# Homework Help: Finding Total Charge from E-field

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1. Jan 1, 2018

### Marcus95

1. The problem statement, all variables and given/known data
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.

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

3. 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 realised 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?

2. Jan 1, 2018

### TSny

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

3. Jan 2, 2018

### 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.

4. Jan 2, 2018

### TSny

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 Qtot.

5. Jan 2, 2018

### J Hann

I tend to agree, but why does R need to go to infinity?

6. Jan 2, 2018

### 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)

7. Jan 2, 2018

### TSny

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

8. Jan 2, 2018

### TSny

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

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.

9. Jan 7, 2018

### Marcus95

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!

10. Jan 7, 2018

### TSny

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

Otherwise, looks good.