Finding Total Charge from E-field

[Marcus95]

Homework Statement

A static charge distribution has a radial electric field of magnitude
$E = \alpha \frac{e^{-\lambda r}}{r}$
where λ and α are positive constants. Calculate the total charge of the distribution.

Homework Equations

Gauss's law $Q/\epsilon_0 = \int \vec{E} \cdot d\vec{S}$
$\rho/\epsilon_0 = \nabla \cdot \vec{E}$

The Attempt at a Solution

I have tried two ways to go about this problem, without success. First I tried using Gauss's law, placing a Gaussian sphere of radius a on the distribution:
$Q/\epsilon_0 = \int \vec{E} \cdot d\vec{S} = 4\pi\alpha (ae^{-\lambda a})$
We should now find the total charge by letting $a \rightarrow \infty$, but $\lim_{a \rightarrow \infty} ae^{-\lambda a} = 0$, giving $Q = 0$ which i thought cannot be right.*

My second method involved finding the charge distribution
$\rho/\epsilon_0 = \nabla \cdot \vec{E} = \alpha \frac{e^{-\lambda r}(1-\lambda r)}{r^2}$ but this gives an integral which is divergent!

*I have done some further thinking and realized that $Q_{tot} = 0$ might be the answer, because the charge density changes sign at $r = 1/\lambda$ so the total net charge could as well be zero.

However, how is it possible that the volume integral of #\rho# is divergent then? Or is it not?

 

[TSny]

My second method involved finding the charge distribution
$\rho/\epsilon_0 = \nabla \cdot \vec{E} = \alpha \frac{e^{-\lambda r}(1-\lambda r)}{r^2}$ but this gives an integral which is divergent!

Make sure you are setting up the integral correctly. Are you using the correct form for the volume element of the integral?

 

[J Hann]

Isn't E = k Q / R^2 (spherical charge distribution)?
Then 2 E R dR = k dQ.
The integral doesn't diverge unless (1 / R) appears in the result at R = 0.

 

[TSny]

Isn't E = k Q / R^2 (spherical charge distribution)?

Yes.

Then 2 E R dR = k dQ.

This doesn't look right. From E = k Q / R² you have k Q = R² E. If you take the differential of this, you need to note that E is a function of R.

But I don't see the point of doing this. kQ = R²E already tells you how much charge is contained in a sphere of radius R. So, taking the limit as R goes to infinity will give you the total charge of the system. This is essentially what @Marcus95 did in his first method of finding Q_tot.

 

[J Hann]

Yes.
This doesn't look right. From E = k Q / R² you have k Q = R² E. If you take the differential of this, you need to note that E is a function of R.

But I don't see the point of doing this. kQ = R²E already tells you how much charge is contained in a sphere of radius R. So, taking the limit as R goes to infinity will give you the total charge of the system. This is essentially what @Marcus95 did in his first method of finding Q_tot.

I tend to agree, but why does R need to go to infinity?
Was this implied in the problem?

 

[J Hann]

Also, doesn't L'Hopital's give a finite value for E as R goes to infinity?
(It does appear to be zero)

 

[TSny]

I tend to agree, but why does R need to go to infinity?
Was this implied in the problem?

The charge density extends to infinity. So, to get the total charge you have to include all the charge out to infinity.

 

[TSny]

Also, doesn't L'Hopital's give a finite value for E as R goes to infinity?
(It does appear to be zero)

$E = \alpha \frac{e^{-\lambda r}}{r}$

L'Hopital gives zero for the limit, but you don't really need L'Hopital. The numerator goes to zero while the denominator goes to infinity as $r$ goes to infinity.

 

[Marcus95]

Make sure you are setting up the integral correctly. Are you using the correct form for the volume element of the integral?

My integral is:
$Q = \int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \alpha\epsilon_0 \frac{e^{-\lambda r}(1-\lambda r)}{r^2} \times r^2sin\theta d\theta dr d\phi$
which gives:
$Q = 4\phi \alpha \epsilon_0 \int_{r=0}^\infty e^{-\lambda r}(1-\lambda r) dr =4\phi \alpha \epsilon_0 [re^{-\lambda r}]_0^\infty = 0$
so it all works out! Thanks a lot!

 

[TSny]

My integral is:
$Q = \int_{r=0}^\infty \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} \alpha\epsilon_0 \frac{e^{-\lambda r}(1-\lambda r)}{r^2} \times r^2sin\theta d\theta dr d\phi$
which gives:
$Q = 4\phi \alpha \epsilon_0 \int_{r=0}^\infty e^{-\lambda r}(1-\lambda r) dr =4\phi \alpha \epsilon_0 [re^{-\lambda r}]_0^\infty = 0$
so it all works out! Thanks a lot!

OK. In the second equation, the $\phi$ should be $\pi$??

Otherwise, looks good.