# Polar Laplace Equation

$$\nabla^2 = \frac{1}{\rho}\frac{\partial}{\partial\rho}\left(\rho\frac{\partial\Phi}{\partial\rho}\right)+\frac{1}{\rho^2}\frac{\partial^2\Phi}{\partial\phi^2} + \frac{\partial^2\Phi}{\partial z^2}=0$$.

## General Solution no Z

Assume we have no $$z$$ dependence. Then we get solution $$Q(\phi) = \exp(\pm im\phi)$$. Then, $$\rho\frac{d}{d\rho}\left(\rho \frac{dR}{d\rho}\right)-m^2R=0$$.

For $$m=0$$, $$\rho \frac{dR}{d\rho}b_0\Rightarrow R(\rho) = a_0+b_0\ln \rho$$.

For $$m\neq 0$$, then, $$\rho^2\frac{d^2R}{d\rho^2} + \rho\frac{dR}{d\rho} = m^2R$$. Let $$R(\rho) = \rho^\alpha$$ as an ansatz. Then, $$(\alpha(\alpha-1) + \alpha - m^2)\rho^\alpha = 0$$. Thus, $$\alpha^2=m^2$$. So, $$\alpha=\pm m$$. Overall, $$\Phi(\rho,\phi) = a_0+b_0\ln\rho + \sum_{m=1}^\infty \left(A_m\rho^m+\frac{B_m}{\rho^m}\right)(c_m\sin m\phi+D_m\cos m\phi)$$.

### Examples

#### Circular Cylinder

Right side of cylinder is at $$+V$$ and the left is $$-0$$. At $$\rho=0$$ then the potential must be zero. Then, $$b_0=0,a_0=0,B_m=0$$. Let $$-\pi\leq \phi\leq \pi$$. So, we get even function for angular solutions. Thus, $$\Phi(\rho,\phi) = \sum_m A_m\rho^m \cos m\phi$$.

Orthogonality, $$\int_{-\pi}^\pi \cos m\phi \cos m'\phi d\phi = \pi\delta_{m'm}$$.

At $$\rho =a$$, $$\int_{-\pi}^\pi\cos m'\phi d\phi = A_m' a^{m'}\pi = 2\left[\int_0^{\frac{\pi}{2}}\cos m'\phi d\phi - \int_{\frac{\pi}{2}}^\pi\cos m'\phi d\phi\right] = \frac{4V}{m'}\sin\frac{m'\pi}{2}$$.

Thus, $$\Phi(\rho,\phi) = \sum_{m\text{ odd}}\frac{4V}{a^m \pi}(-1)^m\rho^m \cos m\phi$$.

#### Corner

Let $$\beta$$ be the angle of the corner, $$\beta\leq 90^\circ$$. And the potential be $$V$$ at the boundary. Let $$0\leq\phi\leq \beta$$ be our region of interest. So, $$\Phi(\rho,\varphi)=R(\rho)Q(\phi)$$. Then, Laplace’s equation can be written,

\begin{align*} &\frac{1}{\rho} \frac{\partial}{\partial\rho} \left(\rho\frac{\partial\Phi}{\partial\rho}\right) + \frac{1}{\rho} \frac{\partial^2\Phi}{\partial\phi^2} \end{align*}

Let the polar equation be equal to $$-\nu^2$$. The radial solution is then $$\rho^\nu,\rho^{-\nu}$$ and for $$\nu=0$$, $$R(\rho)=a_0+b_0\ln\rho$$. For the finite region, the angular solution for $$\nu=0$$ can be $$Q(\phi)=A_0+B_0\phi$$. So, $$\Phi(\rho,\phi) = (a_0+b_0\ln\rho)(A_0+B_0\phi) + \sum_{m=1}^\infty(a_m\rho^m+b_m\rho^{-m})(A_m\sin(\nu_m\phi)+B_m\cos(\nu_m\phi))$$.

For $$\rho=0$$, the potential is finite. So, $$\Phi(\rho,\varphi) = A_0+B_0\phi + \sum_{m=1}^\infty \rho^m(A_m\sin\nu_m\phi+B_m\cos\nu_m\phi)$$.

For $φ=0,β,$$\Phi\to V$$. $$\Phi(\rho,0) = A_0+\sum_m\rho^m B_m$$. So, $$B_m=0$$. $$\Phi(\rho,\beta) = V = A_0 + B_0\beta + \sum_m\rho^m A_m\sin(\nu_m\beta)$$. So,$νm = \frac{m\pi}{\beta}, A0=V,and $$B_0=0$$. Thus, $$\Phi(\rho,\phi) = V + \sum_m A_m\rho^{\frac{m\pi}{\beta}}\sin\left(\frac{m\pi}{\beta}\phi\right)$$. When, $$\rho\to 0$$, $$\Phi(\rho,\phi) \approx = V + A_1\rho^{\frac{\pi}{\beta}}\sin\left(\frac{\pi}{\beta}\phi\right)$$. $$E_\rho = -\frac{\partial\Phi}{\partial\rho} = -A_1\frac{\pi}{\beta}\rho^{\frac{\pi}{\beta}-1}\sin\left(\frac{\pi}{\beta}\phi\right)$$. $$E_\phi = -\frac{1}{\rho}\frac{\partial\Phi}{\partial\phi} = -A_1\frac{\pi}{\beta}\rho^{\frac{\pi}{\beta}-1}\cos\left(\frac{\pi}{\beta}\phi\right)$$. $$\sigma = \varepsilon_0 E_\phi(\phi=0) \propto \rho^{\frac{\pi}{\beta}-1}$$. So, when $$\beta<\pi$$ then $$\sigma\to0$$ as $$\rho\to0$$. Otherwise, the charge density blows up. $$\sigma = \varepsilon_0 E_n = \varepsilon_0\vec{E}\cdot\hat{n}$$. ## General Cylindrical Coordinates \begin{align*} \frac{1}{\rho} \frac{\partial}{\partial\rho} \left(\rho\frac{\partial\Phi}{\partial\rho}\right) + \frac{1}{\rho} \frac{\partial^2\Phi}{\partial\phi^2} + \frac{\partial^2\Phi}{\partial z^2} = 0 \end{align*} Separate, $$\Phi(\rho,\phi,z) = R(\rho)Q(\phi)Z(z)$$. Let $$\frac{d^2Z}{dz^2}=-k^2$$. We then get $$Z(z) = A\exp(ikz) + B\exp(-ikz)$$. Let $$\frac{d^2Z}{dz^2}=k^2$$. We then get $$Z(z) = A\exp(kz) + B\exp(-kz)$$. We get Bessel functions if $$k^2$$ and modified Bessel functions if $$-k^2$$. Consider a Cylinder with a potential $$V(\rho,\phi)$$ at the top and the bottom and sides are grounded. Then, we want $$k^2$$. We then get $$Q(\phi) = \exp(\pm im\phi)$$. The radial function is then Bessel functions, from the differential equation, $$\frac{\rho}{R}\frac{d}{d\rho}\left(\rho\frac{dR}{d\rho}\right) + (k\rho)^2 - m^2 = 0.$$ Or, the modified equation, $$x\frac{d}{dx}\left(x\frac{dR}{dx}\right) + [x^2 - m^2]R = 0.$$ The solutions are $$R(x) = AJ_m(x) + BN_m(x)$$. From the fact we expect a finite potential at the center, we get, $$B=0$$. So, $$R(r) = AJ_m(kr)$$. From the bottom being grounded, we can write, $$Z(z) = C\sinh kz$$ To get zero at the sides, $$R(r) = \sum_n A_n J_m\left(\frac{\beta_{mn}}{a}r\right)$$. This gives an overall solution, $$\sum_{m=0}^\infty\sum_{n=1}^\infty\sinh(k_m^nz)J_m(k_m^nr)(A_{m,n}\sin(m\phi)+B_{m,n}\cos(m\phi))$$ At the top surface, $$z=L$$, $$V=V(\rho,\phi)$$. Then from orthogonality, $$\int_0^{2\pi}\int_0^a\Phi(r,\phi,L)\sin(m'\phi)J_{m'}(k_{m'}^{n'}a) rdr d\phi = \int_0^{2\pi}\int_0^a V(\rho,\phi)\sin(m'\phi)J_{m'}(k_{m'}^{n'}a)rdrd\phi = A_{mn}\sinh(k_{mn}L)\pi\frac{a^2}{2}[J_{m'+1}(k_{m'n'}a)]^2\delta_{mm'}\delta_{nn'}$$. Also from orthogonality for $$n>0$$, $$\int_0^{2\pi}\int_0^a\Phi(r,\phi,L)\cos(m'\phi)J_{m'}(k_{m'}^{n'}a) rdr d\phi = \int_0^{2\pi}\int_0^a V(\rho,\phi)\cos(m'\phi)J_{m'}(k_{m'}^{n'}a)rdrd\phi = A_{mn}\sinh(k_{mn}L)\pi\frac{a^2}{2}[J_{m'+1}(k_{m'n'}a)]^2\delta_{mm'}\delta_{nn'}$$. For $$n=0$$, $$B_{0n}\to \frac{1}{2}B_{on}$$. Orthogonality of sines, $$\int_0^a\sin\frac{n\pi x}{a}\sin\frac{n'\pi x}{a}dx = \frac{a}{2}\delta_{n'n}$$ Orthogonality of the Bessel functions, $$\int J_m(k_{mn}\rho)J_m(k_{mn'}\rho)\rho d\rho = \frac{a^2}{2}[J_{m+1}(k_{m,n}a)^2]\delta_{n'n}$$ #### Bessel Function Notes $$J_m(x) = \left(\frac{x}{2}\right)^{m}\sum_{j=0}^\infty \frac{(-1)^j}{j!(m+j)!}\left(\frac{x}{2}\right)^{2j}$$. $$\rho\to 0$$, $$J_0(x) = 1, J_m(x) = 0, N_m(x)\to-\infty$$. $$\rho\to \infty$$, $$J_m(x) = \sqrt{\frac{2}{\pi x}}\cos(x-m\pi/2-\pi/4), N_m(x)=\sqrt{\frac{2}{\pi x}}\sin(x-m\pi/2-\pi/4)$$. ## Poisson Equation in Spherical Coordinates $$\nabla^2\Phi = -\rho/\varepsilon_0$$. $$\Phi(\vec{x}) = \frac{1}{4\pi\epsilon_0}\int\rho(\vec{x}')G(\vec{x},\vec{x}')dx'^3 - \frac{1}{4\pi}\int\Phi_S(\vec{x}')\frac{\partial G}{\partial n'}da'$$. $$G(\vec{x},\vec{x}') = \frac{1}{|\vec{x}-\vec{x}'|} + F(\vec{x},\vec{x}')$$, $$\nabla^2=-4\pi\delta(\vec{x}-\vec{x}')$$. $$G_D=0$$ 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, $$G(\vec{x},\vec{x}') = -\frac{a}{r'}\frac{1}{|\vec{x}-r''\hat{x}'} + \sum_\ell \frac{r_<^\ell}{r_>^{\ell+1}}P_\ell(\cos\gamma) = -\frac{a}{r'}\frac{1}{|\vec{x}-r''\hat{x}'} + \sum_\ell\frac{4\pi}{2\ell+1} \frac{r_<^\ell}{r_>^{\ell+1}}\sum_{m=-\ell}^\ell Y_{\ell m}^*(\theta',\phi')Y_{\ell m}(\theta,\phi) = \sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1}\left[\sum_{m=-\ell}^{\ell}Y^*Y\right]\left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{a}{r'}\frac{r''^\ell}{r^{\ell+1}}\right) = \sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1} \left[\sum_{m=-\ell}^{\ell}Y^*Y\right] \left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{a^{2\ell+1}}{r^{\ell+1}r'^{\ell+1}}\right)$$. 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_{\ell=0}^\infty\frac{4\pi}{2\ell+1} \left[\sum_{m=-\ell}^{\ell}Y^*(\theta',\phi')Y(\theta,\phi)\right] \left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{a^{2\ell+1}}{r^{\ell+1}r'^{\ell+1}}\right)$$. For our charges inside and image outside, i.e. region of interest is inside, $$G(\vec{x},\vec{x}') = \sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1} \left[\sum_{m=-\ell}^{\ell}Y^*(\theta',\phi')Y(\theta,\phi)\right] \left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{r^\ell r'^\ell}{a^{2\ell+1}}\right)$$. With our Green’s Function, recall, $$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\int\rho(\vec{x}')G(\vec{x},\vec{x}')d^3x' - \frac{1}{4\pi}\Phi_S\cdot\frac{\partial G_D}{\partial n'}da'$$. ### Charged ring inside a grounded sphere Let the ring of charge $$Q$$ have a radius $$a$$ and the sphere have a radius $$b$$. Then, $$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\int\rho(\vec{x}')Gd^3\vec{x}' - \frac{1}{4\pi}\int \Phi\frac{\partial G}{\partial n'}da'$$. Note the second term is zero due to the conductor being grounded at the boundary $$\Phi(b)=0$$. Note, $$\rho(r',\theta',\phi') = \rho(r',\theta') = \frac{Q}{2\pi a}\frac{\delta(r'-a)\delta(\cos\theta'-0)}{a} = \lambda \delta(r'-a)\frac{\delta(\cos\theta')}{r'}$$ $$\Phi(r,\theta) = \frac{1}{4\pi\varepsilon_0}\int \rho(r',\theta')G(r',\theta';r,\theta) d^3x' = \frac{\lambda}{4\pi\varepsilon_0 a}\sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1}\int_0^b \delta(r'-a)\left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)r'^2dr'\int_0^{2\pi}\int_{-1}^1 \delta(\cos\theta)Y_\ell^{m*}(\theta',\varphi')d\cos\theta d\phi = \frac{\lambda}{4\pi\varepsilon_0 a}\sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1}\int_0^b \delta(r'-a)\left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)r'^2dr'(2\pi\sqrt{\frac{2\ell+1}{4\pi}}P_\ell(0)\delta_{m0})Y_\ell^0(\theta,\varphi) = \frac{\lambda}{4\pi\varepsilon_0 a}\sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1}\int_0^b \delta(r'-a)\left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)r'^2dr'(2\pi\sqrt{\frac{2\ell+1}{4\pi}}P_\ell(0))\sqrt{\frac{2\ell+1}{4\pi}}P_\ell(\cos\theta) = \frac{\lambda}{4\pi\varepsilon_0 a}\sum_{\ell=0}^\infty\int_0^b \delta(r'-a)\left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)r'^2dr'(2\pi P_\ell(0))P_\ell(\cos\theta)$$ Let $$R_\ell(r)$$ be defined as: $$r'a: \frac{r^\ell}{a^{\ell+1}}-\frac{r^\ell a^\ell}{b^{2\ell+1}}$$. $$\Phi(r,\theta) = \frac{Q}{4\pi\varepsilon_0}\sum_\ell R_\ell(r)P_\ell(0)P_\ell(\cos\theta)$$. #### Method of Images Consider a charged ring outside the grounded sphere with charge $$Q'=-b/aQ$$. $$dQ' = -b/adQ$$. Radius $$R = b^2/a$$. ### Line Charge in a Conducting Sphere Let the sphere’s radius be $$b$$, the line charge is then $$\lambda = \frac{Q}{2b}$$. Align it along thez-\$axis.

$$G(\vec{x},\vec{x}') = \sum_{\ell,m}\frac{4\pi}{2\ell+1}\left(\frac{r_<^\ell}{r_>^{\ell+1}}-\frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)Y_{\ell}^{m*}(\theta',\phi')Y_\ell^m(\theta,\phi)$$. $$\rho(\vec{x}) = \lambda\delta(x')\delta(y') = \lambda \frac{\delta(\cos\theta'-1)+\delta(\cos\theta')+1}{r'}\frac{1}{2\pi r'} = \frac{\lambda}{2\pi}[\delta(\cos\theta'-1)+\delta(\cos\theta')+1]$$

$$4\pi\varepsilon_0\Phi(\vec{x}) = \int\rho Gd^3x' = \lambda\sum_\ell [P_\ell(1) + P_\ell(-1)]P_\ell(\cos\theta)\int_0^b\left(\frac{r_<^\ell}{r_>^{\ell+1}} - \frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)dr' = \lambda\sum_\ell [1+(-1)^\ell]P_\ell(\cos\theta)\left[\int_0^r\left(\frac{r'^\ell}{r^{\ell+1}} - \frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)dr' + \int_r^b\left(\frac{r^\ell}{r'^{\ell+1}} - \frac{r^\ell r'^\ell}{b^{2\ell+1}}\right)dr'\right] = 2\lambda\left\{\ln\frac{b}{r} + \sum_{n=1}^\infty\frac{4n+1}{2n(2n+1)}\left(1-\left(\frac{r}{b}\right)^{2n}\right)P_{2n}(\cos\theta)\right\} = \frac{Q}{b}\left\{\ln\frac{b}{r} + \sum_{n=1}^\infty\frac{4n+1}{2n(2n+1)}\left(1-\left(\frac{r}{b}\right)^{2n}\right)P_{2n}(\cos\theta)\right\}$$

Note, $$\int_0^{2\pi}\exp(-im\phi')d\phi' = 2\pi\delta_{m0}$$.

$$\sigma = \varepsilon_0\left(\frac{\partial \Phi}{\partial r}\right)_{r=b} = \frac{Q}{4\pi\varepsilon_0}\left(1+\sum_n\left(\frac{4n+1}{2n+1}\right)P_{2n}(\cos\theta)\right)$$.

Created: 2023-06-25 Sun 02:31

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