Polar Laplace Equation

$\nabla^{2} = \frac{1}{ρ} \frac{\partial}{\partial ρ} (ρ \frac{\partial Φ}{\partial ρ}) + \frac{1}{ρ^{2}} \frac{\partial^{2} Φ}{\partial ϕ^{2}} + \frac{\partial^{2} Φ}{\partial z^{2}} = 0$ .

General Solution no Z

Assume we have no $z$ dependence. Then we get solution $Q (ϕ) = \exp (\pm i m ϕ)$ . Then, $ρ \frac{d}{d ρ} (ρ \frac{d R}{d ρ}) - m^{2} R = 0$ .

For $m = 0$ , $ρ \frac{d R}{d ρ} b_{0} \Rightarrow R (ρ) = a_{0} + b_{0} \ln ρ$ .

For $m \neq 0$ , then, $ρ^{2} \frac{d^{2} R}{d ρ^{2}} + ρ \frac{d R}{d ρ} = m^{2} R$ . Let $R (ρ) = ρ^{α}$ as an ansatz. Then, $(α (α - 1) + α - m^{2}) ρ^{α} = 0$ . Thus, $α^{2} = m^{2}$ . So, $α = \pm m$ . Overall, $Φ (ρ, ϕ) = a_{0} + b_{0} \ln ρ + \sum_{m = 1}^{\infty} (A_{m} ρ^{m} + \frac{B_{m}}{ρ^{m}}) (c_{m} \sin m ϕ + D_{m} \cos m ϕ)$ .

Examples

Circular Cylinder

Right side of cylinder is at $+ V$ and the left is $- 0$ . At $ρ = 0$ then the potential must be zero. Then, $b_{0} = 0, a_{0} = 0, B_{m} = 0$ . Let $- π \leq ϕ \leq π$ . So, we get even function for angular solutions. Thus, $Φ (ρ, ϕ) = \sum_{m} A_{m} ρ^{m} \cos m ϕ$ .

Orthogonality, $\int_{- π}^{π} \cos m ϕ \cos m^{'} ϕ d ϕ = π δ_{m^{'} m}$ .

At $ρ = a$ , $\int_{- π}^{π} \cos m^{'} ϕ d ϕ = A_{m}^{'} a^{m^{'}} π = 2 [\int_{0}^{\frac{π}{2}} \cos m^{'} ϕ d ϕ - \int_{\frac{π}{2}}^{π} \cos m^{'} ϕ d ϕ] = \frac{4 V}{m^{'}} \sin \frac{m^{'} π}{2}$ .

Thus, $Φ (ρ, ϕ) = \sum_{m odd} \frac{4 V}{a^{m} π} (- 1)^{m} ρ^{m} \cos m ϕ$ .

Corner

Let $β$ be the angle of the corner, $β \leq 90^{\circ}$ . And the potential be $V$ at the boundary. Let $0 \leq ϕ \leq β$ be our region of interest. So, $Φ (ρ, φ) = R (ρ) Q (ϕ)$ . Then, Laplace’s equation can be written,

\begin{aligned} \frac{1}{ρ} \frac{\partial}{\partial ρ} (ρ \frac{\partial Φ}{\partial ρ}) + \frac{1}{ρ} \frac{\partial^{2} Φ}{\partial ϕ^{2}} \end{aligned}

Let the polar equation be equal to $- ν^{2}$ . The radial solution is then $ρ^{ν}, ρ^{- ν}$ and for $ν = 0$ , $R (ρ) = a_{0} + b_{0} \ln ρ$ . For the finite region, the angular solution for $ν = 0$ can be $Q (ϕ) = A_{0} + B_{0} ϕ$ . So, $Φ (ρ, ϕ) = (a_{0} + b_{0} \ln ρ) (A_{0} + B_{0} ϕ) + \sum_{m = 1}^{\infty} (a_{m} ρ^{m} + b_{m} ρ^{- m}) (A_{m} \sin (ν_{m} ϕ) + B_{m} \cos (ν_{m} ϕ))$ .

For $ρ = 0$ , the potential is finite. So, $Φ (ρ, φ) = A_{0} + B_{0} ϕ + \sum_{m = 1}^{\infty} ρ^{m} (A_{m} \sin ν_{m} ϕ + B_{m} \cos ν_{m} ϕ)$ .

For $φ=0,β, $Φ \to V$ . $Φ (ρ, 0) = A_{0} + \sum_{m} ρ^{m} B_{m}$ . So, $B_{m} = 0$ . $Φ (ρ, β) = V = A_{0} + B_{0} β + \sum_{m} ρ^{m} A_{m} \sin (ν_{m} β)$ . So, $ν_m = \frac{m\pi}{\beta}, A₀=V,$ and $B_{0} = 0$ .

Thus, $Φ (ρ, ϕ) = V + \sum_{m} A_{m} ρ^{\frac{m π}{β}} \sin (\frac{m π}{β} ϕ)$ .

When, $ρ \to 0$ , $Φ (ρ, ϕ) \approx= V + A_{1} ρ^{\frac{π}{β}} \sin (\frac{π}{β} ϕ)$ .

$E_{ρ} = - \frac{\partial Φ}{\partial ρ} = - A_{1} \frac{π}{β} ρ^{\frac{π}{β} - 1} \sin (\frac{π}{β} ϕ)$ .

$E_{ϕ} = - \frac{1}{ρ} \frac{\partial Φ}{\partial ϕ} = - A_{1} \frac{π}{β} ρ^{\frac{π}{β} - 1} \cos (\frac{π}{β} ϕ)$ .

$σ = ε_{0} E_{ϕ} (ϕ = 0) \propto ρ^{\frac{π}{β} - 1}$ . So, when $β < π$ then $σ \to 0$ as $ρ \to 0$ . Otherwise, the charge density blows up.

$σ = ε_{0} E_{n} = ε_{0} \vec{E} \cdot \hat{n}$ .

General Cylindrical Coordinates

\begin{array}{r} \frac{1}{ρ} \frac{\partial}{\partial ρ} (ρ \frac{\partial Φ}{\partial ρ}) + \frac{1}{ρ} \frac{\partial^{2} Φ}{\partial ϕ^{2}} + \frac{\partial^{2} Φ}{\partial z^{2}} = 0 \end{array}

Separate, $Φ (ρ, ϕ, z) = R (ρ) Q (ϕ) Z (z)$ .

Let $\frac{d^{2} Z}{d z^{2}} = - k^{2}$ . We then get $Z (z) = A \exp (i k z) + B \exp (- i k z)$ .

Let $\frac{d^{2} Z}{d z^{2}} = k^{2}$ . We then get $Z (z) = A \exp (k z) + B \exp (- k z)$ .

We get Bessel functions if $k^{2}$ and modified Bessel functions if $- k^{2}$ .

Consider a Cylinder with a potential $V (ρ, ϕ)$ at the top and the bottom and sides are grounded.

Then, we want $k^{2}$ . We then get $Q (ϕ) = \exp (\pm i m ϕ)$ .

The radial function is then Bessel functions, from the differential equation,

\begin{matrix} (1) & \frac{ρ}{R} \frac{d}{d ρ} (ρ \frac{d R}{d ρ}) + (k ρ)^{2} - m^{2} = 0. \end{matrix}

Or, the modified equation,

\begin{matrix} (2) & x \frac{d}{d x} (x \frac{d R}{d x}) + [x^{2} - m^{2}] R = 0. \end{matrix}

The solutions are $R (x) = A J_{m} (x) + B N_{m} (x)$ .

From the fact we expect a finite potential at the center, we get, $B = 0$ . So, $R (r) = A J_{m} (k r)$ .

From the bottom being grounded, we can write, $Z (z) = C \sinh k z$

To get zero at the sides, $R (r) = \sum_{n} A_{n} J_{m} (\frac{β_{m n}}{a} r)$ . This gives an overall solution, $\sum_{m = 0}^{\infty} \sum_{n = 1}^{\infty} \sinh (k_{m}^{n} z) J_{m} (k_{m}^{n} r) (A_{m, n} \sin (m ϕ) + B_{m, n} \cos (m ϕ))$

At the top surface, $z = L$ , $V = V (ρ, ϕ)$ . Then from orthogonality, $\int_{0}^{2 π} \int_{0}^{a} Φ (r, ϕ, L) \sin (m^{'} ϕ) J_{m^{'}} (k_{m^{'}}^{n^{'}} a) r d r d ϕ = \int_{0}^{2 π} \int_{0}^{a} V (ρ, ϕ) \sin (m^{'} ϕ) J_{m^{'}} (k_{m^{'}}^{n^{'}} a) r d r d ϕ = A_{m n} \sinh (k_{m n} L) π \frac{a^{2}}{2} [J_{m^{'} + 1} (k_{m^{'} n^{'}} a)]^{2} δ_{m m^{'}} δ_{n n^{'}}$ . Also from orthogonality for $n > 0$ , $\int_{0}^{2 π} \int_{0}^{a} Φ (r, ϕ, L) \cos (m^{'} ϕ) J_{m^{'}} (k_{m^{'}}^{n^{'}} a) r d r d ϕ = \int_{0}^{2 π} \int_{0}^{a} V (ρ, ϕ) \cos (m^{'} ϕ) J_{m^{'}} (k_{m^{'}}^{n^{'}} a) r d r d ϕ = A_{m n} \sinh (k_{m n} L) π \frac{a^{2}}{2} [J_{m^{'} + 1} (k_{m^{'} n^{'}} a)]^{2} δ_{m m^{'}} δ_{n n^{'}}$ . For $n = 0$ , $B_{0 n} \to \frac{1}{2} B_{o n}$ .

Orthogonality of sines, $\int_{0}^{a} \sin \frac{n π x}{a} \sin \frac{n^{'} π x}{a} d x = \frac{a}{2} δ_{n^{'} n}$ Orthogonality of the Bessel functions, $\int J_{m} (k_{m n} ρ) J_{m} (k_{m n^{'}} ρ) ρ d ρ = \frac{a^{2}}{2} [J_{m + 1} (k_{m, n} a)^{2}] δ_{n^{'} n}$

Bessel Function Notes

$J_{m} (x) = {(\frac{x}{2})}^{m} \sum_{j = 0}^{\infty} \frac{(- 1)^{j}}{j! (m + j)!} {(\frac{x}{2})}^{2 j}$ .

$ρ \to 0$ , $J_{0} (x) = 1, J_{m} (x) = 0, N_{m} (x) \to - \infty$ .

$ρ \to \infty$ , $J_{m} (x) = \sqrt{\frac{2}{π x}} \cos (x - m π / 2 - π / 4), N_{m} (x) = \sqrt{\frac{2}{π x}} \sin (x - m π / 2 - π / 4)$ .

Poisson Equation in Spherical Coordinates

$\nabla^{2} Φ = - ρ / ε_{0}$ .

$Φ (\vec{x}) = \frac{1}{4 π ϵ_{0}} \int ρ ({\vec{x}}^{'}) G (\vec{x}, {\vec{x}}^{'}) d x^{' 3} - \frac{1}{4 π} \int Φ_{S} ({\vec{x}}^{'}) \frac{\partial G}{\partial n^{'}} d a^{'}$ .

$G (\vec{x}, {\vec{x}}^{'}) = \frac{1}{| \vec{x} - {\vec{x}}^{'} |} + F (\vec{x}, {\vec{x}}^{'})$ , $\nabla^{2} = - 4 π δ (\vec{x} - {\vec{x}}^{'})$ .

$G_{D} = 0$ on S.

Recall, the Green’s function gives the potential at $\vec{x}$ due to a point charge at ${\vec{x}}^{'}$ .

For a sphere and a point charge, $q^{'} = - \frac{a}{r^{'}}$ at ${\vec{x}}^{″}$ with $| {\vec{x}}^{″} | = r^{″} = \frac{a^{2}}{r^{'}}$ . Then, $G (\vec{x}, {\vec{x}}^{'}) = \frac{1}{| \vec{x} - {\vec{x}}^{'} |} + F (\vec{x}, {\vec{x}}^{'}) = \frac{1}{| \vec{x} - {\vec{x}}^{'} |} - \frac{a}{r^{'} | \vec{x} - r^{″} \hat{x^{'}}}$ .

So, $G (\vec{x}, {\vec{x}}^{'}) = - \frac{a}{r^{'}} \frac{1}{| \vec{x} - r^{″} {\hat{x}}^{'}} + \sum_{ℓ} \frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} P_{ℓ} (\cos γ) = - \frac{a}{r^{'}} \frac{1}{| \vec{x} - r^{″} {\hat{x}}^{'}} + \sum_{ℓ} \frac{4 π}{2 ℓ + 1} \frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} \sum_{m = - ℓ}^{ℓ} Y_{ℓ m}^{*} (θ^{'}, ϕ^{'}) Y_{ℓ m} (θ, ϕ) = \sum_{ℓ = 0}^{\infty} \frac{4 π}{2 ℓ + 1} [\sum_{m = - ℓ}^{ℓ} Y^{*} Y] (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{a}{r^{'}} \frac{r^{″ ℓ}}{r^{ℓ + 1}}) = \sum_{ℓ = 0}^{\infty} \frac{4 π}{2 ℓ + 1} [\sum_{m = - ℓ}^{ℓ} Y^{*} Y] (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{a^{2 ℓ + 1}}{r^{ℓ + 1} r^{' ℓ + 1}})$ . Note that since $r, r^{'}$ are always on the same side of the sphere and $r^{″}$ is on the other side, for charges outside the sphere \(r''

For our charges outside and the image inside, i.e. region of interest is outside, $G (\vec{x}, {\vec{x}}^{'}) = \sum_{ℓ = 0}^{\infty} \frac{4 π}{2 ℓ + 1} [\sum_{m = - ℓ}^{ℓ} Y^{*} (θ^{'}, ϕ^{'}) Y (θ, ϕ)] (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{a^{2 ℓ + 1}}{r^{ℓ + 1} r^{' ℓ + 1}})$ .

For our charges inside and image outside, i.e. region of interest is inside, $G (\vec{x}, {\vec{x}}^{'}) = \sum_{ℓ = 0}^{\infty} \frac{4 π}{2 ℓ + 1} [\sum_{m = - ℓ}^{ℓ} Y^{*} (θ^{'}, ϕ^{'}) Y (θ, ϕ)] (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{a^{2 ℓ + 1}})$ .

With our Green’s Function, recall, $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int ρ ({\vec{x}}^{'}) G (\vec{x}, {\vec{x}}^{'}) d^{3} x^{'} - \frac{1}{4 π} Φ_{S} \cdot \frac{\partial G_{D}}{\partial n^{'}} d a^{'}$ .

Charged ring inside a grounded sphere

Let the ring of charge $Q$ have a radius $a$ and the sphere have a radius $b$ .

Then, $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int ρ ({\vec{x}}^{'}) G d^{3} {\vec{x}}^{'} - \frac{1}{4 π} \int Φ \frac{\partial G}{\partial n^{'}} d a^{'}$ . Note the second term is zero due to the conductor being grounded at the boundary $Φ (b) = 0$ . Note, $ρ (r^{'}, θ^{'}, ϕ^{'}) = ρ (r^{'}, θ^{'}) = \frac{Q}{2 π a} \frac{δ (r^{'} - a) δ (\cos θ^{'} - 0)}{a} = λ δ (r^{'} - a) \frac{δ (\cos θ^{'})}{r^{'}}$

$Φ (r, θ) = \frac{1}{4 π ε_{0}} \int ρ (r^{'}, θ^{'}) G (r^{'}, θ^{'}; r, θ) d^{3} x^{'} = \frac{λ}{4 π ε_{0} a} \sum_{ℓ = 0}^{\infty} \frac{4 π}{2 ℓ + 1} \int_{0}^{b} δ (r^{'} - a) (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) r^{' 2} d r^{'} \int_{0}^{2 π} \int_{- 1}^{1} δ (\cos θ) Y_{ℓ}^{m *} (θ^{'}, φ^{'}) d \cos θ d ϕ = \frac{λ}{4 π ε_{0} a} \sum_{ℓ = 0}^{\infty} \frac{4 π}{2 ℓ + 1} \int_{0}^{b} δ (r^{'} - a) (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) r^{' 2} d r^{'} (2 π \sqrt{\frac{2 ℓ + 1}{4 π}} P_{ℓ} (0) δ_{m 0}) Y_{ℓ}^{0} (θ, φ) = \frac{λ}{4 π ε_{0} a} \sum_{ℓ = 0}^{\infty} \frac{4 π}{2 ℓ + 1} \int_{0}^{b} δ (r^{'} - a) (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) r^{' 2} d r^{'} (2 π \sqrt{\frac{2 ℓ + 1}{4 π}} P_{ℓ} (0)) \sqrt{\frac{2 ℓ + 1}{4 π}} P_{ℓ} (\cos θ) = \frac{λ}{4 π ε_{0} a} \sum_{ℓ = 0}^{\infty} \int_{0}^{b} δ (r^{'} - a) (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) r^{' 2} d r^{'} (2 π P_{ℓ} (0)) P_{ℓ} (\cos θ)$

Let $R_{ℓ} (r)$ be defined as: $r'a: \frac{r^\ell}{a^{\ell+1}}-\frac{r^\ell a^\ell}{b^{2\ell+1}}$.

$Φ (r, θ) = \frac{Q}{4 π ε_{0}} \sum_{ℓ} R_{ℓ} (r) P_{ℓ} (0) P_{ℓ} (\cos θ)$ .

Method of Images

Consider a charged ring outside the grounded sphere with charge $Q^{'} = - b / a Q$ . $d Q^{'} = - b / a d Q$ . Radius $R = b^{2} / a$ .

Line Charge in a Conducting Sphere

Let the sphere’s radius be $b$ , the line charge is then $λ = \frac{Q}{2 b}$ . Align it along the $z-$axis.

$G (\vec{x}, {\vec{x}}^{'}) = \sum_{ℓ, m} \frac{4 π}{2 ℓ + 1} (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) Y_{ℓ}^{m *} (θ^{'}, ϕ^{'}) Y_{ℓ}^{m} (θ, ϕ)$ . $ρ (\vec{x}) = λ δ (x^{'}) δ (y^{'}) = λ \frac{δ (\cos θ^{'} - 1) + δ (\cos θ^{'}) + 1}{r^{'}} \frac{1}{2 π r^{'}} = \frac{λ}{2 π} [δ (\cos θ^{'} - 1) + δ (\cos θ^{'}) + 1]$

$4 π ε_{0} Φ (\vec{x}) = \int ρ G d^{3} x^{'} = λ \sum_{ℓ} [P_{ℓ} (1) + P_{ℓ} (- 1)] P_{ℓ} (\cos θ) \int_{0}^{b} (\frac{r_{<}^{ℓ}}{r_{>}^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) d r^{'} = λ \sum_{ℓ} [1 + (- 1)^{ℓ}] P_{ℓ} (\cos θ) [\int_{0}^{r} (\frac{r^{' ℓ}}{r^{ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) d r^{'} + \int_{r}^{b} (\frac{r^{ℓ}}{r^{' ℓ + 1}} - \frac{r^{ℓ} r^{' ℓ}}{b^{2 ℓ + 1}}) d r^{'}] = 2 λ {\ln \frac{b}{r} + \sum_{n = 1}^{\infty} \frac{4 n + 1}{2 n (2 n + 1)} (1 - {(\frac{r}{b})}^{2 n}) P_{2 n} (\cos θ)} = \frac{Q}{b} {\ln \frac{b}{r} + \sum_{n = 1}^{\infty} \frac{4 n + 1}{2 n (2 n + 1)} (1 - {(\frac{r}{b})}^{2 n}) P_{2 n} (\cos θ)}$

Note, $\int_{0}^{2 π} \exp (- i m ϕ^{'}) d ϕ^{'} = 2 π δ_{m 0}$ .

$σ = ε_{0} {(\frac{\partial Φ}{\partial r})}_{r = b} = \frac{Q}{4 π ε_{0}} (1 + \sum_{n} (\frac{4 n + 1}{2 n + 1}) P_{2 n} (\cos θ))$ .