The Electric Field

  • Point charges (size of charged body $\ll$ distance away at which we measure),
  • that are not moving,
  • obey Coulomb's Law

We'll delay for later the question of moving charges.

Electrostatics


Coulomb's Law: The force on a test charge $Q$ from a motionless single point charge $q$ a distance $\rr$ away is experimentally found to be $$\myv F = \frac{1}{4\pi \epsilon_0} \frac{qQ}{\rr^2} \uv{\rr}.$$

The point charge $q$ is located at $\myv r'$, and the test charge $Q$ is at $\myv r$, so $$\myv \rr = \myv r - \myv r '.$$

The constant in Coulomb's law is called the permittivity of free space, and in the SI system of units that we'll be using... $$\epsilon_0 = 8.85 \times 10^{-12} \frac{{\text C}^2}{{\text N}\cdot{\text m}^2}.$$

Multiple source charges

We live in a universe such that these electric forces obey a linear superposition principal: The total force acting on a test charge due to several test charges is a simple vector sum of the forces due to each isolated test charge. $$\begineq\myv F_{\text{total}} &= \myv F_1 +\myv F_2 + ...\\ &= \frac{q_1Q}{4\pi \epsilon_0 \rr_1^2}\uv \rr_1 + \frac{q_2Q}{4\pi \epsilon_0 \rr_2^2}\uv \rr_2+...\\ &=Q \myv E,\endeq$$

where I've pulled out the constant $Q$ -- the test charge, leaving a vector sum called...

the electric field: $$\myv E(\myv r) \equiv \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^n \frac{q_i}{\rr_i^2} \uv\rr_i.$$

The electric field $\myv E$ is...

  • a vector function,
  • the force per unit "test" charge due to all of the source charges,
  • depends on the source charges and their positions,
  • depends on the "test position" in space.

Show balloon 'plumb bob' around VdG generator.

Experiential learning! The electric field is not something we typically are aware of. But we *can* have direct physical experience of it using a balloon on a string as a kind of plumb bob--a test charge that we move around and measure the resulting force.

Philosophically--is it just a convenient accounting device for keeping track of the forces? or something more?

Historically Isaac Newton was apologetic about the "Action at a distance" of his theory of gravity. And the electric field is another case of this...

It is inconceivable that inanimate Matter should, without the Mediation of something else, which is not material, operate upon, and affect other matter without mutual Contact... That Gravity should be innate, inherent and essential to Matter, so that one body may act upon another at a distance thro' a Vacuum, without the Mediation of any thing else, by and through which their Action and Force may be conveyed from one to another, is to me so great an Absurdity that I believe no Man who has in philosophical Matters a competent Faculty of thinking can ever fall into it. Gravity must be caused by an Agent acting constantly according to certain laws; but whether this Agent be material or immaterial, I have left to the Consideration of my readers.

Isaac Newton, Letters to Bentley, 1692/3, As quoted in Edward Zalta, Stanford Encyclopedia of Philosophy

Artistically A Sudden Gust of Wind (by Canadian artist Jeff wall), was inspired by Ejiri in Suruga Province, a wood-block print by Hokusai, early 1800s.

Example: Problem 2.2

Two equal charges ($+q$) are a distance $d$ apart. What is the field at a point $P$ which is a distance $z$ away from the midpoint of the two charges?

The fields at $P$ from the two charges have the same magnitude.

$E=\frac{1}{4\pi \epsilon_0}\frac{q}{\rr^2}$

But they are directed $+ \theta$ and $- \theta$ away from the $z$-axis. When added, their horizontal components cancel, and their sum is just the sum of their $z$-components.

You'll need to use at least one trig function, such as $\cos \theta$. Look for right triangles in the diagram, and express $\cos \theta$ in terms of quantities you find labelled in the diagram.

With $\cos \theta = z/\rr$, the total field has magnitude: $$\begineq E_\text{tot} &= 2E \cos \theta= =2\frac{1}{4\pi \epsilon_0}\frac{q}{\rr^2}\frac{z}{\rr}\\ &=\frac{2qz}{4\pi \epsilon_0(z^2+(d/2)^2)^{3/2}}\endeq$$ and points straight up along the $z$-axis.

Changing the right charge to $-q$, what's the field?

Now the vertical components of the two fields cancel, and the total field is the sum of the horizontal components.

$\sin \theta = (d/2)/\rr$ and so: $$E_\text{tot} = 2E \sin \theta=\frac{1}{4\pi \epsilon_0}\frac{2qd/2}{\rr^3}=\frac{qd}{4\pi \epsilon_0(z^2+(d/2)^2)^{3/2}}$$ and points to the right.

Charge density

Instead of discrete point charges, we're usually dealing with a rather more amorphous charge distribution or charge density. Our expression for the electric field needs to change into an integral: $$\myv E(\myv r) = \frac{1}{4\pi \epsilon_0} \int \frac{dq}{\rr^2} \uv \rr ,$$ where $dq$ is some small "chunk" of charge. We shall often deal with...

  • charge spread along a line with $\lambda \equiv$ the charge per unit length, such that $dq= \lambda\, dl$, or,
  • charge spread over an area with $\sigma \equiv$ the charge per unit area, such that $dq= \sigma\,da$, or,
  • charge spread through a volume with $\rho$ the charge per unit volume, such that $dq= \rho \,d\tau$.


A sheet of electric charge has a constant surface charge density of $\sigma$ (Coulombs / meter^2). A strip of tiny, constant width $dw$ is cut out of the sheet. What is the (constant) linear charge density of the strip in terms of $\sigma$ and any differentials shown in the diagram?

In this last case, a volume charge density, Coulomb's law is written $$\myv E(\myv r) = \frac{1}{4\pi \epsilon_0}\int_{\cal V} \frac{\rho(\myv{r}')}{\rr^2(\myv r, \myv r')} \uv{\rr}(\myv r, \myv r')\, d \tau'.$$

 

Example: Problem 2.5

Find the electric field at a position $P$ above the center of a circular ring of radius $r$ with line charge density $\lambda$.

First, note that the parts of the ring across the circle from each other will cancel out any horizontal component of the electric field, so we only need to worry about adding up the vertical ($\uv{z}$) components of the field from different chunks of charge ($dq$). $$\begineq \myv E &=(\myv E)_z \uv z = \uv z \int d\myv E\cdot \uv z =\uv z \int \frac{dq}{\rr^2}\uv\rr\cdot\uv z\\ \endeq$$

The answer is: $$\myv E(z)=\frac{ 1}{4\pi\epsilon_0} \frac{\lambda 2\pi r z}{(r^2+z^2)^{3/2}}\uv z$$ [Does this make sense for $z=0$ and $z \gg r$?]

Problem 2.6

Find the electric field at a position $P$ above the center of a circular disk of radius $R$ with surface charge density $\sigma$. Notice that $R \neq \rr$!

Problem 2.2 gave us the formula for the electric field above the center of a single ring with line charge density $\lambda$ and radius $r$. We can use that result to find the electric field above the center of the disk by thinking of the disk as made up of many rings with different radii and integrating over the the radii. We'll also use our result connecting $\sigma$ and $\lambda$ for a strip cut out of a sheet. consider the disk as an integral over rings of radius $r$ and thickness $dr$. We can use the results of 2.5 to write the field of one of these rings: The total charge of one ring of thickness $dr$ is $\sigma * 2 \pi r * dr=\lambda * 2\pi r$ so the line charge density for one ring is $\lambda=\sigma dr$, and the electric field from one ring is... $$d\myv E_{\text ring}=\frac{1}{4 \pi \epsilon_0} \frac{(\sigma dr) 2\pi r z}{(r^2+z^2)^{3/2}} \uv{z}.$$

Now set up the integral and solve it in CoCalc.... Show that the solution can be written as: $$\myv E = \frac{2\pi \sigma z}{4 \pi \epsilon_0} \left(\frac{ 1}{z} - \frac{ 1}{\sqrt{R^2+z^2}}\right)\uv{z}.$$

Explore $z \gg R$: The solution ought to look like the field of a 'point' charge since it's so far away. It's useful to know about the binomial approximation: If $x$ is small compared to 1, then $(1+x)^p\approx 1+p$.
$(1+R^2/z^2)^{-1/2} \approx 1-\frac{1}{2}\frac{R^2}{z^2}$.

Explore $z \ll R$: close to the disk, the field should look indistinguishable from that of an infinite plane of surface charge.

Problem 2.7

Find the electric field a distance $z$ away from the center of a spherical surface of radius $R$ that has a total charge $q$ on it, spread uniformly over its surface.

Set up an integration of the fields of various rings, which have radii which vary as $r(h)=R\cos(h\pi/2R)$ for $h$ running from $-R$ to $R$...

We've got...

  • $R$ is the radius of the sphere centered on the origin.
  • charge density: $\sigma=q/A=q/(4\pi R^2)$.
  • $z$ is the distance to $P(0,0,z)$, where we want to evaluate the field. By symmetry, $\myv E = E(z)\,\uv z$.
  • $h$ will be our integration variable, whereby we'll cover all the charge on the sphere's surface.
  • There's a ring of charge at height $h$ above the origin.
  • The ring has a circumference of $r$.
  • $r=\sqrt{R^2-h^2}$.
  • The width (thickness) of one ring is $t=dh/\cos\theta=dh\,R/r$.
  • The charge on one ring is $$dq=\sigma\,da=\sigma t 2\pi r=dh\frac{q 2\pi r R}{4\pi R^2r }=dh\frac{q}{2R}.$$
  • The distance of one ring to $P$ is $D=z-h$.
  • The $z-$component of the field of a ring with charge $dq$ a distance $D$ from the center of the ring is (see above): $$dE_z=\frac{ 1}{4\pi\epsilon_0} \frac{dq \,D}{(r(h)^2+D^2)^{3/2}}.$$

So now, with a bit of substitution.... $$\begineq E_z(z)&= \int_{-R}^{+R}\frac{dh}{4\pi\epsilon_0}\frac{q(z-h)}{2R((R^2-h^2)+(z-h)^2)^{3/2}}\\ &= \frac{q}{4\pi\epsilon_0}\frac{\frac{-R+z}{\sqrt{(R-z)^2}}+\frac{R+z}{\sqrt{(R+z)^2}}}{2 z^2} \\ &=\frac{q}{4\pi\epsilon_0 z^2} \endeq$$

The integral was done with Mathematica. Phew!


Brute force vector integration!!