# Finding Total Charge from E-field

• Marcus95
In summary: You're welcome.In summary, a static charge distribution with a radial electric field of magnitude ##E = \alpha \frac{e^{-\lambda r}}{r}##, where λ and α are positive constants, was discussed in terms of calculating the total charge of the distribution. Two methods were attempted, with the first using Gauss's law and the second finding the charge distribution. Ultimately, taking the limit of the charge distribution as the radius goes to infinity gives a total charge of zero. L'Hopital's rule is not needed, as the numerator goes to zero while the denominator goes to infinity. The final integral is correctly set up and gives the expected result of zero.
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?

Marcus95 said:
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?

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.

J Hann said:
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 Qtot.

TSny said:
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.

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

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

J Hann said:
I tend to agree, but why does R need to go to infinity?
The charge density extends to infinity. So, to get the total charge you have to include all the charge out to infinity.

J Hann said:
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.

TSny said:
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!

Marcus95 said:
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.

## 1. How do you calculate total charge from an electric field?

In order to calculate the total charge from an electric field, you can use the formula Q = E x A, where Q is the total charge, E is the electric field strength, and A is the area in which the electric field exists. This formula assumes a uniform electric field and can be used for both positive and negative charges.

## 2. Can you use this formula for non-uniform electric fields?

No, this formula is only valid for uniform electric fields. For non-uniform electric fields, you must use a more complex equation that takes into account the changing electric field strength at different points in space.

## 3. What units are used for electric field and total charge?

The SI unit for electric field is volts per meter (V/m) and the SI unit for total charge is coulombs (C).

## 4. Is there a difference between net charge and total charge?

Yes, there is a difference between net charge and total charge. Net charge refers to the overall charge of a system or object, taking into account both positive and negative charges. Total charge, on the other hand, refers to the sum of all the individual charges in a system or object.

## 5. Can you find the total charge if you only know the electric field strength at a single point?

No, in order to find the total charge, you need to know both the electric field strength and the area in which the electric field exists. If you only have the electric field strength at a single point, you can only find the charge at that specific point, not the total charge.

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