts
NEXT STEPS
  • Review the properties of the Dirac delta function in the context of integrals
  • Study the derivation and applications of Poisson's Equation in electrostatics
  • Examine vector calculus identities relevant to electromagnetism
  • Practice problems from Griffiths' textbook, focusing on precise notation and definitions
USEFUL FOR

Students of electromagnetism, particularly those tackling advanced topics in Griffiths' textbook, as well as educators seeking to clarify the application of mathematical concepts in physics.

ozone
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Sorry guys I have a lot of trouble with proofs and could use your help with this one.

This is right out of Griffiths (problem 2.29) I have a solution manual, but I'd like to try to get a nudge in the right direction before I turn to it.

Poisson's eq:

[itex]\nabla ^2 V = - \frac{ \rho }{\epsilon }[/itex]

Using an identity out of Griffiths we have [itex]\nabla ^2\left(\frac{1}{r}\right) = -4\pi \delta ^3(r)[/itex]

finally we know that [itex]V(r) = \frac{1}{4\pi \epsilon }\int \frac{p(r')}{r} \, dr'[/itex]

distributing the gradient into our function will yield
[itex]V(r) = -\frac{1}{\epsilon }\int p(r')\delta ^3(r)dr'[/itex]

But I do not know what to do with the dirac delta in the final function we have arrived at.
 
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ozone said:
Sorry guys I have a lot of trouble with proofs and could use your help with this one.

This is right out of Griffiths (problem 2.29) I have a solution manual, but I'd like to try to get a nudge in the right direction before I turn to it.

Poisson's eq:

[itex]\nabla ^2 V = - \frac{ \rho }{\epsilon }[/itex]

Using an identity out of Griffiths we have [itex]\nabla ^2\left(\frac{1}{r}\right) = -4\pi \delta ^3(r)[/itex]
The ##r## on the lefthand side is the distance between two points whereas the ##r## in the argument of the delta function should be the vector ##\vec{r}##.

finally we know that [itex]V(r) = \frac{1}{4\pi \epsilon }\int \frac{p(r')}{r} \, dr'[/itex]
The potential V isn't a function of only radial distance. It's a function of all three coordinates. In my edition of Griffiths, he uses the notation V(P) to indicate it's the potential at point P. He doesn't write V(r), which implies something much different. The symbols ##p## and ##\rho## look similar, but they aren't the same symbol. Be consistent. You should also define exactly what you mean by ##r## inside the integral. How is it related to the variables of integration? Finally, ##dr'## should actually be the volume element ##d\tau##.

distributing the gradient into our function will yield
[itex]V(r) = -\frac{1}{\epsilon }\int p(r')\delta ^3(r)dr'[/itex]

But I do not know what to do with the dirac delta in the final function we have arrived at.
The lefthand side should not be the potential V.

Electromagnetism is a fairly math-intensive course, so you should really make an effort to be precise with your notation. The sloppiness that you could get away with in earlier courses will really start to hinder you now. Clean it up so you can avoid making dumb errors and confusing yourself.
 

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