2024-01-18 Poisson and Laplace Equation's

Poisson’s Equation and Laplace’s Equation

• Electric field can be written as the gradient of a scalar potential

$$\mathbf{E}=-\nabla V$$

• What about divergence and curl look like for V?
• The divergence of $\mathbf{E}$ is the Laplacian of $V$. Gauss’s law states:

$${\nabla }^{2}V=-\frac{\rho }{{ϵ}_0}$$

• This is Poisson’s Equation
• Where $\rho =0$ this reduces to Laplace’s equation

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

The Potential of a Localized Charge Distribution

• We found $V$ in terms of $\mathbf{E}$, now find $\mathbf{E}$ from $V$
• Invert Poisson’s Equation!
• For a point charge

$$V(r)=\frac{1}{4\pi {ϵ}_0}\frac{q}{r}$$

• For a continuous distribution

$$V(r)=\frac{1}{4\pi {ϵ}_0}\int \frac{1}{r}dq$$

• For a volume charge

$$V(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \frac{\rho ({\mathbf{r}}^{\prime })}{r}d{\tau }^{\prime }$$

• Note this has no unit vector
• Line charge

$$V=\frac{1}{4\pi {ϵ}_0}\int \frac{\lambda ({\mathbf{r}}^{\prime })}{r}d{l}^{\prime }$$

• Surface charge

$$V=\frac{1}{4\pi ϵ0}\int \frac{\sigma ({\mathbf{r}}^{\prime })}{r}d{a}^{\prime }$$

Boundary conditions

• Typically you have a source charge distribution, and have to calculate $\mathbf{E}$ from it
• Without symmetry, you have to calculate potential first
• Three fundamental quantities of electrostatics
  • $\rho$, volume charge density
  • $\mathbf{E}$, electric charge
  • $V$, potential
• Gauss’s Law for a pillbox

$${\oint }_{\mathcal{S}}\mathbf{E}\cdot da=\frac{1}{{ϵ}_0}{Q}_{enc}=\frac{1}{{ϵ}_0}\sigma A$$

• Where A is the surface area of the pillbox lid Cool diagram of how things interact

Lecture on Electric Potential

Useful math

• The curl of the gradient of the scalar function is 0

$$\vec{\nabla }\times (\vec{\nabla }F)=0$$

• The cross product of the electric field is 0 for electrostatics

$$\vec{\nabla }\times \vec{E}=0$$

• Divergence Theorem

$$\int (\vec{\nabla }\times \vec{E})d\vec{A}=\oint \vec{E}\cdot d\vec{s}$$

Electric potential

• Electric potential function

$$\vec{E}(\vec{r})=-\vec{\nabla }V(\vec{r})$$

• Electric field is a physical concept, while potential is a mathematical one.
• The electric potential function fulfills the electrostatic field condition

$$\vec{\nabla }\times \vec{E}=\vec{\nabla }\times (-\vec{\nabla }V(\vec{r}))=0$$

• Used for electrodynamics and magnetic
• We can use this to derive differential equations in electrodynamics!

$$-{\vec{\nabla }}_{\vec{r}}\frac{1}{\vec{r}}=\frac{r}{r^3}$$ $$\vec{E}(\vec{r})=\frac{1}{4\pi {ϵ}_0}\underset{v^{\prime }}{\int }\frac{\rho {\vec{r}}^{\prime }(\vec{r})}{{\vec{r}}^3}d{V}^{\prime }$$ $$\vec{E}(\vec{r})=\frac{1}{4\pi {ϵ}_0}\underset{V^{\prime }}{\int }\rho ({\vec{r}}^{\prime })\left(-{\vec{\nabla }}_{\vec{r}}\frac{1}{\vec{r}}\right)d{V}^{\prime }$$ $$\vec{E}(\vec{r})=-{\vec{\nabla }}_{\vec{r}}\left[\frac{1}{4\pi {ϵ}_0}\underset{V^{\prime }}{\int }\frac{\rho ({\vec{r}}^{\prime })}{\vec{r}}d{V}^{\prime }\right]=-\vec{\nabla }F(\vec{r})$$

General description of potential from electric charge density at r’

$$V(\vec{r})=\frac{1}{4\pi {ϵ}_0}\underset{V^{\prime }}{\int }\frac{\rho {\vec{r}}^{\prime }}{\vec{r}}d{V}^{\prime }$$

• See cool triangle diagram above! This shows up there Definition of electric potential from electric field

$$\vec{\nabla }\times \vec{E}=0\therefore \oint \vec{E}\cdot d\vec{s}=0\to \underset{{\vec{r}}_a}{\overset{{\vec{r}}_a}{\int }}\vec{E}\cdot d\vec{s}$$ $$V(\vec{r})=-\underset{\vec{r}a}{\overset{{\vec{r}}_b}{\int }}\vec{E}\cdot d\vec{s}=-[v({\vec{r}}_b)-v({\vec{r}}_a)]$$

Electrostatic potential energy (not potential). These are different things!

$$U=-\underset{{\vec{r}}_a}{\overset{{\vec{r}}_b}{\int }}{\vec{F}}_{c}d\vec{s}=-\underset{{\vec{r}}_a}{\overset{{\vec{r}}_b}{\int }}q\vec{E}\cdot d\vec{s}=-q[v({\vec{r}}_b-v({\vec{r}}_a))]$$

To get potential energy, integrate over the electric fields

Boundary Conditions

• Between materials with different electric properties
  • Ex. Plastic and metal, air and water
• Interface of materials with different electric properties
• Start with boundary between metal and vacuum

EX. Spherical Shell with surface charge density $\sigma =\frac{Q}{4\pi R^2}$

$$\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{{ϵ}_0}$$ $$E(r)=0:r<R$$

Electric field inside shell is 0

$$E(r)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}:r\ge R$$

See graph of this: description: Electric field is 0 until point R, where it jumps to a maximum value and decays quadratically?

$$E(r=R)-\frac{1}{4\pi {ϵ}_0}\frac{Q}{R^2}=\frac{4\pi R^2\sigma }{4\pi {ϵ}_0{R}^2}=\frac{\sigma }{{ϵ}_0}$$

For a metal surface (conducting surface)

• the electric field at surface is always perpendicular to surface

$$E_{\perp }=\frac{\sigma }{{ϵ}_0}$$

On a sphere, the electric field is perpendicular to the surface at any point. The parallel component is 0

$$E_{\parallel }=0$$ $$E(r)=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r^2}=\frac{\sigma R^2}{{ϵ}_0{r}^2}$$

The electric potential inside the sphere is a constant

$$\vec{E}=-\vec{\nabla }V\therefore V=\text{constant}$$ $$V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$$

Plot for electric potential:

• inside the sphere it is constant, outside sphere decays at rate $\frac{1}{r}$

$$V_{inside}=V_{outside}:R=r$$ $$\vec{\nabla }V=-\vec{E}$$

The plot of the gradient of V 0 until the sphere boarder, then jumps up and decays

Boundary conditions at electrically charged surfaces

Discontinuous:

$$E_{\perp \text{outside}}-E_{\perp \text{inside}}=\frac{\sigma }{{ϵ}_0}$$ $$\vec{\nabla }{V}_{\text{outside}}-\vec{\nabla }{V}_{inside}⟹\frac{\sigma }{{ϵ}_0}$$

Continuous: see ppt

Poisson’s law and Laplace’s equations for electric potential

$$\vec{\nabla }\times \vec{E}=0$$ $$\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{ϵ}_0}$$ $$\vec{E}=-\vec{\nabla }V$$ $$\vec{\nabla }\cdot \vec{E}=(\vec{\nabla }\cdot (-\vec{\nabla }V))=-{\vec{\nabla }}^{2}V=\frac{\rho }{{ϵ}_0}$$ $${\vec{\nabla }}^2=\vec{\nabla }\cdot \vec{\nabla }=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}$$

Poisson’s Equation Laplace’s equation

In class activities

Required Equations 2.30

$$V(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \frac{\sigma }{r}d{a}^{\prime }$$

2.26

$$V(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\frac{q}{r}$$

For part a. (2 point charges)

$$r=\sqrt{{\left(\frac{d}{2}\right)}^2+z^2}$$ $$V_1=\frac{1}{4\pi {ϵ}_0}\frac{q}{\sqrt{{\left(\frac{d}{2}\right)}^2+z^2}}$$ $$V_{\text{total}}=V_1+V_2$$ $$V_{\text{total}}=\frac{1}{4\pi {ϵ}_0}\frac{2q}{r}$$

Taking the derivative wrt x cancels out, and wrt y gives the same result in Ex 2.1 For part b.

$$V(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \frac{1}{r}dq$$ $$V(\mathbf{r})=k\underset{-L}{\overset{L}{\int }}\frac{\lambda }{\sqrt{{\left(L\right)}^2+z^2}}dL$$ $$V(r)=\lambda \text{arcsinh}\left(\frac{L}{z}\right)$$

the derivative of this is equivalent to the result in Ex 2.2

For part c. Uniform surface charge

$$V(\mathbf{r})=k\int \frac{\sigma }{r}d{a}^{\prime }$$ $$V(\mathbf{r})=k\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}\frac{\sigma }{\sqrt{r^2+z^2}}rdrd\theta$$ $$V(r)=k\underset{0}{\overset{2\pi }{\int }}\sigma \left(\sqrt{z^2+R^2}-z\right)d\theta$$ $$V(r)=k(2\pi \sigma \cdot \left(\sqrt{z^2+R^2}-z\right))$$ $$V(r)=\frac{2\pi \sigma \cdot \left(\sqrt{z^2+R^2}-z\right)}{4\pi {ϵ}_0}$$