Find the electric field produced by the atom at the Bohr radius

[indigojoker]

an electron is distributed around a proton according to the volume charge density $$\rho = A e^{-2r/a_o}$$ where A is a constant, a_o is the Bohr radius and r is the distance from the center of the atom.

Find A:
we know that $$Q=\int \rho dV = -e$$
i was wonder if this was the integral that i set up:
$$-e=\int _0 ^{\inf} A e^{-2r/a_o} 4 \pi r^2 dV$$

Find the electric field produced by the atom at the Bohr radius?
E4 pi a_o^2=-e/epsilon

then solve for E, is this right?

 

[learningphysics]

an electron is distributed around a proton according to the volume charge density $$\rho = A e^{-2r/a_o}$$ where A is a constant, a_o is the Bohr radius and r is the distance from the center of the atom.

Find A:
we know that $$Q=\int \rho dV = -e$$
i was wonder if this was the integral that i set up:
$$-e=\int _0 ^{\inf} A e^{-2r/a_o} 4 \pi r^2 dV$$

Looks right to me... except should have dR in the integral.

Find the electric field produced by the atom at the Bohr radius?
E4 pi a_o^2=-e/epsilon

then solve for E, is this right?

Left side looks right... but on the right side you need the charge enclosed within the bohr radius... not the entire charge...

 

[indigojoker]

the reason why its dr is because:

dV=4pir^2dr

right?For the second part, what is the charge enclosed within the bohr radius? I am not quite sure how that plays a role in the equation for flux.

 

[learningphysics]

the reason why its dr is because:

dV=4pir^2dr

right?

yes.

For the second part, what is the charge enclosed within the bohr radius?

integrate $$\rho dV$$ from 0 to a_o

I am not quite sure how that plays a role in the equation for flux.

what does gauss law say about the flux through the spherical surface at r=a_o

 

[indigojoker]

so the right side should really be:

$$E 4 \pi a_{o}^{2} = \frac{\int_0^{a_o} A e^{-2r/a_o} 4 \pi r^2 dr}{\epsilon_o}$$

 

[learningphysics]

so the right side should really be:

$$E 4 \pi a_{o}^{2} = \frac{\int_0^{a_o} A e^{-2r/a_o} 4 \pi r^2 dr}{\epsilon_o}$$

Right... with dR in the integral on the right side...

 

[indigojoker]

what is the dR integral?

isnt that what i have?

 

[learningphysics]

what is the dR integral?

isnt that what i have?

oops sorry... yes, that's right... I thought I saw dV there before I clicked reply... did you change it?

Anyway, it looks correct now.

 

[indigojoker]

haha yes i changed it right when i saw the dV, you must have hit reply while the system was updating :P

 

[learningphysics]

haha yes i changed it right when i saw the dV, you must have hit reply while the system was updating :P