Solutions to Laplace’s equation

🥖 1 dimensional solution

💼 Case: two infinite parallel-plate conductors held at constant potential

$$ \frac{\text d^2 V}{\text d x^2}=0 \qquad \Rightarrow \qquad V(x)=ax+b $$

Properties: This trivial solution displays properties of solutions to Laplace’s equation

$V(x)=\frac{1}{2} [ V(x+c)+ V(x-c)]$ where $c$ does not have to be infinitesimally small
There can be no local maxima or minima in $V$ </aside>

Proof of statements in 3D

💼 Case: $V$ due to a point charge $q$

Charge $q$ positioned at $\vec r'=z \,\hat z$
Evaluate $V$ over spherical surface at the origin of radius $0<r<z$

Average electrical potential around the sphere

$$ V_{av}=\frac{1}{4\pi r^2}\oint V(\vec r) \, \text da= \frac{1}{4\pi r^2} \frac{q}{4\pi \epsilon_0} \oint \frac{\text da}{R} $$

Which gives us: (expand for details)

$$ V_{av}=\frac{q}{4\pi \epsilon_0} \frac{1}{z} $$

💎 Conclusion: In $\rm 3D$ where $\nabla^2 V=0$ then Condition $1$ is met, ie $V$ is given by average over symmetrically located neighboring points

💎 Conclusion: If there were to be a local minima, $V_o$, all points on the surrounding surface would have $V>V_o$, violating the 1st condition

Earnshaw’s Theorem: A charged particle cannot be held in a position of stable equilibrium by electrostatic forces alone

</aside>

🗒️ Note: Saddle points are allowed: if $\frac{\partial ^2 z}{\partial x^2}>0$ and $\frac{\partial ^2 z}{\partial y^2}<0$ then $\frac{\partial ^2 z}{\partial x^2}+\frac{\partial ^2 z}{\partial y^2}=0$ can be satisfied

Uniqueness theorem

🧔‍♀️ Theorem: The solution to Laplace’s equation in a volume $V$ is uniquely determined if on the boundary surface $S$ we specify either $V$ (Dirichlet) or $\vec \nabla V \cdot \hat n$ where $\hat n$ is normal to the surface $S$ (Newman)

💫 Proof: Suppose there were two different solutions $\nabla^2 V_1=0$ and $\nabla^2 V_2=0$ within a volume, satisfying $V_1=V_2$ on the boundary surface

Since $V_1$ and $V_2$ satisfy Laplace then $f=V_1 -V_2$ also satisfies Laplace’s equation

On the boundary $V_1=V_2$ so $f=0$

Since there can be no maxima or mina within the volume $f=0$ everywhere

Thus we proved that $V_1=V_2$ and the solution is unique

Example of Laplace’s equation in spherical polar coordinates

💼 Case: Conductive sphere with a constant uniform electric field $\vec E=E_0 \hat z$ applied

Setup:

Charges flow with

➕ Charges at the top

➖ charges at the bottom
Surface of the sphere is equipotential

We set:

$V=0$ at surface of sphere
Centre of sphere is at the origin

Boundary conditions:

Potential inside the sphere $=0$
At surface of sphere field must be radial

$$ \vec E(r,\theta,\phi)=\vec E(a,\theta) \,\hat r $$

Screenshot 2024-09-26 181345.png

$E$ inside sphere $=0$
At large distances from sphere

$$ \begin{aligned} V=-E_0 z= - E_0 r\cos\theta \end{aligned} $$
🗒️ Note: this is what happens to the electric field

Solving

Use Laplace equation inside the sphere as there is no charge so

$$ \nabla^2 V=\frac{1}{r^2} \frac{\partial }{\partial r} \left ( r^2 \frac{\partial V}{\partial r} \right )+\frac{1}{r^2\sin \theta}\frac{\partial}{\partial \theta}\left ( \sin \theta \frac{\partial V}{\partial \theta} \right )=0 $$

🗒️ Note: we do not consider $\phi$ due to the symmetry of the setup
We try a separable solution $V(r,\theta)=R(r) \Theta(\theta)$

$$ \underbrace{\frac{1}{R} \frac{\text d}{\text dr} \left ( r^2 \frac{\text dR}{\text dr} \right )}{=l(l+1)} + \underbrace{\frac{1}{\Theta \sin \theta } \frac{\text d}{\text d \theta} \left ( \sin \theta \frac{\text d \Theta}{\text d \theta} \right )}{=-l(l+1)} =0 $$