# 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 $$\myv \rr = \myv r - \myv r '.$$

[BLINK]Sear this fact into your mind for the rest of this course:[/BLINK] $$\myv \rr = \myv \rr(\myv r, \myv r').$$

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 a couple 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.

The electric field is not something we have direct physical experience of, apart from putting a test charge at some point and measuring the force on it.

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

#### 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 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. With $\cos \theta = z/\rr$, the total field is: $$E_\text{tot} = 2E \cos \theta=\frac{1}{4\pi \epsilon_0}\frac{2qz}{\rr^3}=\frac{2qz}{4\pi \epsilon_0(z^2+(d/2)^2)^{3/2}}$$

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 $dq= \rho \,d\tau$.

In this last case, 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 ($dq$). $$\begineq \myv E &=&(\myv E)_z \uv z = \uv z \int d\myv E\cdot \uv z\\ &=& \frac{\uv z}{4\pi\epsilon_0}\int \frac{dq}{\rr^2}\cos\theta\endeq$$

We'll push this forward using...

• $\rr^2 = z^2+r^2$,
• $\cos(\theta) = z/\rr$,
• $dq = \lambda\,dl$, where $l$ is the path length around the ring of radius $r$: That is, $l$ runs from 0 to $2\pi r$.

Substituting into the expression for the electric field above the ring $$\begineq \myv E &=& \frac{\uv z}{4\pi\epsilon_0}\int_0^{2\pi r} \frac{\lambda dl}{(r^2+z^2)}\frac{z}{\sqrt{r^2+z^2}}\\ &=& \frac{\uv z}{4\pi\epsilon_0} \frac{\lambda z}{(r^2+z^2)^{3/2}} \int_0^{2\pi r}dl \\ &=& \frac{ 1}{4\pi\epsilon_0} \frac{\lambda 2\pi r z}{(r^2+z^2)^{3/2}}\uv z . \endeq$$

[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 $R \neq \rr$!

We'll 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... $$d\myv E_{\text ring}=\frac{1}{4 \pi \epsilon_0} \frac{(\sigma dr) 2\pi r z}{(r^2+z^2)^{3/2}} \uv{z}.$$

Then we want to integrate over rings of different radii... $$\begineq\myv E &=& \int d\myv E_{\text ring}=\int_0^R \frac{1}{4 \pi \epsilon_0} \frac{(\sigma dr) 2\pi r z}{(r^2+z^2)^{3/2}}\uv{z} \\ &=&\frac{2\pi \sigma z}{4 \pi \epsilon_0} \int_0^R dr \frac{ r}{(r^2+z^2)^{3/2}}\uv{z}\\ &=&\frac{2\pi \sigma z}{4 \pi \epsilon_0} \left[-\frac{ 1}{(r^2+z^2)^{1/2}}\right]_{r=0}^R\uv{z}\endeq$$

The solution is: $$\myv E = \frac{2\pi \sigma z}{4 \pi \epsilon_0} \left(\frac{ 1}{z} - \frac{ 1}{\sqrt{R^2+z^2}}\right)\uv{z}.$$

Explore $z \gg R$: The solution ought to look like the field of a 'point' charge since it's so far away. Use:
$(1+R^2/z^2)^{-1/2} \approx 1-\frac{1}{2}\frac{R^2}{z^2}$.

Explore $z \ll R$: The field should look like 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!!