# Lectures 2 and 3 - Electric Force and Fields and Continuous Charge Distributions

In this lecture we will discuss the addition of electric forces, introduce the concept of electric fields, and look at how electric field and forces on moving charges and dipoles can be treated using concepts discussed in Mechanics (131/141). We will then learn to infer electric field distributions created by continuous charge distributions.

## Coulomb's law

$F=k\frac{Q_{1}Q_{2}}{r^{2}}$

where $k=8.99\times 10^{9}\mathrm{Nm^{2}/C^{2}}$.

$k$ can also be written $k=\frac{1}{4\pi \epsilon_{0}}$ where $\epsilon_{0}$ is the permittivity of free space, $\epsilon_{0}=8.85\times10^{-12}\mathrm{C^{2}/Nm^{2}}$.

Like all forces, electric forces are vectors. The form of Coulomb's law shown here is appropriate for point charges, but we will learn later on how to deal with distributions of charge using calculus. Note that in any given case a charge exerts an equal and opposite force on another charge to the force that the other charge exerts on it, and thus satisfies Newton's third law.

## Force due to more than one charge

The best approach for finding the force on one point charge due to other point charges is to find the force on the charge due to each charge around it and sum the forces together, paying attention to the direction of the force!.

For the above

$F_{1}=(k\frac{Q_{1}Q_{2}}{l_{12}^{2}}+k\frac{Q_{1}Q_{3}}{(l_{12}+l_{23})^{2}})\,\mathrm{N}$ to the right

$F_{2}=(k\frac{Q_{1}Q_{2}}{l_{12}^{2}}+k\frac{Q_{2}Q_{3}}{l_{23}^{2}})\,\mathrm{N}$ to the left

$F_{3}=(k\frac{Q_{1}Q_{3}}{(l_{12}+l_{23})^{2}}-k\frac{Q_{2}Q_{3}}{l_{23}^{2}})\,\mathrm{N}$ to the left

## Acollinear Charge Configurations

When charges are not lined up nicely in a line the force terms need to be added as vectors, the length of each vector may be found using Coulomb's law:

$F=k\frac{Q_{1}Q_{2}}{r^{2}}$

Here is a visualization of the force in a configuration containing more than two charges not in a straight line.

## Electric Fields

In order to consider the general effect that a charge or distribution of charges will have on charges in their vicinity we can introduce the concept of an electric field.

The electric field $\vec{E}$ is defined as the force $\vec{F}$ a small positive test charge with charge $q$ would experience, divided by that charge, i.e.

$\vec{E}=\frac{\vec{F}}{q}$

The SI units of electric field are $\mathrm{N/C}$ (you will also see fields given in $\mathrm{V/m}$..but we haven't defined the volt yet!)

The magnitude of the electric field due to a single point charge $Q$ at a distance $r$ from the charge is

$E=k\frac{Qq}{r^{2}}\frac{1}{q}=k\frac{Q}{r^{2}}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}$

Given the field $\vec{E}$ at a point in space we can work out the force $\vec{F}$ on an arbitrary charge $q$ using

$\vec{F}=q\vec{E}$

A positive charge experiences a force in the same direction as the field, a negative charge feels a force in the opposite direction to the field.

## Field distributions and field lines

The field due to multiple charges may be found by summing up all the individual fields.

We often represent electric fields by drawing field lines, these lines point in the direction of the electric field, and the strength of the field is represented by the spacing of the lines, the closer the spacing of the lines the stronger the field.

Take a look at the following applet to get a feel for how the electric field vector changes with the strength of each charge and with distance in a setup with two point charges.

We can explore the field distribution around various collections of charges using this applet or in real life using a Geissler tube.

Note the direction of electric field lines, where they originate and terminate, and also the density of field lines and how that relates to the charges responsible for the field and the magnitude of the field at various locations.

We can explore the electric field on a metal sphere by observing the response of the Geissler tube in the vicinity of the field around a Van der Graaf generator; we will demonstrate this next week when we have learnt something about potentials.

## Charge moving in an electric field

The force on a charge $q$ in an electric field is

$\vec{F}=q\vec{E}$

Electric fields can be used to accelerate electrons and control their path, in the lab you will use a device that makes use of this, an oscilloscope based on a cathode ray tube. As in televisions this technology is rapidly being phased out by digital displays, but the physics is still valid!

## Electron gun

In the electron gun, electrons are first generated by a thermal emission process, and then accelerated by an electric field. In a fairly short distance an electron can be accelerated to a very high speed.

If we consider a field of $E=1\times10^{4}\mathrm{N/C}$ over a distance of $1\mathrm{cm}$ we can calculate the velocity using our kinematic equations of motion.

The acceleration produced by the field will be

$a=\frac{F}{m}=\frac{qE}{m}=\frac{1.602\times10^{-19}\times1\times10^{4}}{9.11\times10^{-31}}=1.76\times10^{15}\mathrm{ms^{-2}}$

$v^{2}=v_{0}^{2}+2ax$

$v=\sqrt{2ax}=\sqrt{2\times1.76\times10^{15}\times1\times10^{-2}}=6\times10^{6}\mathrm{ms^{-1}}$

We should note that as we try to accelerate electrons closer to the speed of light it becomes harder to do so. Hence to achieve the acceleration of charged particles to very fast speeds very large electric potentials are required. One way to do this is using a similar concept to the Van der Graaf generator we have here. In fact there is a large Van der Graaff accelerator underneath Stony Brook (between Physics and Chemistry)! This is no longer used for research but is currently being converted in to a teaching tool, through the Center for Accelerator Science (CASE).

## Electron deflection

Suppose we have an electron in the oscilloscope tube that has been accelerated by the electron gun to $6\times10^{6}\mathrm{ms^{-1}}$. We want to deflect it from it's path by $1^{o}$.

How much field needs to be generated between the deflection plates (length $2\mathrm{cm}$) to achieve this?

$v_{x}=6\times10^{6}\mathrm{ms^{-1}}$

$v_{y}=v_{y0}+at$

$t=\frac{x}{v_{x}}=\frac{2\times10^{-2}}{6\times10^{6}}=3.33\times10^{-9}\mathrm{s}$

$1^{o}=\frac{2\pi}{360}=0.0175$

For small angles $\tan\theta\approx\theta$

$\frac{v_{y}}{v_{x}}=0.0175\to v_{y}=0.0175v_{x}$

$a=\frac{v_{y}}{t}$

$a=\frac{F}{m}=\frac{qE}{m}$

$E=\frac{m}{q}0.0175\frac{v_{x}}{t}=\frac{9.11\times10^{-31}}{1.602\times10^{-19}}0.0175\frac{6\times10^{6}}{3.33\times10^{-9}}=179\mathrm{N/C}$

## Electric dipoles

Two equal charges separated by a fixed distance (for example in a molecule) form an electric dipole. To describe a dipole we define a vector called the dipole moment which points from the negative charge to the positive charge. The dipole moment is defined as

$\vec{p}=Q\vec{l}$

A dipole in an electric field, unlike a charge, does not experience a net force. However, if the dipole is not aligned with the field it does experience a torque.

$\tau=QE\frac{l}{2}\sin\theta+QE\frac{l}{2}\sin\theta=pE\sin\theta$

or

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

## Work done in rotating a dipole in a field

From last semester's work on rotational dynamics, we know that if we know the torque that needs to be applied to rotate an object we can find the work done using

$W=\int_{\theta_{1}}^{\theta_{2}}\tau\,d\theta$

Using the result we just derived

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

we can say that

$W=\int_{\theta_{1}}^{\theta_{2}}\tau\,d\theta=-pE\int_{\theta_{1}}^{\theta_{2}}\sin\theta\,d\theta=pE[\cos\theta]_{\theta_{1}}^{\theta_{2}}$

If we say that the potential energy is zero when the dipole is perpendicular to the field and the field is uniform we can define the potential energy as

$U=-W=-pE\cos\theta=-\vec{p}.\vec{E}$

## Field distributions due to continuous charge distributions

As the electric field at a point in space is the sum of the individual fields due to all charges around that point integral calculus is a natural way to calculate the electric field due to continuous charge distributions.

We define an infinitesimal charge element $dQ$ which is a distance r from the point at which we want to know the field. The field due to this charge element $dE$ is

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{r^{2}}$

The electric field is

$\vec{E}=\int d\vec{E}$

We need to be a bit careful here, because the electric field is a vector! Some examples will help demonstrate the point.

## Field due to a ring of charge

We can consider a situation where the symmetry is quite helpful, a thin ring of charge, total charge $Q$ with radius $a$. We can ask what the field is at a point on the axis going through the center of the ring a distance $x$ from the center of the ring.

Our first step is to define a charge element $dQ$ in terms of a length element $dl$.

The magnitude of the field due to this element is then $dE=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{1}{r^{2}}dl=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{1}{x^{2}+a^{2}}dl$

We cannot simply add the magnitude of the field from the whole ring together to get the field, because the field is a vector. However, symmetry makes the problem fairly easily because all the $dE_{\perp}$ components cancel.

Therefore,

$E=\int dE_{x}=\int dE{\cos\theta}=\int_{0}^{2\pi a}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{1}{x^{2}+a^{2}}\cos\theta\,dl$

$=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{x}{(x^{2}+a^{2})^{3/2}}\int_{0}^{2\pi a}dl=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{x}{(x^{2}+a^{2})^{3/2}}2\pi a=\frac{Q}{4\pi\epsilon_{0}}\frac{x}{(x^{2}+a^{2})^{3/2}}$

## Long line of charge

For an infinitely long line of charge with a linear charge density $\lambda$

$dE=\frac{1}{4\pi\epsilon_{0}r^{2}}\,dQ=\frac{1}{4\pi\epsilon_{0}}\frac{1}{x^{2}+y^{2}}\lambda\,dy$

As before we can invoke symmetry to cancel the y components of the field

$E=E_{x}=\int dE\cos\theta=\frac{\lambda}{4\pi\epsilon_{0}}\int_{-\infty}^{\infty}\frac{\cos\theta}{x^{2}+y^{2}}\,dy$

We want to find the field at a fixed distance $x$, so this may be treated as a constant. However, $\theta$ and $y$ are not independent, $y=x\tan\theta$ so we need to convert this into either an integral in terms of either $\theta$ or $y$ only. To do this we need to find an expression for $dy$ in terms of $x$ and $\theta$

$\frac{dy}{d\theta}=\frac{x}{\cos^{2}\theta}$ ⇒ $dy=\frac{x}{\cos^{2}\theta}d\theta$

$E=\frac{\lambda}{4\pi\epsilon_{0}}\int_{-\infty}^{\infty}\frac{\cos\theta}{x^{2}+y^{2}}dy=\frac{\lambda}{4\pi\epsilon_{0}}\int_{-\pi/2}^{\pi/2}\frac{\cos\theta}{(x^{2}+x^{2}\tan^{2}\theta)}\frac{x}{\cos^{2}\theta}d\theta$

$=\frac{\lambda}{4\pi\epsilon_{0}x}\int_{-\pi/2}^{\pi/2}\cos\theta\frac{\cos^{2}\theta}{\cos^{2}\theta+\sin^{2}\theta}\frac{1}{\cos^{2}\theta}\,d\theta$

$=\frac{\lambda}{4\pi\epsilon_{0}x}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta=\frac{\lambda}{4\pi\epsilon_{0}x}[\sin\theta]_{-\pi/2}^{\pi/2}=\frac{\lambda}{2\pi\epsilon_{0}x}$