Conducting shell surronded by, dielectric shell, and an outside $q$ charge

Question

I would like some help with my solution attempt.

I have a conducting shell with radius $R1$, surronded by a dielectric shell with $\varepsilon_1$ and radius $R2$, and on the outside i have a $POINT$ $charge$ $q$ from distance $L$, in Vacuum so $\varepsilon_0$. So its $R1<R2<L$.

The question is the potentials, $\Phi_1,\Phi_2,\Phi_3$.

Edit: My question is how should i choose the coefficents to satisfy the following things:

1.Inside the inner circle the potential should be zero because its a conductor, so there is only potential on its side where $r=R1$, and its a constant. Usually we choose the $B_l r^{-(l+1)}$ part to be zero. But i dont know which one should be zero in a case of conductor with no charge inside

2.The second shell is dielectric, but it doesn't contain the origo it goes from $[R_1,R_2]$, so i also dont know which coefficents to have inside here. \If there would just be a dielectric Sphere with outside charge inside would be the part where $C_l r^{l}$, and in the inside there would be the $D_l r^{-(l+1)}$ part with the point charge $\dfrac{q}{4\pi \varepsilon_0}\dfrac{1}{|\mathbf{r-L}|}$

3.On the outside we have the point charge $$\dfrac{q}{4\pi \varepsilon_0}\dfrac{1}{|\mathbf{r-L}|}=\sum_{n=0}^\infty \dfrac{r^l}{L^{l+1}}P_l{cos\vartheta}+Outer terms from the two shells $$

4.I also have the two types of boundary conditions, which should satisfy

And this is how i attemted it: I write up three cylindrical Laplacians for each section:

$$\Phi_1(r,\vartheta)=\sum_{n=0}^\infty (A_l r^l+B_lr^{-(l+1)})P_l(\cos \vartheta) \\ \Phi_2(r,\vartheta)=\sum_{n=0}^\infty (C_l r^l+D_lr^{-(l+1)})P_l(\cos \vartheta) \\ \Phi_3(r,\vartheta)=\sum_{n=0}^\infty (E_l r^l+F_lr^{-(l+1)})P_l(\cos \vartheta)$$

$B_l$ is zero because $r^{-(l+1)}$ diverges there. I'm not sure if $A_l$ should be zero or not. Because $\Phi_1(r<R_1)=0$, this should give me that $A_l=0$, but on the surface: so $\Phi_1(r=R_1)$, should be a constant value. $\Phi_1(r=R1)=\Phi_2(r=R1)=V1(\text{constans})$.

And there should also be another boundary condition here $$\varepsilon_0 \dfrac{\partial \Phi_1}{\partial r} \bigg|_{r=R_1}=\varepsilon_1 \dfrac{\partial \Phi_2}{\partial r} \bigg|_{r=R_1}$$

When $r=R_1$, the Potential is a constant $V_1$.

So i have these : $$\Phi_2(r=R,\vartheta)=\sum_{n=0}^\infty (C_l R^l+D_lR^{-(l+1)})P_l(\cos \vartheta)=V1$$

$$\sum_{n=0}^\infty (A_l l r^{l-1})P_l(\cos \vartheta)=\sum_{n=0}^\infty (C_l l r^{l-1}+D_l(-l-1)r^{-(l+2)})P_l(\cos \vartheta)$$

I'm not sure in the part where $r\in [R_1,R_2]$. So $\Phi_2$ there is :$$\Phi_2(r,\vartheta)=\sum_{n=0}^\infty (C_l r^l+D_lr^{-(l+1)})P_l(\cos \vartheta) $$. We leave both of the coefficients because there is no divergent part, because it $r$ doesn't go to zero here.

We have the boundary where $r=R2$ and the conditions are: $$\Phi_2(r=R_2)=\Phi_3(r=R_2)\\ \varepsilon_1 \dfrac{\partial \Phi_2}{\partial r} \bigg|_{r=R_2}=\varepsilon_0 \dfrac{\partial \Phi_3}{\partial r} \bigg|_{r=R_2}$$

$$\sum_{n=0}^\infty (C_l R_2^l+D_lR_2^{-(l+1)})P_l(\cos \vartheta) =\sum_{n=0}^\infty (E_l R_2^l+F_lR_2^{-(l+1)})P_l(\cos \vartheta)$$

$$\sum_{n=0}^\infty (C_l l R_2^l+D_l(-l-1)R_2^{-(l+2)})P_l(\cos \vartheta) =\sum_{n=0}^\infty (E_l l R_2^l+F_l(-l-1)R_2^{-(l+2)})P_l(\cos \vartheta)$$

Answer 1

You actually need four different potentials to solve this problem:

• $\Phi_1$, for the range $0<r<R_1$;
• $\Phi_2$, for the range $R_1 < r < R_2$;
• $\Phi_3$, for the range $R_2 < r < L$; and
• $\Phi_4$, for the range $L < r < \infty$.

This is because $\Phi$ does not satisfy Laplace's equation anywhere that there is charge, and in your setup there is a non-zero charge density at $r = R_1$, $R_2$, and $L$.

The relationship between the potentials $\Phi_3$ and $\Phi_4$ would be given by the usual relation: $$ \epsilon_0 \left[ \frac{\partial \Phi_3}{\partial r} - \frac{\partial \Phi_4}{\partial r} \right] = \sigma(\theta) $$ where $\sigma(\theta)$ is the charge density on the sphere $r = L$. For a point charge on the $z$-axis, it can be shown that in spherical coordinates $$ \sigma(\theta) = \frac{q}{L^2} \frac{\delta(\theta)}{\sin \theta} $$ (note that this must be viewed as a distribution, and the factor of $\sin \theta$ will cancel out with the $\sin \theta$ factor in the volume element when this is integrated in spherical coordinates.)

Beyond that, this will be a whole lot of algebra. One piece that you might want to keep in mind is that the potential must remain finite as $r \to \infty$. This means that the coefficients of the $r^l$ terms in $\Phi_4$ must all be zero.