Spherical cavity in a dielectric with a point charge inside

[quasar_4]

Homework Statement

I am trying to find the force acting on a charge placed a distance r0 from the center of a spherical cavity of radius a. The entire cavity is immersed in a dielectric material, epsilon2.

Homework Equations

The Attempt at a Solution

Here's my quandary: I can find the field inside the cavity with no charge. This is fine. But now that there's this point charge floating around, I have

[tex] \nabla \cdot D_{in} = \rho [/tex]
[tex] \nabla \cdot D_{out} = 0 [/tex]

Well, the form of the potential is obviously

[tex] \Phi_{in} = \sum_{l=0}^{\infty} A_l r^l P_l(\cos{\theta}) [/tex]
[tex] \Phi_{out} = -E0 r \cos{\theta}+ \sum_{l=0}^{\infty} B_l r^{-(l+1)} P_l(\cos{\theta}) [/tex]

where E0 is the applied field in the dielectric (without loss of generality, we can choose E0 to be in the z-direction).

I applied boundary conditions -- the normal component of D and the tangential component of E should be continuous at r = a (because there's no charge *there*). You get some nice equations, which you can easily solve for the coefficients A1 and B1 (all other terms in the Legendre polynomials vanish except l =1). I get:

[tex] A_1 = \frac{-3 \epsilon_2 E0}{2 \epsilon_2+\epsilon_0} [/tex]
[tex] B_1 = \frac{E0 a^3(\epsilon_0-\epsilon_2)}{2 \epsilon_2 + \epsilon_0} [/tex]

Great, I can plug these into my potential expressions...

But I don't see how my potential is even affected by this new charge q! I get the same result as for an empty cavity. That can't be right. So, do I need to use a totally different technique? I thought maybe the method of images is the way to go, but I don't understand how this works for a spherical set-up... where's the image charge if q is inside a sphere?

 

[gabbagabbahey]

Homework Statement

I am trying to find the force acting on a charge placed a distance r0 from the center of a spherical cavity of radius a. The entire cavity is immersed in a dielectric material, epsilon2.

Homework Equations

The Attempt at a Solution

Here's my quandary: I can find the field inside the cavity with no charge. This is fine. But now that there's this point charge floating around, I have

[tex] \nabla \cdot D_{in} = \rho [/tex]
[tex] \nabla \cdot D_{out} = 0 [/tex]

Well, the form of the potential is obviously

[tex] \Phi_{in} = \sum_{l=0}^{\infty} A_l r^l P_l(\cos{\theta}) [/tex]
[tex] \Phi_{out} = -E0 r \cos{\theta}+ \sum_{l=0}^{\infty} B_l r^{-(l+1)} P_l(\cos{\theta}) [/tex]
where E0 is the applied field in the dielectric (without loss of generality, we can choose E0 to be in the z-direction).

I don't see any mention of an applied uniform electric field in the problem statement.

As for your point charge, just add an additional [tex]\frac{q}{4\pi\epsilon_0}\frac{1}{|\textbf{r}-\textbf{r}_0|}[/tex] to [itex]\Phi_{in}[/itex] (before applying boundary conditions!).

 

[quasar_4]

I don't know what to make of it that it doesn't mention an applied electric field in the dielectric. Should I then omit the E0*r*cos(theta) from my solution outside the sphere?

 

[gabbagabbahey]

I don't know what to make of it that it doesn't mention an applied electric field in the dielectric. Should I then omit the E0*r*cos(theta) from my solution outside the sphere?

Yes, if there is no externally applied field, then [itex]E_0[/itex] is zero. You can't just copy and paste solutions to similar but different problems and expect to get things right.

 

[quasar_4]

All right, I've rethought this. Now I think it might be best to use images. So here's what I've got:

Inside the sphere:

[tex] \Phi_{in} = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{[r^2 + r_0^2 - 2 r r_0 \cos{\theta}]^{1/2}}+\frac{q'}{[r^2 + {r'}_0^2 - 2 r {r'}_0 \cos{\theta}]^{1/2}} \right) [/tex]

Outside:

[tex] \Phi_{out} = \frac{1}{4 \pi \epsilon_2} \frac{q''}{[r^2 + r_0^2 - 2 r r_0 \cos{\theta}]^{1/2}} [/tex]

Now for D, we can demand that

[tex] \epsilon_0 \frac{\partial \Phi_{in}}{\partial r} |_{r=a} = \epsilon_2 \frac{\partial \Phi_{out}}{\partial r} |_{r=a} [/tex]

while for E,

[tex] \frac{\partial \Phi_{in}}{\partial \theta} |_{r=a} = \frac{\partial \Phi_{out}}{\partial \theta} |_{r=a} [/tex].

These give us two equations, but I can't figure out how to find r0', so I have three unknowns... how can I find q' and q''?

 

[quasar_4]

Update: after using continuity of phi across the boundary I got a third equation, which I used to find r0' = a^2/r0. But using that in my original two sets of equations gives me a q' and a q'' that both depend on theta... can that even be correct? Does it make physical sense that the value of q' would depend on theta? It doesn't seem so to me, in which case I have no idea how to correctly find them... :( any help would be great.