Problem 2.46 Griffiths EM(adsbygoogle = window.adsbygoogle || []).push({});

The electric field of some configuration is given by the expression

[tex] V(\vec{r}) = a \frac{e^{\lambda r}}{r} [/tex]

where A and lambda are constants. Find the electric field E(r) , charge density rho (r) and the total charge Q

i can easily find the E and rho

in fact rho is given by

[tex] \rho = \epsilon_{0} A (4 \pi \delta^3(\vec{r}) - \frac{\lambda^2 e^{-\lambda r}}{r}) [/tex]

now theres the part of finding the total charge Q

i have to integrate roh over all space.. but wait.. the potential will blow up to negative infinity if we included negative value for r. But what about the part with the dirac delta function?? Is it integrated over all space or just the part for which r is valid. On my assignment i integrated the dirac delta from 0 to r and the resulting integral was zero

so my integral looks like

[tex] Q = \epsilon_{0} A \left( 4 \pi \int_{0}^{\infty} \delta^3 (\vec{r}) d\vec{r} - \int_{0}^{\infty} \frac{\lambda^2 e^{-\lambda r}}{r} dr \right) = - 4\pi \epsilon_{0} A [/tex]

but my prof says that the integral for the delta function should for all values of r that is -infty to +infty. But that would yield an answer of zero for Q. how can there be zero charge enclosed??

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# Homework Help: Electric Field Problem 2.46 Griffiths

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