General Physics 26

Field, and Electric Field

A field is something that can be defined anywhere in space. Fields represent physical quantities.

• scalar fields like $T(x,y)$
• vector fields like $\vec{V}(x,y,z;t)$ and gravitation field$\vec{g}$

Physical fields, considered continuous and well-behaved, obey a simple rule and are created by sources. The rule and the sources decide the field.

gravitation field$\vec{g}$ is measured using a test body of small mass $m_0$( does not disturb the mass distribution!). Similarly the electric field is measured by a small test charge $q_0$.

gravitation field	electric field
$\vec{g} = \lim_{m_0\rightarrow0}\frac{\vec{F}}{m_0}$	$\vec{E} = \lim_{q_0\rightarrow0}\frac{\vec{F}}{q_0}$
$\vec{g}(\vec{r}) = -G\frac{M}{r^2}\hat{r}$

Point charge and groups

$$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_0q}{r^2}\hat{r}$$

$$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r}$$

Law of Superposition still works for fields( they’re vectors! ).

$$\vec{E} = \sum \vec{E}_i = \frac{1}{4\pi\epsilon_0}\sum \frac{q}{r_i^2}\hat{r}_i$$

Example: Dipole

Here $Q$ has no sign, just to indicate the magnitude of the charge. Calculate a point $(x,0)$ along $x$ axis, $E_x(x,0) = 0$ for the symmetry.

$$E_y(x,0) = -\frac{2}{4\pi\epsilon_0}\frac{Q}{r^2}\sin\theta = -\frac{1}{4\pi\epsilon_0}\frac{2Qa}{r^3}$$

where $r = \sqrt{a^2 + x^2}$. By Taylor Expansion at $\frac{a}{x} = 0$,

$$E_y(x,0) = -\frac{1}{4\pi\epsilon_0}\frac{2Qa}{(a^2 + x^2)^{\frac{3}{2}}} = \frac{-p}{4\pi\epsilon_0x^3}(1+\frac{a^2}{x^2})^{-\frac{3}{2}} = \frac{-p}{4\pi\epsilon_0x^3}\sum_{n = 0}^{\infty}\frac{(2n+1)!!}{n!(-2)^{2n+1}}(\frac{a}{x})^{2n}$$

$$E_y(x,0) \rightarrow -2\frac{1}{4\pi\epsilon_0}\frac{Qa}{x^3}(\frac{a}{x} \rightarrow 0)$$

The conclusions could be promoted to $xz$ plane. $E \propto \frac{1}{r^3}$ .

Calculate a point $(0,y)$ along $y$ axis, $E_x(0,y) = 0$​.

$$E_y(0,y) = \begin{cases}\frac{1}{4\pi\epsilon_0}(\frac{Q}{(a - y)^2} + \frac{-Q}{(-a-y)^2}) = \frac{Q}{4\pi\epsilon_0}\frac{4ay}{y^4(\frac{a^2}{y^2}-1)^2},\quad(\vert y\vert \geq a)\\ \frac{1}{4\pi\epsilon_0}(-\frac{Q}{(a - y)^2} + \frac{-Q}{(-a-y)^2}) = -\frac{Q}{4\pi\epsilon_0}\frac{2(a^2 + y^2)}{(a^2 - y^2)^2},\quad(\vert y\vert \leq a) \end{cases}$$

$$E_y(0,y) = \frac{Q}{4\pi\epsilon_0}\frac{4ay}{y^4(\frac{a^2}{y^2}-1)^2} \rightarrow 4\frac{1}{4\pi\epsilon_0}\frac{Qa}{y^3},(\frac{a}{y} \rightarrow 0)$$

The electric field magnitude is determined by $Qa$​ at a given point. Thus Dipole Vector $\vec{p}$​ is introduced.

$$p = 2Qa = Ql$$

points from the negative charge to the positive charge.

Continuous distribution

$$\vec{E} = \frac{1}{4\pi\epsilon_0}\int \frac{\rm d\it q}{r^2}\hat{r} = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(r)\rm d\it V}{r^2}\hat{r}$$

Infinite line of charge

A infinite line is of uniform linear charge density $\lambda$ . Calculate the electric field a point at distance $r$ from the line.

\rm d {\it q} = \lambda d{\it x}, \quad \rm d {\it x} = d ({\it r}\tan\theta) = {\it \frac{r}{\cos^\rm 2\theta}}d\theta

d{\it E} = \frac{1}{4\pi\epsilon_0}\frac{\rm d\it q}{\it r'^2} = \frac{1}{4\pi\epsilon_0}\frac{r\lambda\rm d\theta}{ (\cos\theta r')^\rm2} = \frac{1}{4\pi\epsilon_0}\frac{\lambda\rm d\theta}{\it r}

$$E_x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin\theta\rm d{\it E} = 0$$

$$E_y = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos\theta\rm d{\it E} = \frac{1}{4\pi\epsilon_0}\frac{2\lambda}{\it r}$$

The conclusions could also be promoted to plane that is vertical to the line. $E \propto \frac{1}{r}$ .

Uniformly charged Ring

A ring of radius $R$ is uniformly charged $Q$ . Calculate the electric field a point at distance $z$ from the center of the ring.

$$\rm d {\it q} = \lambda d{\it s}, \quad \rm d {\it s} = d ({\it R}\alpha) = {\it R}d\alpha$$

$${\rm d}{\it E} = \frac{1}{4\pi\epsilon_0}\frac{\rm d\it q}{\it r^2} = \frac{1}{4\pi\epsilon_0}\frac{R\lambda\rm d\alpha}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{R\lambda\rm d\alpha}{R^2 + z^2}$$

By the symmetry of ring, it is easy to calculate that $E_x = E_y = 0$

$$E_z = \int_0^{2\pi}\cos\theta\rm d{\it E} = \int_0^{2\pi}\frac{1}{4\pi\epsilon_0}\frac{\it Rz\lambda\rm d\alpha}{(R^2 + z^2)^{\frac{3}{2}}} = \frac{(2\pi{\it R}\lambda)\it z}{4\pi\epsilon_0(R^2 + z^2)^{\frac{3}{2}}} = \frac{\it Qz}{4\pi\epsilon_0(R^2 + z^2)^{\frac{3}{2}}}$$

There are two extreme conditions:

$$E_z = \begin{cases}\frac{\it Qz}{4\pi\epsilon_0R^3},\quad (\frac{R}{z} \rightarrow 0)\\ \quad 0, \quad\quad (z \rightarrow 0) \end{cases}$$

Uniformly charged Disk

A disk of radius $R$ is uniformly charged $Q$ . Calculate the electric field a point at distance $z$ from the center of the disk. Divide the disk into infinitesimal charged rings, which has been calculated above.

$$\rm d {\it q} = \omega\rm d\it A, \quad \rm d {\it A} = 2\pi\omega\sigma\rm d{\omega}$$

$${\rm d}{\it E} = {\rm d}E_z = \frac{z{\rm d}q}{4\pi\epsilon_0(\omega^2 + z^2)^{\frac{3}{2}}} = \frac{z\omega\sigma{\rm d}\omega}{2\epsilon_0(\omega^2 + z^2)^{\frac{3}{2}}}$$

$$E = E_z = \int_0^{R}{\rm d}E_z = \int_0^{R}\frac{z\omega\sigma{\rm d}\omega}{2\epsilon_0(\omega^2 + z^2)^{\frac{3}{2}}} = \frac{\sigma}{2\epsilon_0}(1 - \frac{z}{\sqrt{z^2 + R^2}})$$

Which is consistent with the Coulomb’s Force.

$$E = E_z = \frac{Q}{2\pi\epsilon_0R^2}(1 - \frac{z}{\sqrt{z^2 + R^2}})$$

There are two extreme conditions:

$$E = \begin{cases}\frac{\it Q}{4\pi\epsilon_0z^2},\quad (\frac{R}{z} \rightarrow 0)\\ \frac{\sigma}{2\epsilon_0}, \quad\quad (\frac{z}{R} \rightarrow 0) \end{cases}$$

which gives the electric field magnitudes of point charge and infinite sheet respectively.

Features

Electric field propagates at speed of light.

There’s never a net electric charges inside a conductor. Free charges always move to exactly cancel it out.

Visualization

Field	Vector maps	Field lines	Graphs
direction	direction of vectors	tangent of line	$x,y,z$ or other coordinate systems
magnitude	length of vectors	local density of lines	$E_x - x, E_y-y,E_z-z$,

Note Polar coordinate system

$$\begin{cases}x = r\cos\theta\\ y= r\sin\theta\end{cases} \begin{cases}r =\sqrt{x^2 + y^2}\\ \theta= \arcsin\frac{y}{x}\end{cases}$$

Thus Jacob transform is

$$J = \det(\begin{bmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{bmatrix}) = r$$

Similarly we have Spherical coordinate system.

Field lines

Dipole	Two like charges	Conductors

Example

Deflecting electrode system

Milikan oil drop Experiment

A Dipole in E fields

$$F_{net} = 0, \quad \tau_{net} = Fd\sin\theta = qEd\sin\theta = pE\sin\theta$$

Thus for a dipole $\vec{p}$,

$$\vec{\tau} = \vec{p} \times \vec{E}$$

The work done by external field from $\theta_0$ to $\theta$ is

$$W = \int {\rm d}w = \int_{\theta_0}^{\theta}\vec{\tau}\cdot{\rm d}\vec{\theta} = \int_{\theta_0}^{\theta}-pE\sin\theta{\rm d}\theta = pE(\cos\theta - \cos\theta_0) = -\Delta U= U_0 - U$$

Hence $U(\theta) = -pE\cos\theta = - \vec{p}\cdot\vec{E}$.

Nuclear Model of the Atom

Thomson model

Fig.	Discription
	positive charge: distributed more or less uniformly throughout the entire spherical volume of the atom. electrons: imbedded throughout the diffuse spherical of positive charge.

Rutherford atomic model

$\alpha$ particles scattering experiment

$$E = \begin{cases}\frac{Q}{4\pi\epsilon_0r^2},\quad (r > R)\\ \frac{Qr}{4\pi\epsilon_0R^3},\quad (r < R)\end{cases}$$

To calculate the upper boundary of the deflection,

$$E_{max} =\frac{Q}{4\pi\epsilon_0R^2} = \frac{79\times1.61\times10^{-19}}{4\times3.14\times8.854\times10^{-12}\times10^{-20}} = 1.14\times10^{13} N/C$$

The $\alpha$ particles are accelerated to $v = 1.70 \times 10^7 m/s$.

$$\theta_{max} = \arctan\frac{\Delta v}{v} = \arctan\frac{a\Delta t}{v} = \arctan\frac{2E_{max}qR}{mv^2} = 0.02\deg$$

Rutherford’s model

In the experiment, about $\frac{1}{10^4}$ particles were reversed, which is also the ratio of $r_{nuclear}$ and $r_{atom}$.