#import"@preview/physica:0.9.2":*
#import "template.typ": *

= 静电学中的边值问题 Boundary-Value Problems in Electrostatics

== Method of Images 镜像电荷法

镜像电荷：
- 适当的位置
- 适当的大小
- 相关区域外部
- 能够模拟所需的边界条件
根据静电唯一性定理，镜像电荷法是一种求解边值问题的有效方法。*将有边界的实际问题替换为有镜像电荷但无边界的放大区域。*

#problem[

A point charge located in front of an infinite plane conductor at zero potential.

#figure(
  image("pic/2024-09-27-15-39-01.png", width: 80%),
  numbering: none,
)

左侧是原始电位问题，右侧是等效图像问题。
]

== 接地导体存在时的点电荷 Point Charge in the Presence of a Grounded Conducting Sphere

#problem[

What is the potential produced by a point charge $q$ located at $y$ relative to the origin, around which is centered a grounded conducting sphere of radius $a$? And what is the force acting on the charge $q$?

#figure(
  image("pic/2024-09-27-15-41-20.png", width: 80%),
  numbering: none,
)

Grounded conducting sphere of radius $a$, with charge $q$ and image charge $q'$.

]

#solution[

现在确定镜像电荷的位置和大小：
$
Phi(vb(x)) = q/(4 pi epsilon_0 abs(vb(x)- vb(y))) + q'/(4 pi epsilon_0 abs(vb(x)- vb(y'))) = q/(4 pi epsilon_0 abs(x vu(n)- y vu(n'))) + q'/(4 pi epsilon_0 abs(x vu(n)- y' vu(n')))
$
球面上有
$
Phi(x=a) = q/(4 pi epsilon_0) 1/(a abs(vu(n) - y/a vu(n'))) + q'/(4 pi epsilon_0) 1/(y' abs(vu(n') - a/y' vu(n))) = 0
$
得到
$
q/a = -q'/y', y/a = a/y' => q' = -a/y q , y' = a^2/y
$
这样就可以得到电势
$
Phi(vb(x)) = q/(4 pi epsilon_0) (1/abs(x vu(n) - y vu(n')) - a/y 1/abs(x vu(n) - a^2/y vu(n'))) 
$
#newpara()

进一步，我们可以求出*球体表面感应电荷*
$
sigma = vb(D) dot vu(n) = - epsilon_0 grad Phi dot vu(n) = - epsilon_0 eval(partialderivative(Psi,x))_(x=a)
$

用余弦定理展开$Phi$，有
$
Phi(vb(x)) = q/(4 pi epsilon_0) (1/sqrt(x^2 + y^2 - 2 x y cos gamma) - a/y 1/sqrt(x^2 + a^4/y^2 - 2 x a^2/y cos gamma))
$
从而
$
sigma &= - epsilon_0 eval(partialderivative(Psi,x))_(x=a) \
&= - q/(4 pi a^2) a/y (1- a^2/y^2)/(1 + a^2/y^2 - 2 a/y cos gamma)^(3/2)
$

#newpara()

最后，我们可以求出*电荷$q$所受的力*

可以用镜像电荷对其作用力来计算，有
$
y - y' = y(1 - a^2/y^2)
$
直接利用Coulomb定律，有
$
abs(vb(F)) = 1/(4 pi epsilon_0) q^2/a^2 (a/y)^3 (1 - a^2/y^2)^(-2)
$
另外的方法，利用作用在球面上的总力也可以求解：


*上述结果同样适用于球体内部的电荷$q$，但球体上的总感应电荷并不等同于镜像电荷的大小，而是$-q$*。
]

== 存在带电绝缘导电球体时的点电荷 Point Charge in the Presence of a Charged, Insulated, Conducting Sphere

#problem[

What is the potential produced by a point charge $q$ located at $y$ relative to the origin, around which is centered $a$ *charged, insulated, conducting* sphere with the *total charge of* $Q$ and radius of $a$? And what is the force acting on the charge $q$?

]

#solution[

通过线性叠加求解电势：
- 从接地导电球开始
- 断开接地线，在球体中加入一定量的电荷$(Q-q')$，添加的电荷会均匀地分布在球面上
$
q' = - a/y q, y' = a^2/y 
$
有电势
$
Phi(vb(x)) = 1/(4 pi epsilon_0) (q/(abs(vb(x) - vb(y))) - (a q)/(y abs(vb(x) - a^2/y^2 vb(y))) + (Q + a/y q)/(abs(vb(x) )))
$
作用力相当于镜像电荷的作用力加上球体上的总感应电荷的作用力
$
vb(F) &= (1/(4 pi epsilon_0) q(Q+a/y q)/y^2 - 1/(4 pi epsilon_0) q^2/a^2 (a/y)^3 (1 - a^2/y^2)^(-2)) vb(y)/y\
&= 1/(4 pi epsilon_0) q/y^2 (Q- (q a^3(2y^2 - a^2))/(y (y^2 - a^2)^2)) vb(y)/y
$
当$y>>a$的时候，可以有近似
$
vb(F) = 1/(4 pi epsilon_0) q/y^2 (Q - (2q a^3)/y^3) vb(y)/y = 1/(4 pi epsilon_0) (q Q)/y^2 vb(y)/y
$
当$Q>>q$时，为零力位置（不稳定平衡点）：
$
Q = (q a^3(2y^2 - a^2))/(y (y^2 - a^2)^2) approx (q a^3 dot a^2)/(a dot 4a dot (y-a)^2 ) = (q a^2)/(4(y-a)^2)\
=> y approx a(1+1/2sqrt(q/Q))
$

#figure(
  image("pic/2024-10-07-21-28-00.png", width: 40%),
  numbering: none,
  caption: [半径为$a$的绝缘导电球体携带总电荷$Q$时对点电荷$q$的作用力。]
)


]

== 固定电位导电球附近的点电荷 Point Charge Near a Conducting Sphere at Fixed Potential

#problem[

What is the potential produced by a point charge $q$ located at $y$ relative to the origin, around which is centered a conducting sphere held at a *fixed potential* $V$ with the radius of $a$? And what is the force acting on the charge $q$? Find the potential by linear superposition. 

]

#solution[

相当于在接地导电球的基础上，加上一个电势为$V$的电荷
$
Q = 4 pi epsilon_0 a V
$

]

== Conducting Sphere in a Uniform Electric Field by Method of Images 静电场中的导电球

#problem[

What is the potential produced by a conducting sphere of radius a in a uniform electric field $E_0$?

]

#solution[

均匀电场：位于$z=-R$和$+R$位置的两个电荷$+Q$和$-Q$之间的电场近似恒定，极限为$R,Q➔∞$，$Q/R^2$恒定。


#figure(
  image("pic/2024-10-07-22-37-45.png", width: 80%),
  numbering: none,
)

#figure(
  image("pic/2024-10-07-22-37-52.png", width: 80%),
  numbering: none,
)

势能可以写成
$
Phi(vb(x)) = (Q/(4 pi epsilon_0))/(r^2 + R^2 + 2 r R cos theta)^(1/2) - (Q/(4 pi epsilon_0))/(r^2 + R^2 - 2 r R cos theta)^(1/2) \
- (a Q)/(4 pi epsilon_0)/(R (r^2 + a^4/R^2 + (2 a^2 r)/R cos theta)^(1/2)) + (a Q)/(4 pi epsilon_0)/(R (r^2 + a^4/R^2 - (2 a^2 r)/R cos theta)^(1/2))
$
考虑到$R>>r$：
$
Phi(vb(x)) =& Q/(4 pi epsilon_0 R) (1-r/R cos theta) - Q/(4 pi epsilon_0 R) (1+r/R cos theta) \
&- (a Q)/(4 pi epsilon_0 r R) (1-a^2/(r R) cos theta) + (a Q)/(4 pi epsilon_0 r R) (1+a^2/(r R) cos theta)\
=& 1/(4 pi epsilon_0) (- (2 Q)/R^2 r cos theta + (2 Q)/R^2 a^3/r^2 cos theta)\
=&- E_0 (r - a^3/r^2) cos theta\
=&- E_0 z + E_0 a^3/r^3 z
$
诱导表面电荷密度
$
sigma = - epsilon_0 eval(partialderivative(Phi,x))_(x=a) = 3 epsilon_0 E_0 cos theta
$
1. $-E_0 z$： 就是均匀场$E_0$的电势
2. 电荷密度的表面积分为0，接地球和绝缘球没有区别。
]

== Green Function for the Sphere; General Solution for the Potential of a Sphere 球体的Green函数；球体电势的一般解

=== Green Function

$
nabla^('2) G(vb(x), vb(x')) = - 4 pi delta(vb(x) - vb(x'))\
G(vb(x), vb(x')) = 1/abs(vb(x) - vb(x')) + F(vb(x), vb(x'))\
nabla^('2) F(vb(x), vb(x')) = 0
$
为满足同质边界条件而选择的单位源及其镜像（或多个镜像）所产生的电势，只是适合于 Dirichlet 或 Neumann 边界条件的 Green 函数。

=== Green Function for the Sphere

根据接地导电球体时的点电荷：
$
Phi(vb(x)) = q/(4 pi epsilon_0) 1/abs(x vu(n) - y vu(n')) + q'/(4 pi epsilon_0) 1/abs(x vu(n) - y' vu(n'))\
q' = -a/y q, y' = a^2/y
$
其中
- $x$：势能被计算出的$P$点
- $x'$：单位源的位置$P'$
将$y->x',q->4 pi epsilon_0$，得到Green函数
$
G(vb(x), vb(x')) = 1/abs(vb(x) - vb(x')) - a/(x'abs(vb(x) - a^2/x'^2 vb(x')))
$

#grid(columns: (1fr, 1fr))[
#figure(
  image("pic/2024-10-08-00-09-59.png", width: 80%),
  numbering: none,
)
][
#figure(
  image("pic/2024-10-08-00-10-08.png", width: 80%),
  numbering: none,
)
]

将点$P$, $P'$写成球坐标形式
$
P' &: (x' sin theta' cos phi', x' sin theta' sin phi', x' cos theta')\
P &: (x sin theta cos phi, x sin theta sin phi, x cos theta)
$
根据
$
(P_x - P'_x)^2 + (P_y - P'_y)^2 + (P_z - P'_z)^2 = x^2 + x'^2 - 2 x x' cos gamma
$
得到
$
cos gamma = cos theta cos theta' + sin theta sin theta' cos(phi - phi')
$
从而最终的Green函数为
#text(fill:red)[
$
G(vb(x), vb(x')) &= 1/abs(vb(x) - vb(x')) - a/(x'abs(vb(x) - a^2/x'^2 vb(x')))\
&= 1/abs(vb(x) - vb(x')) + F(vb(x), vb(x'))\
G(vb(x), vb(x')) &= 1/sqrt(x^2 + x'^2 - 2 x x' cos gamma) - 1/(sqrt((x^2 x'^2)/a^2 + a^2 - 2 x x' cos gamma))
$
]
- 对称性：
  $
  G(vb(x), vb(x')) = G(vb(x'), vb(x))
  $
  源点和观测点的物理互换性
- $G=0$如果$x$或者$x'$在球表面

=== General Solution for the Potential

Dirichlet 边界条件@DbGreen 下静电边界值问题的形式解法
$
Phi(vb(x)) = 1/(4 pi epsilon_0) integral_V G_D(vb(x),vb(x')) rho(vb(x')) dd(vb(x'), 3) - 1/(4 pi) integral.cont_S Phi(vb(x')) partialderivative(G_D,n') dd(vb(a)')\
$
其中
$
G_D (vb(x), vb(x')) = 0
$
对于$vb(x')$在球面$S$上
$
G(vb(x), vb(x')) = 1/sqrt(x^2 + x'^2 - 2 x x' cos gamma) - 1/(sqrt((x^2 x'^2)/a^2 + a^2 - 2 x x' cos gamma))
$
有边界条件
$
eval(pdv(G_D, n'))_(x'=a) = - eval(pdv(G_D, x'))_(x'=a) = - (x^2-a^2)/(a(x^2+a^2-2 x a cos theta)^(3/2))
$
球面上的电势是指定的球的Laplace方程
$
rho(vb(x')) = 0
$
从而
$
Phi(vb(x)) &= 1/(4pi) integral.cont_S Phi(vb(x')) partialderivative(G_D,n') dd(vb(a)')\
&= 1/(4pi) integral.cont_S Phi(a,theta',phi') (x^2 - a^2)/(a(x^2 + a^2 - 2 x a cos gamma)^(3/2)) dd(a')\
&= 1/(4pi) integral.cont_S Phi(a,theta',phi') (a(x^2 - a^2))/(a(x^2 + a^2 - 2 x a cos gamma)^(3/2)) dd(Omega')\
$
对于Posson方程，该项代入@DbGreen，即可。

对于内部问题，$pdv(G_D,n')$的符号正好相反。

=== Conducting Spheres with Hemispheres at Different Potentials具有不同电位半球的导电球体

#problem[

What is the potential produced by one conducting sphere of radius $a$ made up of two hemispherical shells separated by a small insulating ring? The hemispheres are kept at different potentials of $±V$.

]

#solution[

#figure(
  image("pic/2024-10-08-00-32-00.png", width: 30%),
  numbering: none,
)

$
Phi(vb(x)) &= 1/(4pi) integral.cont_S Phi(a,theta',phi') (a(x^2 - a^2))/(a(x^2 + a^2 - 2 x a cos gamma)^(3/2)) dd(Omega')\
$
注意到
$
dd(Omega') = sin theta' dd(theta',phi')\
cos gamma = cos theta cos theta' + sin theta sin theta' cos(phi - phi')
$
有
$
Phi(vb(x)) &= V/(4pi) integral_0^(2pi) dd(phi') (integral_0^1 dd(cos theta') - integral_(-1)^0 dd(cos theta')) (a(x^2 - a^2))/(a(x^2 + a^2 - 2 x a cos gamma)^(3/2))\
$
做代换
$
theta' -> pi - theta', phi' -> phi' + pi\
integral_(-1)^0 dd(cos theta') -> - integral_1^0 dd(cos theta'), cos gamma -> - cos gamma
$
得到
$
Phi(vb(x)) &= (V a (x^2-a^2))/(4pi) integral_0^(2pi) dd(phi') integral_0^1 dd(cos theta')\ & ((x^2 + a^2 - 2 x a cos gamma)^(-3/2) - (x^2 + a^2 + 2 x a cos gamma)^(-3/2))\
$
考虑$z$轴正半轴上的电势
$
theta = 0 => cos gamma = cos theta'\
Phi(vb(x)) &= V(1-(z^2-a^2)/(z sqrt(z^2+a^2)))
$
当$z=a$时，$Phi = V$, $z>>a$时，$Phi = (3V a^2)/(2z^2)$

*电位的进一步推导*

记
$
alpha = (a x)/(x^2 + a^2)
$
有
$
(1-2alpha cos gamma)^(-3/2) - (1+2alpha cos gamma)^(-3/2) =  6 alpha cos gamma + 35 alpha^3 cos^3 gamma + ...
$
再结合积分
$
integral_0^(2 pi) dd(phi') integral_0^1 dd(cos theta') cos gamma = pi cos theta\
integral_0^(2 pi) dd(phi') integral_0^1 dd(cos theta') cos^3 gamma = pi/4 cos theta (3 - cos^2 theta)
$
从而
$
Phi(vb(x)) &= (V a (x^2-a^2))/(4pi) integral_0^(2pi) dd(phi') integral_0^1 dd(cos theta')\ & ((x^2 + a^2 - 2 x a cos gamma)^(-3/2) - (x^2 + a^2 + 2 x a cos gamma)^(-3/2))\
&approx (3V a^2)/(2x^2) (x^3(x^2-a^2))/(x^2 + a^2)^(5/2) cos theta (1 + 35/24 (a^2 x^2)/(x^2 + a^2)^2 (3 - cos^2 theta) + ...)
$
如果$z>>a$，有
$
(x^3(x^2-a^2))/(x^2 + a^2)^(5/2) approx (1- a^2/x^2)/(1 + a^2/x^2)^(5/2) approx 1 - 7/2 a^2/x^2
$
$
Phi(x, theta, phi) = (3V a^2)/(2 x^2) (cos theta - 7/12 a^2/x^2 (5/2 cos^3 theta - 3/2 cos theta)+ ...) 
$<shpere>
考虑$z$轴正半轴上的电势
$
theta = 0\
Phi = (3V a^2)/(2 z^2)
$
事实上，中的两个角度因子是$P_1(cos θ)$和$P_3(cos θ)$的 Legendre 多项式。

Legendre 多项式是普通 Legendre 微分方程的解：
$
dv(,x) (1-x^2 dd(,x)P_n (x)) + n(n+1) P_n(x) = 0
$
]

== Orthogonal Functions and Expansions 正交函数与展开

=== Orthogonal Functions

一组实函数或复函数$U_n (xi) ,n=1,2,...,n$ 在 $(a,b)$ 上平方可积，称这组函数是标准正交的，如果
$
integral_a^b U_m^* (xi) U_n (xi) dd(xi) = delta_(m n)
$
其中$delta_(m n)$是Kronecker delta。

=== Expansions in Orthonormal Functions

对于任意在$(a,b)$上平方可积的函数$f(xi)$，可以用正交函数$U_n (xi)$展开
$
f(xi) <-> sum_(n=1)^N a_n U_n (xi)
$

系数的选择应使均方误差最小：
$
M_N = integral_a^b abs(f(xi) - sum_(n=1)^N a_n U_n (xi))^2 dd(xi)\
pdv(M_N, a_n) = 0 => a_n = integral_a^b f(xi) U_n^* (xi) dd(xi)
$

证明：
$
M_N &= integral_a^b abs(f(xi) - sum_(n=1)^N a_n U_n (xi))^2 dd(xi)\
&= integral_a^b (f(xi) - sum_(n=1)^N a_n U_n (xi)) (f^*(xi) - sum_(n=1)^N a_n^* U_n^* (xi)) dd(xi)\
pdv(M_N, a_n) &= - integral_a^b f^*(xi) U_n^* (xi) dd(xi) + a_n^* = 0\
&=> a_n = integral_a^b f(xi) U_n^* (xi) dd(xi)
$

=== Completeness of the set of orthonormal functions

完备性：存在一个有限数$N_0$，对于$N>N_0$，均方误差$M_N$可以小于任意小的正数。

$
f(xi) = sum_(n=1)^oo a_n U_n (xi)
$

_数学物理中通常出现的所有正交函数集已被证明是完整的。_

=== Completeness relation (or closure relation)

完备性条件是
$
f(xi) &= sum_(n=1)^oo a_n U_n (xi) = sum_(n=1)^oo U_n (xi)integral_a^b f(xi') U_n^* (xi') dd(xi') \
&= integral_a^b f(xi') sum_(n=1)^oo U_n (xi) U_n^* (xi') dd(xi')
$
注意到
$
integral_a^b delta(xi - xi') f(xi') dd(xi') = f(xi)
$
则有完备性条件
$
sum_(n=1)^oo U_n^* (xi') U_n (xi) = delta(xi' - xi)
$

*Fourier series.*

定义在$(-a/2,a/2)$上的函数可以被基函数
$
f_m (x) = sqrt(2/a) sin((2 pi x m)/a)\
g_m (x) = sqrt(2/a) cos((2 pi x m)/a)
$
展开为
$
F(x) = 1/2 A_0 + sum_(m=1)^oo A_m (cos((2 pi x m)/a) + B_m sin((2 pi x m)/a))
$
其中
$
A_m = 2/a integral_(-a/2)^(a/2) f(x) cos((2 pi x m)/a) dd(x)\
B_m = 2/a integral_(-a/2)^(a/2) f(x) sin((2 pi x m)/a) dd(x)
$
该函数族也有正交归一关系
$
integral_(-a/2)^(a/2) f_m (x) f_n (x) dd(x) = delta_(m n)\
integral_(-a/2)^(a/2) g_m (x) g_n (x) dd(x) = delta_(m n)\
integral_(-a/2)^(a/2) f_m (x) g_n (x) dd(x) = 0
$


=== Generalization to two-dimensional

对于任意函数$f(ξ,η)$，取值为$(a,b) times (c,d)$，且每个维度上的正交函数分别为$U_n (ξ)$和$V_n (η)$，可展开为
$
f (ξ,η) = sum_(n=1)^oo sum_(m=1)^oo a_(n m) U_n (ξ) V_m (η)
$
其中
$
a_(n m) = integral_a^b integral_c^d f(ξ,η) U_n^* (ξ) V_m^* (η) dd(ξ) dd(η)
$

*Fourier integral.*

在$(-a/2,a/2)$的正交函数族
$
U_m (x) = 1/sqrt(a) e^((2 pi i m x)/a)
$
展开为
$
f(x) = 1/sqrt(a) sum_(m=-oo)^(oo) A_m e^((2 pi i m x)/a)
$
其中
$
A_m = 1/sqrt(a) integral_(-a/2)^(a/2) f(x') e^((-2 pi i m x')/a) dd(x')
$
该函数族也有正交归一关系
$
integral_(-a/2)^(a/2) U_m (x) U_n^* (x) dd(x) = delta_(m n)
$
以及完备性条件
$
sum_(m=-oo)^(oo) 1/a e^((2 pi i m x)/a) e^((-2 pi i m x')/a) = delta(x'-x)
$
当$a->oo$的时候，得到Fourier变换。
$
(2 pi m)/a -> k, sum_m -> integral dd(m) -> a/(2pi) integral dd(k)
$
则有
$
f(x) = 1/sqrt(2pi) integral_(-oo)^oo A(k) e^(i k x) dd(k)
$
其中
$
A(k) = 1/sqrt(2pi) integral_(-oo)^oo f(x) e^(-i k x) dd(x)
$
也有正交归一关系
$
integral_(-oo)^oo e^(i k x) e^(-i k' x) dd(x) = 2pi delta(k-k')
$
以及完备性条件
$
1/(2pi) integral_(-oo)^oo e^(i k x) e^(-i k x') dd(k) = delta(x-x')
$

== Separation of Variables; Laplace Equation in Rectangular Coordinates 分离变量；矩形坐标系下的Laplace方程


=== Laplace Equation in Rectangular Coordinates

$
laplacian Phi = 0\
pdv(Phi,x,2) + pdv(Phi,y,2) + pdv(Phi,z,2) = 0
$

=== Separation of Variables

偏微分方程可以用三个常微分方程来求解。 

电势可以用三个函数的乘积来表示：
$
Phi(x,y,z) = X(x) Y(y) Z(z)
$
代入有
$
Y(y) Z(z) dv(X,x,2) + X(x) Z(z) dv(Y,y,2) + X(x) Y(y) dv(Z,z,2) = 0
$
则有
$
1/X dv(X,x,2) + 1/Y dv(Y,y,2) + 1/Z dv(Z,z,2) = 0
$
令
$
1/X dv(X,x,2) =- alpha^2 , 1/Y dv(Y,y,2) =  - beta^2 , 1/Z dv(Z,z,2) = gamma^2 = alpha^2 + beta^2
$
分离变量的解为
$
Phi = (A e^(i alpha x) + B e^(-i alpha x))(C e^(i beta y) + D e^(-i beta y))(E e^(sqrt(alpha^2 + beta^2) z) + F e^(-sqrt(alpha^2 + beta^2) z))
$

#problem[
Consider a rectangular box with dimensions $(a,b,c)$ in the $(x,y,z) $directions. All surfaces of the box are kept at zero potential, except the surface z=c, which is at potential $V(x,y)$. Find the potential everywhere inside the box.
]
#solution[
#figure(
  image("pic/2024-10-09-10-56-19.png", width: 80%),
  numbering: none,
)

#figure(
  image("pic/2024-10-09-10-57-23.png", width: 80%),
  numbering: none,
)

#figure(
  image("pic/2024-10-09-10-57-46.png", width: 80%),
  numbering: none,
)

#figure(
  image("pic/2024-10-09-10-58-04.png", width: 80%),
  numbering: none,
)
]

分离变量的解对于边界条件会给出参数的限制条件，例如本例
$
X_n (x) = 2 A sin((n pi x)/a)
$
这里的${X_n}$就构成了一个完备的正交函数集，可以用来展开$Phi$。


== A Two-Dimensional Potential Problem, Summation of Fourier Series 二维电势问题，Fourier级数求和

=== Two-Dimensional Potential Problem

$
laplacian Phi = 0\
pdv(Phi,x,2) + pdv(Phi,y,2) = 0
$
可以分离变量
$
Phi(x,y) = X(x) Y(y)\
1/X dv(X,x,2) + 1/Y dv(Y,y,2) = 0
$
令
$
1/X dv(X,x,2) =- alpha^2 , 1/Y dv(Y,y,2) =  alpha^2
$
从而分离变量的解为
$
Phi = (A e^(i alpha x) + B e^(-i alpha x))(C e^(alpha y) + D e^(-alpha y))
$

#problem[

Consider a two-dimensional potential problem in the region of $0≤x≤a$, $y≥0$.

]

#solution[

#figure(
  image("pic/2024-10-09-11-13-43.png", width: 80%),
  numbering: none,
)
#figure(
  image("pic/2024-10-09-11-14-07.png", width: 80%),
  numbering: none,
)
#figure(
  image("pic/2024-10-09-11-15-16.png", width: 80%),
  numbering: none,
)
#figure(
  image("pic/2024-10-09-11-15-35.png", width: 80%),
  numbering: none,
)
#figure(
  image("pic/2024-10-09-11-15-47.png", width: 80%),
  numbering: none,
)
]

== Field and Charge Densities in Two-Dimensional Corners and Along Edges 二维角和边上的场和电荷密度

=== Field in Two-Dimensional Polar Coordinates

二维极坐标中的Laplace方程
$
laplacian Phi = 0\
1/rho pdv(,rho)(rho pdv(Phi,rho)) + 1/rho^2 pdv(Phi,phi,2) = 0
$
可以分离变量
$
Phi = R(rho) Psi(phi)
$
得到
$
rho/R dv(,rho)(rho dv(R,rho)) + 1/Psi dv(Psi,phi,2) = 0
$
令
$
1/R dv(R,rho,2) = v^2, 1/Psi dv(Psi,phi,2) = -v^2
$
从而分离变量的解为：
当$v=0$时
$
R = a_0 + b_0 ln rho\
Psi = A_0 + B_0 phi
$
当$v≠0$时
$
R = a_v rho^v + b_v rho^(-v)\
Psi = A_v cos(v phi) + B_v sin(v phi)
$
从而分离变量的解为
$
Phi = a_0 + b_0 ln rho + sum_(n=1)^oo a_n rho^n sin(n phi + alpha_n) + b_n rho^(-n) sin(n phi + beta_n)
$
如果原点处不发散，有
$
b_n = 0, n=1,2,...
$

#problem[

Field and Charge Densities in Two-Dimensional Corners
]

#solution[

#figure(
  image("pic/2024-10-11-01-48-04.png", width: 80%),
  numbering: none,
)

#figure(
  image("pic/2024-10-11-01-49-43.png", width: 80%),
  numbering: none,
)

#figure(
  image("pic/2024-10-11-01-51-17.png", width: 80%),
  numbering: none,
)

]