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

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 $$\Phi_1(r, this should give me that $$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: $$\Phi_2(r=R_2)=\Phi_3(r=R_2)\\ \varepsilon_1 \dfrac{\partial \Phi_2}{\partial r} \bigg|_{r=R_2}=\varepsilon_0 \dfrac{\partial \Phi_3}{\partial r} \bigg|_{r=R_2}$$

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

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

• I've read your description of the setup twice now and I can't figure out where the charge $q$ is. Is it a point charge? A spherical shell? A diagram would be really helpful. May 17 at 18:12
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. May 17 at 18:13
• Thanks for the responses! There is a point charge,and its outside of both shells. from a distance L from the origo which is greater than both the radiuses. The second shell is directly on the inner one so they touch. I added my questions about the coefficients i hope its more clear now what is my problem May 18 at 5:59

• $$\Phi_1$$, for the range $$0;
• $$\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.