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Griffitsh electrostatics problem

  • #1
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griffiths electrostatics problem

if you have the text on you, i'm looking at question 2.46 (in the third edition).

i am given

[tex]
V(\mathbf{r}) = A\exp (-\lambda r)/r
[/tex]


i found that

[tex]
\mathbf{E} (\mathbf{r}) = A\exp (-\lambda r)(\lambda /r + 1/r^2) \hat{r}
[/tex]

the solution for [tex]\rho[/tex] given in the text is

[tex]
\rho = \epsilon _0 A [4\pi \delta^3 (\mathbf{r}) - \lambda^2 \exp (-\lambda r)/r]
[/tex]


however, in my solution, i got that exponential funtcion multiplying both terms in the answer.

i don't think i made a mistake in my steps, either. :/

i used the following properties:

[tex]
\nabla \cdot (f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot (\nabla f)
[/tex]

and

[tex]
\nabla \cdot (\mathbf{A} + \mathbf{B}) = \nabla \cdot \mathbf{A} + \nabla \cdot \mathbf{B}
[/tex]



from that first property, the first term on the right hand side seems to me to make it pretty clear that the exponential function should be in both terms in the answer. :/

thank you.
 
Last edited:

Answers and Replies

  • #2
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hold on... gotta get the latex straightened out...
 
  • #3
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0
could someone also help me with getting the fractions in latex working?

i needed to put something in fractions that itself was something in latex, and i'm not sure what i did incorrectly. :/
 
  • #4
sic
9
0
basically, there is no meaning to the function Dirac delta is multiplied by (only to its value at 0), because in any point except for 0, delta is 0 by itself anyway..
 
  • #5
429
0
sic said:
basically, there is no meaning to the function Dirac delta is multiplied by (only to its value at 0), because in any point except for 0, delta is 0 by itself anyway..
oh, i think that's it.

yeah, in this case the function is equal to 1 at r=0, so it wouldn't really matter if it is included or not.


i'll go to my teacher's office hours, just to make sure.
 
  • #6
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yeah, a quick check back to ch. 1 in griffiths straightened that out.


now... a new problem! :biggrin:


the next step is to integrate rho to find the total charge, q.

i used spherical coordinates (obviously!), so the delta function term disappears (as it will be multiplied by r-squared).

my limits of integration for the radius were from 0 to infinity.

however, i get a negative answer.

[tex]
Q = -A 4\pi\epsilon _0 .
[/tex]


i find this disturbing. would it be allowed to switch the limits of integration, or is that verboten?
 
  • #7
quasar987
Science Advisor
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What would you want to change the limits of int. to?! Anyway, you've probably made a mistake in integrating.

[tex]\int \rho d\tau = \epsilon _0 A \left(\int 4\pi \delta^3 (\mathbf{r}) d\tau - \lambda^2 \int \exp (-\lambda r)/r d\tau \right)[/tex]

the integral with the delta fucntion is just 4pi (because of equation (1.97), or more generally, because of equation (1.98)). The other one, you'll have to find a way to evaluate. Plz tell me if you find how to evaluate the limit that ensues. :grumpy:
 
  • #8
quasar987
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Ah, L'Hospital.
 
  • #9
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quasar987 said:
Ah, L'Hospital.
yep, that's how to do it.
 
  • #10
429
0
quasar987 said:
What would you want to change the limits of int. to?! Anyway, you've probably made a mistake in integrating.

[tex]\int \rho d\tau = \epsilon _0 A \left(\int 4\pi \delta^3 (\mathbf{r}) d\tau - \lambda^2 \int \exp (-\lambda r)/r d\tau \right)[/tex]

the integral with the delta fucntion is just 4pi (because of equation (1.97), or more generally, because of equation (1.98)). The other one, you'll have to find a way to evaluate. Plz tell me if you find how to evaluate the limit that ensues. :grumpy:

ah, googly-moogly!

i thought that the first integral would be zero since

[tex]
d\tau = r^2 \sin (\theta) drd\theta d\phi.
[/tex]

and then i figured that the integral would be equal to [tex]0^2[/tex], but now that i'm looking at example 1.16 on pg. 51, i'm seeing that that's not the case.

thanks!

now let's see what the answer is. (post yours, too! you taking EM this semester, as well? :biggrin: )
 
  • #11
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ok, now i'm getting an answer of 0.

that's... unsatisfying. is it even physical? :eek:
 
  • #12
quasar987
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Of course. It just means that the total charge is 0. there are as much "+" as the are "-"; very possible.

P.S. The EM fun was last semester for me.
 
  • #13
quasar987
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I'm very puzzled by all this right now though. Both the integrator and Mapple say that

[tex]\int r e^{-\lambda r}dr = \frac{1}{2}r^2 e^{-\lambda r}[/tex]

But when I differentiate the answer, I don't get the integrand. When mapple differentiate the answer, it DOES get the integrand. Where am I mistaken ?!?!

[tex]\frac{d}{dr}\frac{1}{2}r^2 e^{-\lambda r} = r e^{-\lambda r} + \frac{-\lambda}{2}r^2 e^{-\lambda r} [/tex]
 
  • #14
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quasar987 said:
I'm very puzzled by all this right now though. Both the integrator and Mapple say that

[tex]\int r e^{-\lambda r}dr = \frac{1}{2}r^2 e^{-\lambda r}[/tex]

But when I differentiate the answer, I don't get the integrand. When mapple differentiate the answer, it DOES get the integrand. Where am I mistaken ?!?!

[tex]\frac{d}{dr}\frac{1}{2}r^2 e^{-\lambda r} = r e^{-\lambda r} + \frac{-\lambda}{2}r^2 e^{-\lambda r} [/tex]

looks like the programs are treating the exponential function as a constant! that's not good!

i evaluated the integral using integration by parts.




yeah, the + charge balancing the - charge makes sense to me. i'm just trying to wrap my head around what it would look like in space! maybe i should give that up. :tongue:
 

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