Solving Electrostatic Problems

From Gauss's Theorem, we know that $$\nabla \cdot E = \frac{\rho}{\epsilon}$$ Substitute the definition of electric potential, we have $$\nabla^2 V = \frac{\rho}{\epsilon}$$ This is known as the Laplace's equation.

First Uniqueness Theorem

If boundary conditions are given, then the solution to the Laplace's equation is uniquely determined.
Proof
Suppose \(V_1\) and \(V_2\) are both solution to Laplace's equation satisfying the same boundary condition.
Let \(V_3 = V_2 - V_1\) $$\nabla^2V_3 = \nabla^2V_2 - \nabla^2V_1 = 0$$ i.e. no local maxima or minima everywhere except boundaries. As \(V_2\) and \(V_1\) have the same boundary conditions, all boundaries must be zero for their difference. Thus, \(V_3\) is zero everywhere. $$V_2 = V_1$$

Example (Griffiths Third Edition Ex 3.2)

A point charge \(q\) is situated a distance \(a\) from the center of a grounded conducting sphere of radius \(R\). Find the potential outside the sphere.

Solution:

Choose an image charge \(q'\) at a distance \(a\) from the centre of the sphere such that the potential on the surface of the sphere is 0 \begin{align} \frac{1}{4\pi\epsilon_0}\left[\frac{q}{d}+\frac{q'}{d'}\right]&=0\\ \frac{d}{d'}&=-\frac{q}{q'} \end{align} Applying cosine law, $$\frac{q^2}{q'^2} = \frac{d^2}{d'^2} = \frac{R^2+s^2-2Rs\cos\theta}{R^2+a^2-2Ra\cos\theta}$$ $$q^2(R^2+a^2-2Ra\cos\theta) = q'^2(R^2+s^2-2Rs\cos\theta)$$ At \(\theta=0\), $$q^2(R^2+a^2-2Ra)=q'^2(R^2+s^2-2Rs)$$ At \(\theta=\pi\), $$q^2(R^2+a^2+2Ra)=q'^2(R^2+s^2+2Rs)$$ Subtracting the two equations, $$4Raq^2 = 4Rsq'^2 \implies q^2a=q'^2s$$ At \(\theta=\pi/2\), $$q^2(R^2+a^2)=q'^2(R^2+s^2)$$ Then, $$R^2 + a^2 = \frac{a}{s}(R^2+s^2)$$ $$a^2-\frac{R^2+s^2}{s}a+R^2=0$$ So, \(a=s\) or \(a=R^2/s\). As \(a<s\), $$a=\frac{R^2}{s}$$ and the image charge is $$q'=-q\sqrt{\frac{a}{s}}=q\frac{R}{s}$$ The potential is $$V(\vec{r}) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{d}+\frac{q'}{d'}\right)$$ Applying cosine law, $$V(r,\theta) = \frac{1}{4\pi\epsilon_0}\left[\frac{q}{\sqrt{r^2+a^2-2ra\cos\theta}}-\frac{q}{\sqrt{R^2+(ra/R)^2-2ra\cos\theta}}\right]$$ Since the boundary conditions are the same, by the first uniqueness theorem, the solution is unique, hence equivalent to the situation in the question.

Checkpoint (Griffiths Third Edition Q3.10)

Two semi-infinite grounded conducting planes meet at right angles. In the region between them, there is a point charge \(q\). Find the potential in this region.

Set up the image charge as in the image above. The potential then is $$V(x,y) = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{\sqrt{(x-a)^2+(y-b)^2+z^2}} \\+ \frac{1}{\sqrt{(x+a)^2+(y+b)^2+z^2}} -\frac{1}{\sqrt{(x+a)^2+(y-b)^2+z^2}} \\-\frac{1}{\sqrt{(x-a)^2+(y+b)^2+z^2}}\right)$$

Second Uniqueness Theorem

If a region with a charge density \(\rho\) contains only conductors and the total charge of each conductors are given, then the electric field of this region is uniquely determined.
Proof
Let \(E_1\) and \(E_2\) be two solutions satisfying the conditions of the region $$\nabla \cdot \textbf{E}_1 = \frac{\rho}{\epsilon}$$ and $$\nabla \cdot \textbf{E}_2 = \frac{\rho}{\epsilon}$$ For outer boundary of the whole region, $$\int_{outer} \textbf{E}_1 \cdot d\textbf{a} = \frac{Q_{tot}}{\epsilon}$$ and $$\int_{outer} \textbf{E}_2 \cdot d\textbf{a} = \frac{Q_{tot}}{\epsilon}$$ Let \(\textbf{E}_3 = \textbf{E}_2 -\textbf{E}_1\), we have $$\int_{outer} \textbf{E}_3 \cdot d\textbf{a} = 0$$ Let \(V_3\) be the electric potential, i.e. \(\textbf{E}_3 = \nabla V_3\) By Green's identity, $$\int(V_3\nabla^2V_3 + \nabla V_3\cdot\nabla V_3)d\tau = \oint V_3 \nabla V_3 \cdot d\textbf{a}$$ Since \(\nabla^2V_3 = \nabla^2V_2 - \nabla^2V_1 = 0\), $$\int E_3^2 d\tau = \oint V_3 \textbf{E}_3 \cdot d\textbf{a}$$ As \(V_3\) is constant over each (conductor) surface, $$\int E_3^2 d\tau = V_3\oint \textbf{E}_3 \cdot d\textbf{a} = 0$$ Since \(E_3\) is zero everywhere, $$\textbf{E}_2 =\textbf{E}_1$$ If not all are conductors but either \(V\) or \(\frac{\partial V}{\partial n}\) of each boundary is given, then for each surface
if \(V\) is given, \(V_2 = V_1\) on that surface (by 1st Uniqueness Theorem, solution is unique), then \(V_3 = 0\) $$\int E_3^2 d\tau = 0 $$ if \(\frac{\partial V}{\partial n}\) is given, \(E_{2\bot} = E_{1\bot}\) on that surface, then \(\textbf{E}_3 \cdot d\textbf{a} = 0\) $$\int E_3^2 d\tau = 0 $$ Since \(E_3\) is zero on every surface, $$\textbf{E}_2 =\textbf{E}_1$$

Separation of Variables

If the potential or charge density on the boundaries of a region is given, find a solution that is products of functions that each depends only on one of the coordinate. If the solution satisfies the boundary conditions and solves the Laplace's equation, by first uniqueness theorem, the solution is the electric potential of the region.

Example (Griffiths Third Edition Ex 3.5)

An infinitely long rectangular metal pipe (sides \(a\) and \(b\)) is grounded, but one end, at x=0, is maintained at a specified potential \(V_0(y,z)\). Find the potential inside the pipe.

Solution: As there is no charge, the Laplace's equation is $$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2V}{\partial y^2} + \frac{\partial^2V}{\partial z^2} = 0$$ The boundary conditions are \begin{cases} V=0 \text{ when } y=0\\ V=0 \text{ when } y=a\\ V=0 \text{ when } z=0\\ V=0 \text{ when } z=b\\ V\to 0 \text{ when } x\to \infty\\ V=V_0(y,z) \text{ when } x=0 \end{cases} Let $$V(x,y,z)=X(x)Y(y)Z(z)$$ $$\frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2}+\frac{1}{Z}\frac{d^2Z}{dz^2} = 0$$ Then, $$\frac{1}{X}\frac{d^2X}{dx^2}=C_1, \frac{1}{Y}\frac{d^2Y}{dy^2}=C_2, \frac{1}{Z}\frac{d^2Z}{dz^2} = C_3$$ where \(C_1+C_2+C_3=0\) . As \(X\) should decay at infinity hence cannot be sinusoidal, \(C_1\) should be positive while \(C_2\) and \(C_3\) are negative. Let \(C_2=-k^2\) and \(C_3=-l^2\). So, \(C_1=k^2+l^2\) \begin{cases} \frac{d^2X}{dx^2}=(k^2+l^2)X\\ \frac{d^2Y}{dy^2}=-k^2Y\\ \frac{d^2Z}{dz^2}=-l^2Z \end{cases} The solutions are \begin{align} X(x) &= Ae^{\sqrt{k^2+l^2}x}+Be^{-\sqrt{k^2+l^2}x}\\ Y(y) &= C\sin ky + D\cos ky\\ Z(z) &= E\sin lz + F\cos lz \end{align} By the boundary conditions, \(A=D=F=0\) and \(k=n\pi/a\) and \(l=m\pi/b\), where \(n\) and \(m\) are positive integers. $$V(x,y,z) = Ce^{-\pi\sqrt{(n/a)^2+(m/b)^2}x}\sin(n\pi y/a)\sin(m\pi z/b)$$ The general solution is $$V(x,y,z) = \sum^{\infty}_{n=1}\sum^{\infty}_{m=1}C_{n,m}e^{-\pi\sqrt{(n/a)^2+(m/b)^2}x}\sin(n\pi y/a)\sin(m\pi z/b)$$ At \(x=0\), $$\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}C_{n,m}\sin(n\pi y/a)\sin(m\pi z/b)=V_0(y,z)$$ Let \(m'\) and \(n'\) be some arbitrary positive integers. Multiply by \(\sin(n'\pi y/a)\sin(m'\pi z/b)\) on both sides then integrate $$\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}C_{n,m}\int^a_0\sin(\frac{n\pi y}{a})\sin(\frac{n'\pi y}{a})dy\int^b_0\sin(\frac{m\pi z}{b})\sin(\frac{m'\pi z}{b})dz \\= \int^a_0\int^b_0V_0(y,z)\sin(\frac{n'\pi y}{a})\sin(\frac{m'\pi z}{b})dydz$$ The integral on the left side is $$\int^a_0\sin(n\pi y/a)\sin(n'\pi y/a)dy = \begin{cases} 0, \text{ if }n'\neq n,\\ a/2, \text{ if }n'=n \end{cases} $$ So, $$C_{n,m}\left(\frac{a}{2}\right)\left(\frac{b}{2}\right)=\int^a_0\int^b_0V_0(y,z)\sin(\frac{n\pi y}{a})\sin(\frac{m\pi z}{b})dydz$$ \begin{align} C_{n,m} &= \frac{4V_0}{ab}\int^a_0\sin(\frac{n\pi y}{a})dy\int^b_0\sin(\frac{m\pi z}{b})dz \\ &= \begin{cases} 0, \text{ if }n\text{ or }m\text{ is even,}\\ \frac{16V_0}{\pi^2nm},\text{ if }n\text{ and }m\text{ are odd,} \end{cases} \end{align} Therefore, $$V(x,y,z)=\frac{16V_0}{\pi^2}\sum^{\infty}_{n,m=1,3,5...}\frac{1}{nm}e^{-\pi\sqrt{(n/a)^2+(m/b)^2}}\sin(\frac{n\pi y}{a})\sin(\frac{m\pi z}{b})$$

Checkpoint (Griffiths Third Edition Q3.15)

A cubical box (sides of length \(a\)) consists of five metal plates, which are welded together and grounded. The top is made of a separate sheet of metal, insulated from the others, and held at a constant potential \(V_0\). Find the potential inside the box.

The boundary conditions are \begin{cases} V=0, x=0\\ V=0, x=a\\ V=0, y=0\\ V=0, y=a\\ V=0, z=0\\ V=V_0, z=a \end{cases} Let $$V(x,y,z)=X(x)Y(y)Z(z)$$ $$\frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2}+\frac{1}{Z}\frac{d^2Z}{dz^2} = 0$$ Then, $$\frac{1}{X}\frac{d^2X}{dx^2}=C_1,\text{ } \frac{1}{Y}\frac{d^2Y}{dy^2}=C_2,\text{ } \frac{1}{Z}\frac{d^2Z}{dz^2} = C_3$$ where \(C_1+C_2+C_3=0\) . We want \(z\) to be exponential and \(x\) and \(y\) to be sinusoidal Let \(C_1=-k^2\) and \(C_2=-l^2\). So, \(C_3=k^2+l^2\) \begin{cases} \frac{d^2X}{dx^2}=-k^2X\\ \frac{d^2Y}{dy^2}=-l^2Y\\ \frac{d^2Z}{dz^2}=(k^2+l^2)Z \end{cases} The solutions are \begin{cases} X(x) = A\sin(kx)+B\cos(kx)\\ Y(y) = C\sin(ly)+D\cos(ly)\\ Z(z) = Ee^{\sqrt{k^2+l^2}z}+Ge^{-\sqrt{k^2+l^2}z} \end{cases} By the boundary conditions, \(B=D=0\) and \(E+G=0\) and \(k=n\pi/a\) and \(l=m\pi/b\), where \(n\) and \(m\) are positive integers. So, $$Z(z) = 2E\sinh(\pi\sqrt{n^2+m^2}z/a)$$ Then, the general solution is $$V(x,y,z)=\sum^{\infty}_0\sum^{\infty}_0C_{n,m}\sin\left(\frac{n\pi x}{a}\right)\sin\left(\frac{m\pi y}{a}\right)\sinh(\frac{\pi\sqrt{n^2+m^2}z}{a})$$ At \(z=a\), $$V_0 = \sum^{\infty}_0\sum^{\infty}_0C_{n,m}\sin\left(\frac{n\pi x}{a}\right)\sin\left(\frac{m\pi y}{a}\right)\sinh(\pi\sqrt{n^2+m^2})$$ Let \(m'\) and \(n'\) be some arbitrary positive integers. Multiply by \(\sin(n'\pi y/a)\sin(m'\pi x/a)\) on both sides then integrate $$\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}C_{n,m}\sinh(\frac{\pi\sqrt{n^2+m^2}z}{a}) \\\int^a_0\sin(\frac{n\pi y}{a})\sin(\frac{n'\pi y}{a})dy\int^b_0\sin(\frac{m\pi x}{a})\sin(\frac{m'\pi x}{a})ds \\= \int^a_0\int^b_0V_0(y,z)\sin(\frac{n'\pi y}{a})\sin(\frac{m'\pi x}{a})dydx$$ The integral on the left side is $$\int^a_0\sin(n\pi y/a)\sin(n'\pi y/a)dy = \begin{cases} 0, \text{ if }n'\neq n,\\ a/2, \text{ if }n'=n \end{cases} $$ So, $$C_{n,m}\sinh(\frac{\pi\sqrt{n^2+m^2}z}{a})\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)=\int^a_0\int^b_0V_0(y,z)\sin(\frac{n\pi y}{a})\sin(\frac{m\pi x}{a})dydx$$ \begin{align} C_{n,m}\sinh(\frac{\pi\sqrt{n^2+m^2}z}{a}) &= \frac{4V_0}{a^2}\int^a_0\sin(\frac{n\pi y}{a})dy\int^b_0\sin(\frac{m\pi x}{a})dx \\ &= \begin{cases} 0, \text{ if }n\text{ or }m\text{ is even,}\\ \frac{16V_0}{\pi^2nm},\text{ if }n\text{ and }m\text{ are odd,} \end{cases} \end{align} Therefore, $$V(x,y,z) = \frac{16V_0}{\pi^2}\sum_{n=1,3,5...}\sum_{m=1,3,5...}\frac{1}{nm}\sin(\frac{n\pi x}{a})\sin(\frac{m\pi y}{a})\frac{\sinh(\pi\sqrt{n^2+m^2}z/a)}{\sinh(\pi\sqrt{n^2+m^2})}$$

Legendre Polynomials

For spherical coordinate, the Laplace's equations become $$\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial V}{\partial r}) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}(\sin\theta\frac{\partial V}{\partial \theta}) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2V}{\partial \phi^2} = 0$$ If the problem is azimuthal symmetry, \(V\) is independent of \(\phi\), so $$\frac{\partial}{\partial r}(r^2\frac{\partial V}{\partial r}) + \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}(\sin\theta\frac{\partial V}{\partial \theta}) = 0$$ Using separation of variables, $$V(r, \theta) = R(r)\Theta(\theta)$$ The general solution of \(R(r)\) is $$R(r) = Ar_l + \frac{B}{r^{l+1}}$$ and \(\Theta(\theta)\) is $$\Theta(\theta) = P_l(\cos\theta)$$ where \(P_l(x)\) is the Rodrigues formula $$P_l(x) = \frac{1}{2_ll!}\frac{d^l}{dx^l}(x^2 -1 )^l$$ So, $$V(r, \theta) = (Ar_l + \frac{B}{r^{l+1}})P_l(\cos\theta)$$

Example (Griffiths Third Edition Ex 3.9)

A specified charge density \(\sigma_0(\theta)\) is glued over the surface of a spherical shell of radius \(R\). Find the resulting potential inside and outside the sphere.

Solution: For \(r \leq R\), the \(B_l\) terms must be 0 so that the potential will not diverge at origin $$V(r,\theta) = \sum^{\infty}_{l=0}A_lr^lP_l(\cos\theta)$$ For \(r > R\), the \(A_l\) terms must be 0 so that the potential will not diverge at infinity $$V(r,\theta) = \sum^{\infty}_{l=0}\frac{B_l}{r^{l+1}}P_l(\cos\theta)$$ At \(r=R\), $$\sum^{\infty}_{l=0}A_lR^lP_l(\cos\theta) = \sum^{\infty}_{l=0}\frac{B_l}{R^{l+1}}P_l(\cos\theta)$$ As for each coefficients of the Legendre polynomials are equal, we have $$B_l=A_lR^{2l+1}$$ There is a discontinuity in radial derivative of \(V\) at the surface, i.e. \(r=R\) $$\left(\frac{\partial V_{\text{out}}}{\partial r}-\frac{\partial V_{\text{in}}}{\partial r}\right) = -\frac{1}{\epsilon_0}\sigma_0(\theta)$$ $$-\sum^{\infty}_{l=0}(l+1)\frac{B_l}{R^{l+2}}P_l(\cos\theta)-\sum^{\infty}_{l=0}lA_lR^{l-1}P_l(\cos\theta)=-\frac{1}{\epsilon_0}\sigma_0(\theta)$$ $$\sum^{\infty}_{l=0}(2l+1)A_lR^{l-1}P_l(\cos\theta)=\frac{1}{\epsilon_0}\sigma_0(\theta)$$ As the Legendre polynomials are orthogonal functions \begin{align} \int^1_{-1}P_l(x)P_{l'}(x)dx &= \int^{\pi}_0P_l(\cos\theta)P_{l'}(\cos\theta)\sin\theta d\theta \\ &= \begin{cases} 0, \text{ if }l'\neq l,\\ \frac{2}{2l+1}, \text{ if }l'=l \end{cases} \end{align} So, the coefficients are given by \begin{align} \sum^{\infty}_{l=0}(2l+1)A_lR^{l-1} \int^{\pi}_0P_l(\cos\theta)P_{l'}(\cos\theta)\sin\theta d\theta &=\frac{1}{\epsilon_0}\int^{\pi}_0\sigma_0(\theta)P_{l'}(\cos\theta)\sin\theta d\theta\\ (2l+1)A_lR^{l-1}\frac{2}{2l+1} &= \frac{1}{\epsilon_0}\int^{\pi}_0\sigma_0(\theta)P_{l}(\cos\theta)\sin\theta d\theta\\ A_l&=\frac{1}{2\epsilon_0R^{l-1}}\int^{\pi}_0\sigma_0(\theta)P_l(\cos\theta)\sin\theta d\theta \end{align}

Checkpoint (Griffiths Third Edition Q3.18)

The potential at the surface of a sphere (radius \(R\)) is given by $$V_0=k\cos 3\theta$$ where \(k\) is a constant. Find the potential inside and outside the sphere. (Assume there's no charge inside or outside the sphere.)

The potential at the surface can be express as the \(P_1\) and \(P_3\) of the Legendre polynomials \begin{align} V_0(\theta) &= k\cos(3\theta)\\ &= k(4\cos^3\theta-3\cos\theta) \\ &= k[\alpha P_3(\cos\theta) + \beta P_1(\cos\theta)] \end{align} The coefficients \(\alpha\) and \(\beta\) are \begin{align} 4\cos^3\theta - 3 \cos\theta &= \alpha\left[\frac{1}{2}(5\cos^3\theta-3\cos\theta)\right] + \beta\cos\theta \\ &= \frac{5\alpha}{2}\cos^3\theta+\left(\beta - \frac{3}{2}\alpha\right)\cos\theta \\ \end{align} So, \(\alpha = \frac{8}{5}\) and \(\beta=-\frac{3}{5}\) Then, $$V_0(\theta)=\frac{k}{5}[8P_3(\cos\theta)-3P_1(\cos\theta)]$$ The general solution of the potential is $$V(r,\theta) = \begin{cases} \sum^{\infty}_{l=0}A_lr^lP_l(\cos\theta),\text{ for }r\leq R\\ \sum^{\infty}_{l=0}\frac{B_l}{r^{l+1}}P_l(\cos\theta),\text{ for }r\geq R \end{cases} $$ Using the orthogonal properties, \begin{align} A_l &= \frac{2l+1}{2R^l}\int^{\pi}_0V_0(\theta)P_l(\cos\theta)\sin\theta d\theta\\ &= \frac{2l+1}{2R^l}\frac{k}{5}\left[8\int^{\pi}_0P_3(\cos\theta)P_l(\cos\theta)\sin\theta d\theta - 3\int^{\pi}_0P_1(\cos\theta)P_l(\cos\theta)\sin\theta d\theta\right]\\ &= \frac{k}{5}\frac{2l+1}{2R^l}\left[\frac{8(2)}{2l+1}\delta_{l3}-\frac{(3)(2)}{2l+1}\delta_{l1}\right]\\ &= \frac{kl}{5R^l}(8\delta_{l3}-3\delta_{l1})\\ &= \begin{cases} 8k/5R^3, \text{ if }l=3\\ -3k/5R, \text{ if }l=1\\ 0, \text{ otherwise} \end{cases} \end{align} So, for \(r\leq R\), \begin{align} V(r,\theta) &= -\frac{3k}{5R}rP_1(\cos\theta) + \frac{8k}{5R^3}r^3P_3(\cos\theta)\\ &= \frac{k}{5}\left[8\left(\frac{r}{R}\right)^3P_3(\cos\theta)-3\left(\frac{r}{R}\right)P_1(\cos\theta)\right]\\ &= \frac{k}{5}\left[8\left(\frac{r}{R}\right)^3\frac{1}{2}[5\cos^3\theta-3\cos\theta]-3\left(\frac{r}{R}\right)\cos\theta\right]\\ &= \frac{kr}{5R}\cos\theta\left[4\left(\frac{r}{R}\right)^2(5\cos^2\theta-3)-3\right] \end{align} for\(r\leq R\), as \(B_l=A_lR^{2l+1}\) $$B_l= \begin{align} 8kR^4/5, \text{ if }l=3\\ -3kR^2/5, \text{ if }l=1\\ 0, \text{ otherwise} \end{align} $$ So, \begin{align} V(r,\theta) &= \frac{-3kR^2}{5}\frac{1}{r^2}P_1(\cos\theta) + \frac{8kR^4}{5}\frac{1}{r^4}P_3(\cos\theta)\\ &= \frac{k}{5}\left[8\left(\frac{R}{r}\right)^4P_3(\cos\theta)-3\left(\frac{R}{r}\right)^2P_1(\cos\theta)\right]\\ &=\frac{k}{5}\left(\frac{R}{r}\right)^2\cos\theta\left[4\left(\frac{R}{r}\right)^2(5\cos^2\theta-3)-3\right] \end{align}

Multipole Expansion

At a point \(\vec{r}\) from the origin, the potential set up by a charge distribution is given by $$V(\textbf{r}) = \frac{1}{4\pi\epsilon}\int\frac{1}{d}\rho(\textbf{r}')d\tau',$$ where \(\vec{r'}\) is the displacement vector from origin to the \(dq\) charge element (see figure). Express \(d\) by cosine law, $$d^2 = r^2 + (r')^2 -2rr'\cos\theta = r^2(1 + (\frac{r'}{r})^2 - 2(\frac{r'}{r}\cos\theta))$$ By Taylor expansion, \begin{align} \frac{1}{d} &= \frac{1}{r}[1 - \frac{1}{2}(\frac{r'}{r}(\frac{r'}{r} - 2\cos\theta) + \frac{3}{8}(\frac{r'}{r})^2(\frac{r'}{r} - 2\cos\theta)^2 +...] \\ &= \frac{1}{r}[1 + (\frac{r'}{r}\cos\theta + (\frac{r'}{r})^2(3\cos^2\theta - 1)/2 +...] \\ &= \frac{1}{r}\sum_{n=0}^{\infty}P_n(\cos\theta) \\ \end{align} So, $$V(\textbf{r}) = \frac{1}{4\pi\epsilon}\sum_{n=0}^{\infty}\frac{1}{r^{n+1}}\int(r')^nP_n(\cos\theta)\rho(\textbf{r}')d\tau'.$$ Expand to see the first few terms, $$V(\textbf{r}) = \frac{1}{4\pi\epsilon_0}\left[\frac{1}{r}\int\rho(\textbf{r}')d\tau'+\frac{1}{r^2}\int r'\cos\theta\rho(\textbf{r}')d\tau'+...\right].$$

The nth term is inversely proportional to the n power of \(r\). So, when \(r\) is far away while keeping \(r'\) not that far away from the origin, we may neglect the higher order terms (as they are less significant) to estimate the electric potential.

When so far away that the second or higher order terms are negligible,

$$V(\textbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{1}{r}\int\rho(\textbf{r}')d\tau' = \frac{1}{4\pi\epsilon_0}\frac{Q}{r},$$ that is to say, we treat it like a point charge.