2024-01-11 Fundamental Electrostatics

Coulomb’s Law

• The force on a point charge $Q$ from another point charge $q$ at rest at a distance $r$ away is given by Coulomb’s Law

$$F=\frac{1}{4\pi {ϵ}_0}\frac{qQ}{r^2}\hat{r}$$

• $e_0$ is the permittivity of free space, ${ϵ}_0=8.85E-12$

The Electric Field

• Given several point charges the total force on another point charge is the summation of forces from the individual point charges

$$\mathbf{F}={\mathbf{F}}_1+{\mathbf{F}}_2+\cdots =\frac{1}{4\pi {ϵ}_0}\left(\frac{q_{1}Q}{r_1^2}\widehat{r_1}+\frac{q_{1}Q}{r_2^2}\widehat{r_2}+\dots \right)$$

• This sucks. Lets use the electric field instead.
• We define the electric field as:

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\sum _{i=1}^n\frac{q_i}{r_i^2}{\hat{r}}_i$$

• And the force from the electric field on a point charge Q is given by

$$\mathbf{F}=Q\mathbf{E}$$

• The electric field is a function of position.
• It is a vector quantity.

Continuous Charge Distributions

• Generalizing electric field to a continuous distribution over a region

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \frac{1}{r^2}dq$$

• Common forms include line charges, surfaces charges, and volume charges
• Volume charge distribution is also sometimes referred to as Coulomb’s law.

$$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \frac{\rho ({\mathbf{r}}^{\prime })}{r^2}\hat{r}d{\tau }^{\prime }$$

The Divergence of $\mathbf{E}$

• Calculating the divergence of $\mathbf{E}$ from $\mathbf{E}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}\int \frac{\rho ({\mathbf{r}}^{\prime })}{r^2}\hat{r}d{\tau }^{\prime }$ gives us the following

$$\nabla \cdot \mathbf{E}=\frac{1}{4\pi {ϵ}_0}\int \nabla \cdot \left(\frac{\hat{r}}{r^2}\right)\rho (r^{\prime })d{\tau }^{\prime }$$

• The generic solution of a vector field is given by

$$\nabla \cdot \left(\frac{\hat{r}}{r^2}\right)=4\pi {\delta }^3(r)$$

• Where $\delta$ is the delta function which is 0 everywhere except for 0, where its $\infty$
• This can derive the differential form of Gauss’s law

$$\nabla \cdot \mathbf{E}=\frac{1}{{ϵ}_0}\rho (r)$$

• To convert from the differential form to the integral form we can use the following

$${\int }_v\nabla \cdot \mathbf{E}d\tau ={\oint }_{\mathcal{S}}\mathbf{E}\cdot d\mathbf{a}=\frac{1}{{ϵ}_0}{\int }_{\mathcal{S}}\rho d\tau =\frac{1}{{ϵ}_0}{Q}_{enc}$$

The Curl of E

• The integral around a closed path is zero

$$\oint \mathbf{E}\cdot d\mathbf{l}=0$$

• Applying Stoke’s theorem

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

• These two equations hold true for any static charge distribution whatever.

Lecture 1/11 Electrostatic fields

• Maxwell’s equations in differential/integral forms
  • Maxwells Equations in Integral form
    • Gauss’ Law
    • Gauss’ Law for magnetic field
    • Faraday’s law
    • Ampere-Maxwell Law
• Fundamental definitions of Electrostatic Fields
• Generalized description of electric field of point charges and charge distributions
• Useful math

Why do we bother with differential forms?

• Derivatives are easy
• Integrals are hard
• Makes some computations possible

Vector Calculus:

• Divergence Theorem
• Stokes Theorem aka Curl Theorem
• These have been proven correct.

Maxwell Equations converted from Integral to Differential Form

$$\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{{ϵ}_0}=\frac{1}{{ϵ}_0}\int \rho dV$$

Use Divergence Theory to convert Surface integral to a volume integral over the divergence:

$$\int \vec{\nabla }\cdot \vec{E}dV=\frac{1}{{ϵ}_0}\int \rho dv$$ $$\int \left[\vec{\nabla }\cdot \vec{E}-\frac{\rho }{{ϵ}_0}\right]dV=0$$

Therefore, to make the R.H.S = 0, the divergence of the electric field has to be equal to $\frac{\rho }{{ϵ}_0}$

$$\vec{\nabla }\cdot \vec{E}-\frac{\rho }{{ϵ}_0}=0⟹\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{ϵ}_0}$$

For the Magnetic Gauss law

$$\oint \vec{B}\cdot d\vec{A}=0=\int (\vec{\nabla }\cdot \vec{B})dV⟹\vec{\nabla }\cdot \vec{B}=0$$

For Faraday’s Law using Stokes Theorem

$$\oint \vec{E}\cdot d\vec{s}=-\frac{d{\Phi }_B}{dt}=\frac{d}{dt}\int \vec{B}\cdot d\vec{A}$$ $$\oint \vec{E}\cdot d\vec{s}=\int (\vec{\nabla }\times \vec{E})\cdot d\vec{A}=-\frac{d}{dt}\int \vec{B}\cdot d\vec{A}$$ $$\int (\vec{\nabla }\times \vec{E}+\frac{d\vec{B}}{dt})d\vec{A}=0$$ $$\therefore \vec{\nabla }\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$

For Ampere-Maxwell Law using Stokes Theorem

$$\oint \vec{B}\cdot d\vec{s}=\int (\vec{\nabla }\times \vec{B})\cdot d\vec{A}={\mu }_0\int \vec{j}d\vec{A}+{\mu }_0{ϵ}_0\frac{d}{dt}\int \vec{E}\cdot d\vec{A}$$ $$\int \left(\vec{\nabla }\times \vec{B}-{\mu }_0\vec{j}-{\mu }_0{ϵ}_0\frac{d\vec{E}}{dt}\right)d\vec{A}=0$$ $$\therefore \vec{\nabla }\times \vec{B}={\mu }_0\vec{j}+{\mu }_0{ϵ}_0\frac{d\vec{E}}{dt}$$

Electrodynamics vs Electrostatics and Magnetostatics

• Electrodynamics
  • Utilizes coupled differential equations
• Electrostatics and Magnetostatics
  • Nothing depends on time. (static)
  • The time derivative is 0
  \nabla \times \mathbf{E} = 0 $$ - Definition of electrostatic fields and potentials - Gauss' law and Faraday's law for static electric fields in differential form are the mathematical description of electric field lines

Gauss’s Law in Differential form

$$\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{ϵ}_0}=constant$$

• Field lines go outwards for positive charge, inwards for negative
• In an electrostatic field,

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

• Electrostatic fields do not form closed loop field lines
• The curl of the magnetic field $\ne$ 0
• An electrodynamic field does not follow this rule

Notes on charge, force and field

• Coulomb force attracts opposite forces and separates like charges
  • This is also one of the fundamental forces of nature
• Why use electric field?
  • Dont have to work with many particles
  • Given by $\vec{F}=q\vec{E}$

General and Special cases for point charge electric field

Special case where the point charge is the origin

$$\vec{E}(\vec{r})=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^2}\hat{r}=\frac{1}{4\pi {ϵ}_0}\frac{q}{r^3}\hat{r}$$

General case for all points

$$\vec{E}(\vec{r})=\frac{1}{4\pi {ϵ}_0}\frac{q}{|\vec{r}-{\vec{r}}^{\prime }{|}^2}=\frac{q}{4\pi {ϵ}_0}\vec{r}-\frac{{\vec{\mathrm{R}}}^{\prime }}{\vec{r}}$$

…

Electric Charge Distribution

• Use electric charge density to make the Gauss’ integral solvable

Useful Math Tricks

$$\frac{\vec{r}-{\vec{r}}^{\prime }}{|\vec{r}-{\vec{r}}^{{}^{prime}}|}=-\nabla$$

DELTA FUNCTION 0 for all values where x $\ne$ 0 $\infty$ for x=0

$$\frac{dF}{dx}=0$$

because the slope has no slope Aka Derivative of step function

Example with divergence

$$\vec{\nabla }\cdot \vec{E}(\vec{r})$$ $$\vec{\nabla }{)}_{\vec{r}}\vec{E}={\vec{\nabla }}_{\vec{r}}\left(\frac{1}{4\pi {ϵ}_0}\int (\rho (\vec{F}))\frac{\vec{\tau }-{\vec{\tau }}^{\prime }}{|\vec{\tau }-{\vec{\tau }}^{\prime }{|}^3}d{V}^{\prime }\right)$$ $$k=\frac{1}{4\pi {ϵ}_0}$$ $$k{\int }_{V^{\prime }}\rho (F^{\prime })(\nabla )$$

finish later

Example with Curl

$$\vec{E}(\vec{\tau })=k{\int }_{v^{\prime }}\rho (F^{\prime })\frac{{\vec{\tau }}^{\prime }-{\vec{\tau }}^{\prime }}{|{\vec{\tau }}^{\prime }-{\vec{\tau }}^{\prime }{|}^3}d{v}^{\prime }$$ $$\vec{E}(\vec{\tau })=k\int \rho ({\vec{\tau }}^{\prime })\left[-{\vec{\nabla }}_F\frac{1}{|\vec{r}-{\vec{r}}^{\prime }|}\right]d{v}^{\prime }$$ $${\vec{\nabla }}_F\times \vec{E}(\vec{r})=k\int \rho ({\vec{r}}^{\prime })\left[{\vec{\nabla }}_F\times \left(-{\vec{\nabla }}_{\vec{r}}\frac{1}{|\vec{r}-{\vec{r}}^{\prime }|}\right)\right]dv=0$$

Self-Worked Problems

Find the electric field a distance z above the center of a circular loop of radius r (Fig. 2.9) that carries a uniform line charge λ.

• We can find the electric field in the z direction from many infinitesimal segments of the loop.

$$d=\sqrt{z^2+r^2}$$ $$\mathbf{E}(\mathbf{r})=\frac{1}{4\pi {ϵ}_0}{\int }_0^{2\pi }\frac{\lambda r}{{\left[z^2+\frac{d^2}{4}\right]}^{3/2}}d\theta$$ $$E=\frac{16\pi q\lambda r}{[d^2+4z^2{]}^{3/2}}$$