Gauss' Law: Charge of a Hydrogen Atom

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The discussion centers on applying Gauss' Law to a hydrogen atom model, where a positive charge is at the center and a negative charge is distributed around it according to a specific exponential function. Participants emphasize the need to correctly calculate the charge density and the electric field, noting that the charge density is not uniform and requires integration. The integration approach is discussed, particularly in spherical coordinates, to derive the total charge. There is also a reminder to use a homework template for clarity in future posts. The conversation highlights the importance of precise mathematical representation and integration techniques in physics problems.
Lucas Silva
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New user has been reminded to fill out the Homework Help Template when starting a new schoolwork thread.
Suppose the hydrogen atom consists of a positive point charge (+e), located in the center of the atom, which is surrounded by a negative charge (-e), distributed in the space around it.

The space distribution of the negative charge changes according to the law p=Ce^(−2r/R), where C is a constant, r is the distance from the center of the atom, and R is Bohr's radius.

Find the value of the constant C by using the electrical neutrality of the atom.

Find the electrical field for r<R

Find the eltrical field for r > R

So what I've done is:

Since p stands for charge density, p = Q/V , where V = 4/3 Pi R^3 and Q = -Q

Thus, p = C*exp^(−2r/R) = -Q / 4/3 Pi R^3

Solving for C... C=[ -3*Q* r^2 * exp(2R/r)] / 4*Pi* R^3

For the electrical field I couldn't think of anything...

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Sorry for the portuguese image... It was the only I could find. Note Ao = R
 

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Your equation for finding the charge assumes uniform charge density. The density is not uniform, so you will have to do an integral. For the electric field you need to use Gauss's Law.
 
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kuruman said:
Your equation for finding the charge assumes uniform charge density. The density is not uniform, so you will have to do an integral. For the electric field you need to use Gauss's Law.
But if I integrate considering Q = ∫ ρ dV

Considering a Gaussian Surface with thickness of dr' ; the volume would be 4 * pi * r' * dr'
Where dV = 4 Pi r' dr' ... integrating from 0 to R I will get the same result as before.
 
Not quite.
$$Q=\int{\rho(r) dV}=\int{\frac{C}{r^2}~e^{-\frac{2r}{a_0}}dV}$$
What is ##dV## in spherical coordinates?

Also, for future reference please use the homework template when posting homework problems or your post may be deleted.
 
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kuruman said:
Not quite.
$$Q=\int{\rho(r) dV}=\int{\frac{C}{r^2}~e^{-\frac{2r}{a_0}}dV}$$
What is ##dV## in spherical coordinates?

Also, for future reference please use the homework template when posting homework problems or your post may be deleted.
Thank you ! Sorry for not using the homework template, I was really in a rush.

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Sorry, I cannot read the pictures. Your writing is small and faint and lighting is poor. If you do not have the time to learn and use LaTeX, please write bigger use blank ink and a lot of light before taking another picture.
 
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