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S2.12 Electrostatic forces

Preamble

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Preamble

The mathematical form of the equations in this section are very similar to those in S2.11. But here we explore what happens if we have interacting particles where the force can be attractive or repulsive. And we’ll delve slightly deeper in the world of vectors to look at the idea of a vector field.
 

[A] About the force

Key Point 2.5

The electrostatic (Coulomb) force of interaction between two point charges q1 and q2 separated by distance r:

  • has magnitude
    \[ F_E = \frac{K\mid q_1q_2 \mid }{r^2} \]
Graphic - No title - elforcezero

with K a fundamental constant of electrostatics

Commentary

Commentary

K is a collection of constants, such that

$$K = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 N\, m^2\,C^{-2}$$

The quantity ϵ0 is known as the Permittivity Constant of Free Space with an experimentally measured value of

$$ \epsilon_0 = 8.85418\times10^{-12}\,\rm C^2\,N^{-1}\,m^{-2} $$

And a first look at quantisation

And a first look at quantisation

Charge is quantised: it can only take certain values and these are integer multiples of the fundamental unit of charge, that on an electron.

So any net charge q can be written as

q=N e
where N is the number of number of unbalanced charges (N is positive if there are more protons than electrons, and negative if vice versa) and e is the charge of one electron, being $1.609\times10^{-19}\,\rm C$.

The quantisation of charge was first experimentally verified by Robert Millikan in 1913 by the famous Oil Drop experiment, for which combined with his work on the Photoelectric Effect, he won the Nobel prize in 1923.

The charge on an electron is very small and in many conditions we will be able to ignore the quantisation and treat charge as a continuous variable.

 
 

[B] Example problem

Estimate the ratio of the gravitational force between two electrons and the electrostatic (Coulomb) force between two electrons.

Solution

Solution

Don’t we need the separation?

Don’t we need the separation?

It looks like the question should have specified the separation of the electrons. But it doesn’t need to. Keep the separation as r - don’t choose an arbitrary value. What we are about to demonstrate holds for all distances.....
 

The gravitational force between two equal masses, m1=m2=m, separated by distance r is of magnitude (Key Point 2.4)

\[ F_G =G \frac{m^2}{r^2} \]
where G is the universal gravitational constant with value
G=6.67×10-11Nm2kg-2

The electrostatic (Coulomb) force between two equal charges, q1=q2=e, separated by distance r is of magnitude (Key Point 2.5)

\[ F_E = K\frac{e^2}{r^2} \]
where K has the value:
K=8.99×109Nm2C-2

Both forces are of inverse-square form: their strength varies inversely as the square of the separation.

Taking the ratio of the two forces the separation r cancels to leave:

\[ \frac{F_G}{F_E} = \frac{G}{K} \times \left( \frac{e}{m}\right)^{-2} = \frac{6.67 \times 10^{-11}}{9.0 \times 10 ^9 \times (1.76 \times 10^{11})^2} =2.4 \times 10^{-43} \hspace{0.5cm}\mbox{\rm (2sf)}\hspace{0.5cm} \]

You might want to contemplate the implications of what we have just calculated, in particular, this result seems to imply that gravity is so small as to be irrelevant and what it means for collections of like charged species in close proximity to one another, such as the protons in the nuclei that make up you, me and everything around us......

 

[C] Dealing with multiple charges - the principle of superposition

In the same way that we have been adding forces as vectors throughout this section, electrostatic forces are no exception and serve as an illustration of the Principle of Superposition in action.

Let’s start with the vector form of the force due on one charge from another. If we have two point charges q1 and q2 seperated by a vector $\vec{r}_{1\,2} = r\,\hat{r}_{1\,2}$ as shown below

3.1

then the force on q1 as aresult of q2 is a vector giving,

Key Point 2.6

Coulomb’s Law in vector form, being:
$$ \vec{F}_{1\,2} = - {1\over4\pi\,\epsilon_0} {q_1\,q_2\over r^2}\, \hat{r}_{1\,2} $$
where $\hat{r}_{1\,2}$ is a unit vector from q1 in the direction of q2.

Note that this force is directed away from q1 if both charges have the same sign and towards q2 if opposite. Generalising to a series of charges, for example q1q6

Then the force on q1 as a result of the charges q2q6 will be a vector sum, being

Key Point 2.7

$$ \vec{F}_1 = \vec{F}_{1\,2} + \vec{F}_{1\,3} + \cdots = \sum_{j= 2}^N \vec{F}_{1\,j} $$

where each force $\vec{F}_{1\,j}$ is the force on q1 as a result of the $j^{\rm th}$ charge being given by Key Point 2.6.

[D] The Electric Field

We have just seen how to calculate the force on a charged particle due to the presence of a second charge. But if the two charges are nowhere near each other how is that force ’felt’ by the charge? How can there be this action at a distance?

We can explain this by saying that particle 2 sets up an electric field in the space all around it. Paticle 1 is affected by this field. Thus particle 2 exerts a force on particle 1 not by touching it but by the electric field its presence has created.

Scalar vs Vector fields

Scalar vs Vector fields

Imagine I asked you to measure the temperature of all points in the room you are currently in. You would divide up the space and make a measurement at each point. The collection of values is said to be a scalar field, as temperature is a scalar quantity.

But imagine I wanted you to measure the ’breeziness’ at your location and gave you something that would measure both wind speed and direction. You would return with a series of vectors at various points - values of the wind speed and direction.

An electric field is thus analagous to this ’wind-field’.

 

We can define the electric field $\vec E$ at a point, P, in the vicinity of a charge by considering the electrostatic force acting on a test positive charge of q0 at P.

Key Point 2.8

$$ \vec{F} = q_0\,\vec{E} $$

The magnitude of the field at this point is given by $e = \frac{F}{q_0}$ and the direction is that of the force that acts on the test positive charge.

If we then extend this idea to a collection of charges as we did previously for the electrostatic force

3.2

Then we can write the force on this charged particle as,

$$ \vec{F}_1 = q_1\,\vec{E}_T $$
where $\vec{E}_T$ is the total Electric Field resulting from the other charges. Since the force on the particle F1 is a linear sum of contributions from all other particles, we can show that the Electric Field

Key Point 2.9

$$ \vec{E}_T = \vec{E}_2 + \vec{E}_3+\cdots = \sum_{j=2}^N\vec{E}_j $$
where $\vec{E}_j$ is the Electric Field due to the $j^{\rm th}$ charge, so being,
$$ \vec{E}_j = {1\over4\pi\epsilon_0} {q_j\over r_j^2} \hat{r}_{1\,j} $$

So the Electric Field seen by charge q1 is a Linear Vector Summation of the Electric Fields due to the other charges; the principle of superposition again.

[E] Field lines

Field lines provide a convenient way for us to visualise the vector nature of an electric field. They are lines of force which show the direction of the field at a point in space, with the separation between the lines indicating the magnitude or strength.

3.3

Analysis

Analysis

You can sort out the direction once and for all by considering what the field lines will look like if you have a test positive charge at some distance away from a positively charged particle.

The direction of the electrostatic force will be oriented away from the positively charged particle, along the line conneting it to the test positive charge. Likewise if your particle was negatively charged, the field lines would point towards it.

 

Now consider two charges of +q and -q separated by a fixed distance d as shown. This configuration is known as an Electric Dipole.

3.4

Mathematical analysis

Mathematical analysis

We want to calculate the vector Electric Field at the position (0,a). There are two charges, so two contributions to the electric field, which according to the the Superposition principle will be the vector sum of the electric fields from each of the charges.

For the positive charge, located at $({d\over2},0)$, then at the point (0,a) we have that the vector from the positive charge is

$$ \vec{r}_+ = r_+\,\hat{r}_+\quad\mbox{where}\quad r_+^2 = a^2 + {d^2\over4}\quad\mbox{and}\quad \hat{r}_+ = -{d\over2r_+}\,\hat{\imath} + {a\over r_+}\,\hat{\jmath} $$
You should check for yourself that $|\hat{r}_+| = 1$.

The Electric Field from the positive charge is thus,

$$ \vec{E}_+ = {1\over4\pi\epsilon_0}{q\over r_+^2}\left[ - {d\over2r_+}\,\hat{\imath} + {a\over r_+}\,\hat{\jmath}\right] $$
Now for the negative charge located at $(-{d\over2},0)$, then at (0,a) we have that the vector from the negative charge is
$$ \vec{r}_- = r_-\,\hat{r}_-\quad\mbox{where}\quad r_-^2 = a^2 + {d^2\over4}\quad\mbox{and}\quad \hat{r}_- = {d\over2r_-}\hat{\imath} + {a\over r_-}\hat{\jmath} $$
so that the Electric Field from the negative charge is,
$$ \vec{E}_- = {1\over4\pi\epsilon_0}{-q\over r_-^2}\left[ {d\over2r_-}\,\hat{\imath} + {a\over r_-}\,\hat{\jmath}\right] $$
We have that r+=r-=r, and total field is $\vec{E}_T =\vec{E}_+ + \vec{E}_-$. The $\hat{\jmath}$ components cancel, giving the result below....

 

Key Point 2.10

The total Electric Field is given by
$$ \vec{E}_T = - {1\over4\pi\epsilon_0} {q\,d\over r^3}\,\hat{\imath} $$
which is parallel to the x-axis and pointing from the positive towards the negative charge, which is what you expect from the diagram.

Considering Key Point 2.10 we see that the Electric Dipole is characterised by the vector quantity

$$ \vec{p} = q\,d\,\hat{\imath} $$
which is known as the Electric Dipole Moment, which has units of $\rm C\,m$.

Note the direction of $\vec{p}$, it is from the negative towards the positive end of the dipole. So in terms of the Electric Dipole Moment the Electric Field a distance a perpendicular to the dipole axis is given by,

Key Point 2.11

$$ \vec{E}_T = - {1\over4\pi\epsilon_0} {\vec{p}\over r^3} $$

We can repeat this calculation at all point in space and form the field lines from a dipole as shown

3.5
AForces and field simulation

Learning Resources

Textbook: The treatment of electrostatics starts in HRW in Chapter 21, and much of the next 11 chapters is concerned with developing towards a treatment of Maxwell’s Equations. However, we will start (and remain) in the foothills, covering Chapter 21 and selected bits of Chapter 22.
Course Questions:
Self-Test Questions: