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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 3.47

Here's the layout of our problem:

Finding the potential

Let's start by recalling that the general solution of Laplace's equation in spherical coordinates (see section 3.3.2: Spherical Coordinates, page 139)

$$\begin{array}{}\text{(Eq. 3.65, page 141)}& V(r,\theta )=\sum _{l=0}^{\mathrm{\infty }}(A_l{r}^l+\frac{B_l}{r^{l+1}})P_l(\cos \left(\theta \right))\end{array}$$

Where the $P_l$'s are Legendre Polynomials. To keep this solution self-contained, here's a table of them (as seen on page 140):

Where in our case, $x\to \cos \left(\theta \right)$.

We need to subject Eq. 3.65 to the Boundary conditions that the physical setup has as well as common potential properties.

Outside the sphere of radius $b$, we must have $A_l=0$ for all $l$. This is because we need $V\to 0$ as $r\to \mathrm{\infty }$. So far this reduces the general solution to

$$V_{r\ge b}(r,\theta )=\frac{B_l}{r^{l+1}}{P}_l(\cos \theta )$$

In the yellow region we can't say much for now. To make things clearer, let's denote the solution to the yellow region ($a\le r\le b$) with coefficients $A^{\prime }$ and $B^{\prime }$ (rather than $A$ and $B$ as in the outside solution):

$$V_{a\le r\le b}(r,\theta )=\sum _{l=0}^{\mathrm{\infty }}(A_l^{\prime }{r}^l+\frac{B_l^{\prime }}{r^{l+1}})P_l(\cos \left(\theta \right))$$

Of course, since the inner sphere is a conductor, we must have:

$$V_{r\le a}(r,\theta )=V_0$$

Where $V_0$ is constant.

Let's continue imposing constraints and requirements. Recall that potentials must be continuous, meaning that:

$$\begin{array}{rl}{V}_{r\ge b}(b,\theta )& =V_{a\le r\le b}(b,\theta )\\ \text{and}{V}_{a\le r\le b}(a,\theta )& =V_{r\le a}(a,\theta )\end{array}$$

From continuity at $r=b$ we have:

$$\begin{array}{rl}{V}_{r\ge b}(b,\theta )& =V_{a\le r\le b}(b,\theta )\\ \sum _{l=0}^{\mathrm{\infty }}\frac{B_l}{b^{l+1}}{P}_l(\cos \theta )& =\sum _{l=0}^{\mathrm{\infty }}(A_l^{\prime }{b}^l+\frac{B_l^{\prime }}{b^{l+1}})P_l(\cos \left(\theta \right))\\ \text{(*)}\frac{B_l}{b^{l+1}}& =(A_l^{\prime }{b}^l+\frac{B_l^{\prime }}{b^{l+1}})\\ B_l& =b^{2l+1}{A}_l^{\prime }+B_l^{\prime }\end{array}$$

Where the step (*) follows because the Legendre Polynomials are a system of complete and orthogonal polynomials, so the coefficients must match on a 1-to-1 correspondence.

We can see that the continuity at $r=b$ gave us a relation between the unknown coefficients in the outside region and those in the middle region.

From continuity at $r=a$ we have:

$$\begin{array}{rl}{V}_{a\le r\le b}(a,\theta )& =V_{r\le a}(a,\theta )\\ \sum _{l=0}^{\mathrm{\infty }}(A_l^{\prime }{a}^l+\frac{B_l^{\prime }}{a^{l+1}})P_l(\cos \left(\theta \right))& =V_0\end{array}$$

We can't use the same argument as in the case for continuity at $r=b$, where we got rid of the Legendre polynomials through a completeness/orthogonality argument, since the right hand side doesn't depend on the Legendre polynomials.

Still, notice the following: The right hand side is a constant. The left hand side is a function of $\theta$. The only way to make these two ideas match is to make the coefficients of all the $\theta$-dependent polynomials zero, and the rest will be equal to the right hand side: $V_0$:

$$\begin{array}{rl}\text{(**)}{A}_l^{\prime }{a}^l+\frac{B_l^{\prime }}{a^{l+1}}& =V_0\text{when}l=0\\ \text{(***)}{A}_l^{\prime }{a}^l+\frac{B_l^{\prime }}{a^{l+1}}& =0\text{when}l\ne 0\end{array}$$

Again, the first Legendre polynomial is the only constant polynomial. From (**) it follows that (we can now change $l\to 0$):

$$B_0^{\prime }=a{V}_0-a{A}_0^{\prime }$$

And from (***) it follows that

$$B_l^{\prime }=-a^{2l+1}{A}_l^{\prime }\text{when}l\ne 0$$

At this point, the three potentials can be simplified to (where by simplified we mean having less unknown coefficients):

$$\begin{array}{rl}{V}_{r\le a}(r,\theta )& =V_0\\ V_{a\le r\le b}(r,\theta )& =\sum _{l=0}^{\mathrm{\infty }}(A_l^{\prime }{r}^l+\frac{B_l^{\prime }}{r^{l+1}})P_l(\cos \left(\theta \right))\\ & \to A_0^{\prime }+\frac{a{V}_0-a{A}_0^{\prime }}{r}+\sum _{l=1}^{\mathrm{\infty }}(A_l^{\prime }{r}^l+\frac{B_l^{\prime }}{r^{l+1}})P_l(\cos \left(\theta \right))\\ V_{r\ge b}(r,\theta )& =\sum _{l=0}^{\mathrm{\infty }}\frac{B_l}{r^{l+1}}{P}_l(\cos \theta )\\ & \to \sum _{l=0}^{\mathrm{\infty }}\frac{b^{2l+1}{A}_l^{\prime }+B_l^{\prime }}{r^{l+1}}{P}_l(\cos \theta )\end{array}$$

Let's now impose the condition having to do with $\sigma$:

$$\begin{array}{}\text{(Eq. 2.36, page 88)}& \frac{\partial V_{\text{above}}}{\partial n}-\frac{\partial V_{\text{below}}}{\partial n}=-\frac{1}{{ϵ}_0}\sigma \end{array}$$

Where in this problem

$$\begin{array}{rl}\sigma & =k\cos \left(\theta \right)\\ & \equiv k{P}_1(\cos \left(\theta \right))\\ V_{\text{above}}& =V_{r\ge b}\\ V_{\text{below}}& =V_{a\le r\le b}\end{array}$$

The fact that $\sigma$ can be written in terms of one of the Legendre polynomials (with argument $\cos \left(\theta \right)$) will allow us to generate further constraints. To keep things simple, I will use the potentials without substituting the constraints found through continuity:

$$\begin{array}{rl}\frac{\partial V_{r\ge b}}{\partial r}& =\sum _{l=0}^{\mathrm{\infty }}{B}_l\frac{-(l+1)}{r^{l+2}}{P}_l(\cos \theta )\\ \frac{\partial V_{a\le r\le b}}{\partial r}& =\sum _{l=0}^{\mathrm{\infty }}(A_l^{\prime }l{r}^{l-1}+B_l^{\prime }\frac{-(l+1)}{r^{l+2}})P_l(\cos \theta )\end{array}$$

Equation 2.36 then says:

$$\sum _{l=0}^{\mathrm{\infty }}(B_l\frac{-(l+1)}{b^{l+2}}-A_l^{\prime }l{b}^{l-1}+B_l^{\prime }\frac{(l+1)}{b^{l+2}})P_l(\cos \theta )=-\frac{k}{{ϵ}_0}{P}_1(\cos \theta )$$

(Notice that I evaluated $r\to b$).

If $l=1$, then:

$$\begin{array}{rl}-B_1\frac{2}{b^3}-A_1^{\prime }+B_1^{\prime }\frac{2}{b^3}& =-\frac{k}{{ϵ}_0}\\ \text{so}{A}_1^{\prime }+\frac{2}{b^3}(B_1-B_1^{\prime })& =\frac{k}{{ϵ}_0}\end{array}$$

Where, as above, we had the match the coefficients of $P_1$ on both sides.

If $l\ne 1$, then the coefficient of $P_l$ must be zero, meaning

$$B_l\frac{-(l+1)}{b^{l+2}}-A_l^{\prime }l{b}^{l-1}+B_l^{\prime }\frac{(l+1)}{b^{l+2}}=0$$

Which we further simplify to a more useable form:

$$(l+1)(B_l-B_l^{\prime })+l{b}^{2l+1}{A}_l^{\prime }=0$$

Let's summarize our findings so far:

(f1) From continuity at $r=b$:

$$\begin{array}{}\text{(i)}& B_l=b^{2l+1}{A}_l^{\prime }+B_l^{\prime }\end{array}$$

(f2) From continuity at $r=a$:

$$\begin{array}{}\text{(ii)}& B_0^{\prime }& =a{V}_0-a{A}_0^{\prime }\text{(iii)}& \text{With }l\ne 0B_l^{\prime }& =-a^{2l+1}{A}_l^{\prime }\end{array}$$

(f3) From surface charge:

$$\begin{array}{}\text{(iv)}& A_1^{\prime }+\frac{2}{b^3}(B_1-B_1^{\prime })& =\frac{k}{{ϵ}_0}\text{(v)}& \text{With }l\ne 1(l+1)(B_l-B_l^{\prime })+l{b}^{2l+1}{A}_l^{\prime }& =0\end{array}$$

Use (ii) and (iii) to write (i) in terms of $A_l^{\prime }$ only:

$$\begin{array}{rl}{B}_l& =(b^{2l+1}-a^{2l+1})A_l^{\prime }\\ B_0& =(b-a)A_0^{\prime }+a{V}_0\text{With }l\ne 0\end{array}$$

Using this and (iii) in (v) for any case where $l\ne 0\ne 1$:

$$(l+1)((b-a)A_l^{\prime }+a{V}_0+a^{2l+1}{A}_l^{\prime })+l{b}^{2l+1}{A}_l^{\prime }=0$$

Since we can factor out $A_l^{\prime }$, it follows that

$$A_l^{\prime }=0$$

For any $l>1$. This implies

$$B_l^{\prime }=0$$

For any $l>1$ (were I looked at (iii)). And looking at (i), it follows that

$$B_l=0$$

For any $l>1$. At this point it must be clear that we haven't found any specific values for any of the constants, but we did find that out of the infinite ones we had to find, only 6 might be non-zero: $B_l,A_l^{\prime },B_l^{\prime }$ with $l=0$ or $l=1$.

From (v) with $l=0$ it follows that

$$B_0=B_0^{\prime }$$

And with $A_0^{\prime }=0$, it follows that

$$B_0=B_0^{\prime }=a{V}_0$$

For $l=1$, it is immediate that

$$A_1^{\prime }=\frac{k}{3{ϵ}_0}$$

Giving (since we can relate $A_1^{\prime }$ and $B_1^{\prime }$):

$$B_1^{\prime }=-\frac{a^{3}k}{3{ϵ}_0}$$

Finally:

$$B_1=\frac{(b^3-a^3)k}{3{ϵ}_0}$$

So

$$V(r,\theta )=\{\begin{array}{l}\begin{array}{cc}a{V}_0/r+(b^3-a^3)k\cos \left(\theta \right)/3r^2{ϵ}_0& r\ge b\\ a{V}_0/r+(r^3-a^3)k\cos \left(\theta \right)/3r^2{ϵ}_0& a\le r\le b\end{array}\end{array}$$

Finding the induced surface charge

$$\begin{array}{rl}{\sigma }_{\text{induced}}(\theta )& =-{ϵ}_0\frac{\partial V}{\partial r}{|}_{r=a}\\ & =-{ϵ}_0(-\frac{2k(r^3-a^3)\cos (\theta )}{3ϵr^3}-\frac{a{V}_0}{r^2}+\frac{k\cos (\theta )}{ϵ}){|}_{r=a}\\ & =\frac{V_0{ϵ}_0}{a}-k\cos \left(\theta \right)\end{array}$$

Finding the total charge of the system

$$\begin{array}{rl}{q}_{\text{induced}}& =\int {\sigma }_{\text{induced}}da\\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{2\pi }(\frac{V_0{ϵ}_0}{a}-k\cos \left(\theta \right))a^2\sin \left(\theta \right)d\theta d\varphi \\ & =4\pi a{ϵ}_0{V}_0\end{array}$$

Which also corresponds to the total charge of the system.

The behavior of $V$ at large $r$.

Consider the $V$ for $r\ge b$. For large $r$, the 2nd term can be ignored and we are left with

$$V(r,\theta )\approx \frac{a{V}_0}{r}$$

Notice that this is the same as putting the total charge found on part (b) in the point-particle formula.

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