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How to find the charge density given by a Tricky Potential?

  • #1
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Homework Statement


I am trying to solve Problem 2.45 in Electrodynamics by Griffiths, however, my answers were different from those in the book, I am suspect I got a missing step but I could not find it, so here is the Given Problem

Find the charge density [itex]\rho[/itex] given by a potential [tex]V=A\frac{e^{-\lambda r}}{r}[/tex] where [itex]A[/itex] and [itex]\lambda[/itex] are constant.

Homework Equations



[tex]E=-\nabla V[/tex] [tex]\nabla\cdot E = \frac{\rho}{\varepsilon_0}[/tex]

The Attempt at a Solution


So what I did is, first, find for the E-field using: [tex]E=-\nabla V,[/tex] so [tex]E=-\hat{r}\frac{\partial }{\partial r} A\frac{e^{-\lambda r}}{r}[/tex] by employing chain rule, the E-field is now given by [tex]E= A\lambda\frac{ e^{-\lambda r}}{r}\hat{r}+A\frac{ e^{-\lambda r}}{r^2}\hat{r}.[/tex]

Now, the charge density [itex]\rho[/itex] can be obtained thru [tex] \rho=\varepsilon_0\nabla\cdot E,[/tex] substituting the E-field to the equation, we will get [tex] \rho=\varepsilon_0\nabla\cdot \left [A\lambda\frac{ e^{-\lambda r}}{r}\hat{r}+A\frac{ e^{-\lambda r}}{r^2}\hat{r} \right ],[/tex] evaluating again using chain rule, we will get [tex]\rho=\varepsilon_0 \left [A\lambda e^{-\lambda r}\nabla\cdot\frac{\hat{r}}{r} +\frac{1}{r}\left( \frac{1}{r^2}\frac{\partial }{\partial r} r^2 A\lambda e^{-\lambda r} \right ) + A\lambda e^{-\lambda r}\nabla\cdot\frac{\hat{r}}{r^2} +\frac{1}{r^2} \left( \frac{1}{r^2}\frac{\partial }{\partial r} r^2 A e^{-\lambda r} \right ) \right ],[/tex] this will eventually become a bit more messy but the charge density will become [tex]\rho=A \varepsilon_0 e^{-\lambda r} [ 4\pi \delta^{3}(\mathbf{r})-\frac{\lambda^2}{r} ][/tex] however, this a bit different from Griffiths' answer of [tex]\rho=A \varepsilon_0 [ 4\pi \delta^{3}(\mathbf{r})-\frac{\lambda^2 e^{-\lambda r}}{r}][/tex]

where did I went wrong?
 

Answers and Replies

  • #2
vela
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Use the delta function property ##f(x)\delta(x) = f(0)\delta(x)##.
 
  • #3
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OMG! That was just the best!
 

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