Computing dipole moment from charge density

[laser1]

Homework Statement

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Relevant Equations

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For this question, I get that $$\mathbf{p} = \int \mathbf{r}\,\rho(\mathbf{r})\,d^3r$$ but for the bounds, does r go from 0 to $r_0$ or from 0 to $\infty$? I would think infinity, but then how do I use the "hint"? I also don't get the significance of shifting the electron charge density. Thank you!

 

[kuruman]

For this question, I get that $$\mathbf{p} = \int \mathbf{r}\,\rho(\mathbf{r})\,d^3r$$ but for the bounds, does r go from 0 to $r_0$ or from 0 to $\infty$? I would think infinity, but then how do I use the "hint"? I also don't get the significance of shifting the electron charge density. Thank you!

What is $r_0$? The density extends to infinity therefore the integral should go to infinity.

Don't try to do the integral in your head. Set it up and perhaps you will see how to use the hint.

As for the significance of shifting the electron charge density, what do you think is the dipole moment of the hydrogen atom in zero applied electric field?

 

[PhDeezNutz]

The hint about the integral deals with the fact that the differential volume element of a sphere is

$r^2 \sin \theta \, dr \, d\theta \, d\phi$ notice the $r^2$

Also note that your dipole moment should also include the charge of the proton which should be represented by some kind of delta function.

 

[laser1]

What is r0?

How far away the proton and electron are from each other.

As for the significance of shifting the electron charge density, what do you think is the dipole moment of the hydrogen atom in zero applied electric field?

0 because equal charge.

Also note that your dipole moment should also include the charge of the proton which should be represented by some kind of delta function.

Good point. I tried this out though and got the integral of r delta(r) will turns out to be 0.

Regarding the "hint" about the integral, I still don't see it! $$\mathbf{p} =\int\mathbf{r}\,\rho_{\text{electron}}(\mathbf{r}-\mathbf{r}_0)\,d^3r$$ and by u sub $R=r-r_0$$$\mathbf{p} = \mathbf{r}_0 \int \rho_{\text{electron}}(\mathbf{R})\,d^3R = \mathbf{r}_0(-e) = -e\,\mathbf{r}_0.$$

 

[laser1]

The hint about the integral deals with the fact that the differential volume element of a sphere is

r2sin⁡θdrdθdϕ notice the r2

Maybe instead of using the fact that integral of rho over all space is -e, I integrate it explicitly? I get the same answer, though.

 

[kuruman]

How far away the proton and electron are from each other.

Not quite. You don't need a proton to answer this. Initially you have a spherically symmetric charge distribution $$\rho(\mathbf r)=-\frac{e}{a^3\pi}e^{-\frac{r}{2a}}.$$ The center of the spherical distribution is the origin of coordinates. Quantity $\mathbf r_0$ is the new position of this center when it is shifted away from the origin.

0 because equal charge.

What equal charge? You need to show that $$\int \mathbf r \rho(\mathbf r)d^3=-\frac{e}{a^3\pi}\int \mathbf r ~e^{-\frac{r}{2a}}d^3r=0$$ to understand how this works. The origin is at the proton and there is no proton contribution to the dipole moment. When the external electric field is applied, the proton is assumed to still be at the origin whilst the spherical negative charge distribution shifts with respect to it.

Maybe instead of using the fact that integral of rho over all space is -e, I integrate it explicitly? I get the same answer, though.

Please show your work. We cannot help you if we don't know what you did. I suspect that you did not take into account the fact that the dipole moment is a vector which means that you need to do three integrals, one for each Cartesian direction.

Your answer to part (a), $\mathbf p=-e \mathbf r_0$ is correct. What else is the problem asking you do? Please post the entire problem.

Also note that your dipole moment should also include the charge of the proton which should be represented by some kind of delta function.

I believe that the origin of coordinates coincides with the proton assumed to be a point charge at the origin of coordinates. Therefore the positive proton charge does not contribute to the dipole moment. It makes no sense to put the origin at any point other than at the proton.

 

[PhDeezNutz]

@kuruman and @laser1 you are absolutely right. The proton at the origin will contribute 0 to the dipole moments. I was just being pedantic. Apologies.

I will go one step further. I think there is an even stronger conclusion; the dipole moment is independent of choice of origin if the monopole is zero…….as it is supposed to be in a hydrogen molecule (I think).

 

[kuruman]

I will go one step further. I think there is an even stronger conclusion; the dipole moment is independent of choice of origin if the monopole is zero…….as it is supposed to be in a hydrogen molecule (I think).

Right. You have a pure dipole in the sense that it doesn't matter where the origin is because the dipole moment depends on the difference between the position vectors, $$\mathbf p=(+q)\mathbf r_2+(-q)\mathbf r_1=q(\mathbf r_2-\mathbf r_1).$$