Multipole Expansion for Two Charged Hemispheres

[the_fater]

Homework Statement

Consider two homogeneously charged solid hemispheres with radius R separated in the x-y-plane by a negligible slit. The upper hemisphere has a total charge of +Q, and the lower hemisphere has a total charge of −Q.
1. Write down an expression for the volume charge density ρ of the two hemispheres.
2. Calculate the spherical and the Cartesian dipole moment.

Relevant Equations

electric dipole moment ##\vec P = \int \rho(\vec r^{'}) r^{'} \, dv^{'}##

My thought: First of all, I find the upper hemisphere (with a total charge +Q): ##ρ(\vec r)=\frac {V} {Q}## where V is the volume of the upper hemisphere = ## \frac {2} {3} \pi R^3##. Secondly, find the lower hemisphere (with a total charge −Q): ##ρ(\vec r)=\frac {V} {Q}## where V is the volume of the lower hemisphere = ## \frac {2} {3} \pi R^3##. I suggest that these expressions represent the volume charge densities for the upper and lower hemispheres. However, I am not entirely sure about it, so I am uncertain if I can find the correct dipole moments in Cartesian and spherical coordinates. Thanks for any help!

 

[PhDeezNutz]

Isn’t density Q/V and not V/Q?

Where as you said V is volume.

 

[the_fater]

Isn’t density Q/V and not V/Q?

Where as you said V is volume.

Thank you for noticing. It's Q/V, sorry for the typo.

 

[the_fater]

I'm considering the calculation of the volume charge density ##\rho## for the two hemispheres. It can be obtained by dividing the total charge of each hemisphere by its volume. The volume of a hemisphere is given by ##\frac {2}{3} \pi R^3##. Therefore, the volume charge density for the upper hemisphere is ##\rho = \frac {3Q}{2 \pi R^3} ##, and for the lower hemisphere, it is ##\rho = \frac {-3Q}{2 \pi R^3} ##.
Now, considering the dipole moment, which is the product of the charge and the distance between the charges. In spherical coordinates, the dipole moment is zero because the charges are symmetrically distributed about the origin. Similarly, in Cartesian coordinates, the dipole moment is also zero due to the symmetric distribution of charges about the origin.
Please correct me if I have misunderstood or made mistakes anywhere. Thank you!

 

[PhDeezNutz]

I am most definitely not getting a dipole moment of 0. Intuitively it should be anything BUT 0. You have 2 opposite charges separated by negligible distance........this is practically the definition of a dipole.

I'll give you a hint

##d_{volume} = r^2 \sin \theta \, dr \, d\theta \, d \phi##

for a sphere

for both of the top and the bottom ##r## goes from ##0## to ##R##

for both and the and the bottom ##\phi## goes from ##0## to ##2 \pi##

Here's two questions:

For the top what is the range of ##\theta##?

For the bottom what is the range of ##\theta##?

 

[kuruman]

However, I am not entirely sure about it, so I am uncertain if I can find the correct dipole moments in Cartesian and spherical coordinates.

Your equation ##\vec P = \int \rho(\vec r^{'}) r^{'} \, dv^{'}## is inconsistent and will not take you very far. You show a vector on the LHS and a scalar on the RHS. Since the charge density is uniform in each hemisphere, you might as well write ##\rho(\vec r')=\pm \rho=\text{const.}## Then the correct expression would be (I prefer to use bold characters for vectors) $$\mathbf{P} = \int \rho~\mathbf{r}' \, dv^{'}.$$That's 3 equations in 1:$$\begin{align}
& P_x = \int \rho~x'\, dv^{'} \nonumber \\
& P_y = \int \rho~y'\, dv^{'} \nonumber \\
& P_z = \int \rho~z'\, dv^{'} \nonumber \\
\end{align}$$Once you have the Cartesian components, it should be easy to write this in spherical coordinates using the standard unit vector transformations from one to the other.

This is in addition to what @PhDeezNutz said with which I concur.

 

[haruspex]

$$\mathbf{P} = \int \rho~\mathbf{r}' \, dv^{'}.$$

I guess that can be arrived at by notionally pairing each small region with its diametrically opposite equivalent. From the symmetry, we could instead pair it with its reflection in the XY plane.

 

[the_fater]

I am most definitely not getting a dipole moment of 0. Intuitively it should be anything BUT 0. You have 2 opposite charges separated by negligible distance........this is practically the definition of a dipole.

I'll give you a hint

##d_{volume} = r^2 \sin \theta \, dr \, d\theta \, d \phi##

for a sphere

for both of the top and the bottom ##r## goes from ##0## to ##R##

for both and the and the bottom ##\phi## goes from ##0## to ##2 \pi##

Here's two questions:

For the top what is the range of ##\theta##?

For the bottom what is the range of ##\theta##?

Thanks for the hint! Here's my idea below. I hope I'm on the right track!
Spherical Dipole Moment = ##\int \rho(\vec r) \vec r \, dV##. For the upper hemisphere = ##\int_{0}^{\pi/2} \int_{0}^{2\pi} \int_{0}^{R} \frac {3Q} {2 \pi R^3} \vec r r^2 \sin(\theta) \, dr \, d\phi \, d\theta ##. For the lower hemisphere = ##\int_{\pi/2}^{\pi} \int_{0}^{2\pi} \int_{0}^{R} \frac {-3Q} {2 \pi R^3} \vec r r^2 \sin(\theta) \, dr \, d\phi \, d\theta ##. In Cartesian coordinates, the dipole moment for the upper hemisphere = ##\int_0^R \int_0^R \int_0^R \frac {3Q} {2 \pi R^3} \vec r \, dx \, dy \, dz ##. The dipole moment for the lower hemisphere = ##\int_{-R}^{0} \int_0^R \int_0^R \frac {-3Q} {2 \pi R^3} \vec r \, dx \, dy \, dz ##.

 

[PhDeezNutz]

I have no idea what a spherical dipole moment is. To me a dipole moment is a dipole moment regardless of what basis it is expressed in.

Read the post quoted below more carefully.

Also you seem to think that to find the dipole moment in Cartesian you must integrate in Cartesian…….that is not correct and I would highly advise against it. Integrate in spherical coordinates when dealing with a sphere.

$$\mathbf{P} = \int \rho~\mathbf{r}' \, dv^{'}.$$That's 3 equations in 1:$$\begin{align}
& P_x = \int \rho~x'\, dv^{'} \nonumber \\
& P_y = \int \rho~y'\, dv^{'} \nonumber \\
& P_z = \int \rho~z'\, dv^{'} \nonumber \\
\end{align}$$Once you have the Cartesian components, it should be easy to write this in spherical coordinates using the standard unit vector transformations from one to the other.

This is in addition to what @PhDeezNutz said with which I concur.

@the_fater what is x in spherical coordinates?

What is y in spherical coordinates?

What is z in spherical coordinates?

 

[kuruman]

I guess that can be arrived at by notionally pairing each small region with its diametrically opposite equivalent. From the symmetry, we could instead pair it with its reflection in the XY plane.

I always thought that it is the definition of the dipole term in general. It does not have to be a "pure" dipole that has the symmetry you mentioned. In a multipole expansion, point charge ##q## at position vector ##\mathbf{a}## from the origin has a monopole potential due to charge ##q## at the origin plus a dipole term due to a dipole moment ##\mathbf{p}=q~\mathbf{a}## at the origin plus higher order terms.