2024-01-25 Laplace Via Direct Integration

Electrostatics in a nutshell

• Field lines start on positive, end on negative
• Divergence of electric field is prop. to charge density
• Electrostatic field is gradient of scalar function (potential)

General description of electrostatic field and potential

• Vector ${\vec{r}}^{\prime }$ points to charge
• Vector $\vec{r}$ points to a test point

Boundary conditions

Perpendicular boundary charges are discontinuous. Parallel boundary conditions are continuous.

• Charged hollow sphere example.
  • Inside the sphere, $E=0$
    • Can be determined with Gauss’s law
  • Outside the sphere, similar to point charge sitting at the origin
  • Discontinuity between inside and outside sphere

Methods of solving Laplace’s equation

Direct integration of the 1-d Laplace and Poisson equations

Hollow Sphere example: Given a hollow sphere with total charge $Q$ This gives us $\sigma =\frac{Q}{4\pi R^2}$ Find potential by solving Laplace’s equation

$${\vec{\nabla }}^{2}V=0$$

We will use spherical coordinates. Need to find Laplacian operator for spherical coordinates from the textbook

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

Potential is equal around the sphere in both azimuthal and polar angles

$$\frac{\partial V}{\partial \theta }=0,\frac{\partial V}{\partial \varphi }=0$$ $$\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial V}{\partial r}\right)=0$$ $$\frac{\partial }{\partial r}\left(r^2\frac{\partial V}{\partial r}\right)=0⟹r^2\frac{\partial V}{\partial r}=c_1$$ \implies \frac{\partial V}{\partial r} = \frac{c_{1}}{r^{2}} \implies V = \int \frac{c_{1}}{r^{2}} \, dr = -\frac{c_{1}}{r} + c_{2} $$We have 2 integration constants, $c_{1} \text{ and } c_{2}$. These are given by boundary conditions.

V_{in} =(r=R) = V_{out} (r=R)

-\frac{c_{1}}{R} + c_{2}= \frac{1}{4\pi\epsilon_{0}} \frac{Q}{R} + c_{2}

c_{2} = 0, V_{in}(t) =\frac{Q}{4\pi\epsilon_{0}R} = \text{ constant}

$$Inside:$$

\vec{E}_{in} = -\vec{\nabla}V = 0

$$TLDR:Iftheproblemisreducedtoa1-Dproblem,itcanbesolvedusingdirectintegration.Youcancheck(atleastforthehollowsphere)using[[Gaus{s}^{\prime }Law]].$$

\vec{E} = -\vec{\nabla}V

If the field $\vec{E}$ is 0, then the potential has to be **constant** (inside of hollow sphere). Can be solved for with:

V_{in} = const. = \frac{Q}{4\pi \epsilon_{0}R}

$$Thepotentialoutsidethehollowsphereis$$

V_{out} = \frac{1}{4\pi\epsilon_{0}} \frac{Q}{r}

$$Youcanprovethiswithlaplac{e}^{\prime }sequation.Theyagree!Boundaryconditionsisthereforeasfollows:$$

V_{in} = V_{out} ; \ \ r=R

# Methods of solving [[Poisson's Equation]] Start with general methods of solving for the electric potential:

V(\vec{r}) = \int _{V’} \frac{\rho(\vec{r})1}{\mathscr{r}} , dV’

\vec{\nabla}^{2}{\vec{r}} V(\vec{r}) = -\frac{\rho}{\epsilon{0}} \implies \nabla^{2}{r} V(\vec{r}) = \frac{1}{4\pi\epsilon{0}} \int \rho(\vec{r})\vec{\nabla}^{2} \frac{1}{|\vec{r}- \vec{r}’|} , dV’

$$RecallHomework1,question13$$

-\vec{\nabla} \frac{1}{|\vec{r}-\vec{r}’|} = \frac{\vec{r}-\vec{r}’}{|\vec{r}-\vec{r}’|^{3}}

\vec{\nabla}{\vec{r}}\left( \vec{\nabla}{\vec{r}} \frac{1}{\vec{r} -\vec{r}’} \right)

\vec{\nabla} _{\vec{r}} \left( - \frac{\vec{r}-\vec{r}’}{|\vec{r}-\vec{r}’|} \right)

$$UseDeltafunction$$

4\pi\delta(\vec{r}-\vec{r}’)

### General comments on Laplace and Poisson Solving the Laplace and Poisson equations is solving a 2nd order diffeq 2nd order diffeq's are elliptic differential equations and are unique Trust the math All extrema in Laplace equation occurs at the boundary This makes it unique ## In class problems 3.3 Find the general solution to Laplace’s equation in spherical coordinates, for the case where V depends only on r. Do the same for cylindrical coordinates, assuming V depends only on s. Spherical case (see above)

-\frac{c_{1}}{R} + c_{2}

$$Cylindricalcase$$

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

\frac{d^{2}V}{d\theta^{2}} = 0, \ \frac{d^{2}V}{dz^{2}} =0

\frac{1}{r} \frac{d}{dr}\left( r \frac{dV}{dr} \right)

r \frac{dV}{dr} = c_{1}

\frac{dV}{dr} = \frac{c_{1}}{r}

dV = \frac{c_{1}}{r}dr

V = \int \frac{c_{1}}{r} , dr

V = c_{1}\ln(r) + c_{2}

$$2.53a.[[Poisso{n}^{\prime }sEquation]]$$

\vec{\nabla}^{2} V = -\frac{\rho}{\epsilon_{0}}

\nabla^2 V = \frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} + \frac{\partial^2 V}{\partial z^2} = \frac{\rho}{\epsilon_{0}}

-\frac{\rho}{\epsilon_{0}} = \frac{\partial^2 V}{\partial x^2}

$$b.$$

T=qV = \frac{1}{2} mv^{2}

\sqrt{ \frac{2qV(x)}{m}}=v

$$c.$$

I = \rho v

\frac{I}{\rho} = v

$$d.$$

-\frac{\rho}{\epsilon_{0}} = \frac{\partial^2 V}{\partial x^2}

\sqrt{ \frac{2qV(x)}{m}}=v = \frac{aI}{\rho}

\frac{aI\sqrt{ \frac{2qV(x)}{m}}}{v}=\rho

-\frac{aI\sqrt{ \frac{2qV(x)}{m}}}{v\epsilon_{0}} = \frac{\partial^2 V}{\partial x^2}

\beta \sqrt{ V(x) } = \frac{\partial^2 V}{\partial x^2}

\beta V^{1/2} =\frac{\partial^2 V}{\partial x^2}