Up until now we have discussed how objects move due to forces, but we have not discussed how forces are generated nor discussed energy at all.
Everyone knows that positive charges attract negative charges and repel positive ones. However, let's look more into understanding how much force this attraction and repulsion generates. What we won't be covering here is the specifics of what a field actually is. For our purposes a field is anything which affects a space and generates forces on other objects in that space.
The formula of the magnitude of a force in an electrostatic field is given by
\[F = \frac{Qq}{4 \pi \epsilon d^2} \]
Where \(F\) is the force, \(Q\) is the charge of on particle, \(q\) is the charge of the other particle, \(\epsilon\) is the permittivity of the medium and \(d\) is the distance between the charges
We can abstract this by defining an electric field strength
\[E = \frac{Q}{4 \pi \epsilon d^2}\]
Thus a small test charge \(q\) placed inside this field will experience force
\[F = qE\]
However, this analysis is a little thin. It has the flaw of not working for non-uniform fields. It also isn't a vector, and really we should try and put all relationships into 3 dimensional space. Therefore, let's bring this equation into the third dimension.
\[\mathbf{F}(\mathbf{r}) = q \mathbf{E}(\mathbf{r})\]
The equation has been given an upgrade! Now force and electric field are vectors, and we introduce \(r\), a vector for the position of the test charge within the electric field.
We can actually do a very similar thing for gravity
\[\mathbf{F}(\mathbf{r}) = m \mathbf{g}(\mathbf{r})\]
The fields behave more or less the same. The only difference of course being that gravity acts upon mass rather than charge.
We are also now prepared to make a definition of a new concept: energy. We define the work done as the distance travelled when a force is applied. As an equation we can simply write this as
\[W = F \times s\]
But this form has some flaws. For one, it is only a scalar, and we prefer vectors where possible. Also, it doesn't allow us to work with non-uniform fields. To fix this, as usual, we introduce an integral.
\[\Delta W = \int \mathbf{F} d\mathbf{s}\]
Let's consider a falling object again. This is an object falling due to gravity, starting at rest. Earlier chapters have shown us that the velocity will be given by \(y = at = gt\) and \(y = \frac{1}{2} g t^2\). Let's begin by eliminating \(t\) from these equations.
If we rearrange \(y = gt\) into \(t = \frac{y}{g}\) and substitute into \(y = \frac{1}{2} g t^2\) we get
\[y = \frac{1}{2} g \frac{v^2}{g^2} = \frac{1}{2} \frac{v^2}{g}\]
Now, let's try and think about this in another way. Consider this problem in terms of average velocity. Since the object starts from rest, \(u = 0\). Therefore, if the distance travelled is the average velocity times time, then \(y = \frac{1}{2} (v - 0) t = \frac{1}{2}vt\). Also consider that velocity is given by \(v = gt\), since we are starting from rest. A simple substitution will give us
\[y = \frac{1}{2} v \frac{v}{g} = \frac{1}{2} \frac{v^2}{g}\]
which is the same result from a moment ago.
Now consider if this object had an initial downward velocity, \(u\). Let's repeat the same analysis. Well, now the distance will be given by \(y = \frac{1}{2} (v + u)t \), still simply the average velocity times time. The final velocity will now be given by \(v = u + gt\). Now if we rearrange for \(t\) we find \(t = \frac{v - u}{g}\). Now let's try substituting for \(t\) to find
\[y = \frac{1}{2} \frac{(v - u)(v+u)}{g} = \frac{1}{2} \frac{v^2 - u^2}{g}\]
Now, let's find an expression for the product of the force (\(mg\)) with distance.
\[mgy = \frac{1}{2} mg \frac{(v - u)(v+u)}{g} = \frac{1}{2} m v^2 - \frac{1}{2} m u^2\]
I will very suggestively not cancel the \(m\).
Finally, let's plug some "numbers" in. Consider the object falls from \(h_1\) to \(h_2\) such that \(y = h_1 - h_2\).
\[mg (h_1 - h_2) = \frac{1}{2} m v^2 - \frac{1}{2} m u^2\]
And then let's group together the start terms and end terms
\[mgh_1 + \frac{1}{2} m u^2 = mgh_2 + \frac{1}{2} m v^2\]
What we have made here is an equation that is independent of time, showing some values that are conserved. We call these conserved values energy. Total energy in a system, much like with momentum, is always conserved. It can change between different "kinds" though. In this instance there are two "kinds" of energy - potential (represented by the \(mgh\) terms) and kinetic (represented by the \(\frac{1}{2} mv^2\) terms). We will be exploring potential and kinetic energy more in future, but for now I want to quickly introduce a new concept - work done. Work done represents this conversion from one kind of energy to other by a force.
Work done can crudely be described as the force times the distance travelled. This holds for uniform forces in one dimension - we will explore three dimension and non-uniform forces soon.
Fields in Three Dimensions
Let's consider the three dimensional version of these problems. If we consider that in three dimensions acceleration due to gravity for a falling object is given by
\[\mathbf{g} = \frac{\mathbf{v} - \mathbf{u}}{t}\]
and the displacement given by
\[\frac{1}{2} (\mathbf{v} + \mathbf{u}) t\]
We can eliminate \(t\) from these equations by taking the scalar product. Earlier I was not quite candid on why we would want to eliminate \(t\) so badly. Remember, by eliminating the time-evolving part of a system we arrive at the time-invariant parts.
\[\mathbf{g} \cdot \mathbf{s} = \frac{1}{2} (\mathbf{v} + \mathbf{u}) (\mathbf{v} - \mathbf{u}) = \frac{1}{2} (v^2 - u^2)\]
Remembering our definition of work done as the force applied over a distance, it maps easily into three dimensions by
\[W = \mathbf{F} \cdot \mathbf{s}\]
And let's use our result to figure out what the work done is in this instance.
\[W = m \mathbf{g} \cdot \mathbf{s} = \frac{1}{2} m v^2 - \frac{1}{2} m u^2\]
By definition, \(\frac{1}{2} m v^2 - \frac{1}{2} m u^2\) is the change in kinetic energy, so again, the work done by the gravitational field is equal to the change in kinetic energy.
In short, uniform fields in three dimensions pose little difficulty. For symmetry reasons, we can always rotate our coordinate system, and the laws of physics hold. In other words, we can replace all of the variables in our equations with vectors and follow the maths through without much difficulty.
Non-uniform Fields
As you might have guessed, moving from uniform to non-uniform solutions generally requires us to use integrals. Writing the equations we already have as integrals we find
\[\Delta W = \int \mathbf{F} \cdot d \mathbf{s}\]
\[\Delta \phi = - \int \mathbf{g} \cdot d \mathbf{s}\]
\[\Delta \phi = - \int \mathbf{E} \cdot d \mathbf{s}\]
These are very special kinds of integrals called line integrals - we really do add up every value along the path \(\mathbf{s}\). Let's do some example problems using line integrals to see how to work with them.
Consider the following field
\[\mathbf{E} = \frac{F}{a^2} \colvec{yz\\xz\\xy}\]
Let's evaluate the line integral to find the change in potential when moving from the point (0,0,0) to (1,1,1) in a straight line in two ways. First we will use a substitution \(\mathbf{s}\), and then we will do it in three stages.
Using a substitution, since \(x=y=z\) at all points along the line, we can rewrite this field as
\[\mathbf{E} = \frac{F}{a^2} \colvec{x^2\\x^2\\x^2}\]
So, the line integral is
\[\Delta \phi = - \int \mathbf{E} ds = -\int^1_0 E_x dx -\int^1_0 E_y dy -\int^1_0 E_z dz \]
But since \(E_x = E_y = E_z\) at all points, we can write
\[\Delta \phi = -3 \int^1_0 \frac{F}{a^2} x^2 dx = -\frac{F}{a^2}\]
Now, let's try in stages. First from (0,0,0) to (1,0,0)
\[\Delta \phi_1 = -\int^1_0 \frac{F}{a^2} yz dx = 0\]
Since \(y=z=0\). Now from (1,0,0) to (1,1,0)
\[\Delta \phi_2 = -\int^1_0 \frac{F}{a^2} xz dy = 0\]
Since \(z=0\). Now from (1,1,0) to (1,1,1)
\[\Delta \phi_3 = -\int^1_0 \frac{F}{a^2} xy dz = -\frac{F}{a^2}\]
Field Due to a Ring
Consider a thin ring of radius \(R\). It has an even linear charge density of \(\lambda C/m\) (Coulombs per metre). We want to calculate the electric field at a particular point \(P\) along the z axis.
The first step is to split the ring into many small parts, such that for each of the smaller parts any curvature is irrelevant. This straight length of ring will be \(ds\), so that the small amount of charge on that ring segment is
\[dq = \lambda ds\]
Now, let's consider the electric field from this one element. The vector from \(ds\) to \(P\) we will call \(\mathbf{r}\), so the magnitude of the small electric field is given by
\[ dE = \frac{1}{4 \pi \epsilon_0} \frac{dq}{r^2} = \frac{1}{4 \pi \epsilon_0 \frac{\lambda ds}{r^2}} \]
\(r\) is the hypotenuse of a triangle between \(P\), \(ds\) and the centre of the ring though, so we can rewrite this as
\[dE = \frac{1}{4 \pi \epsilon_0} \frac{\lambda ds}{z^2 + R^2}\]
Since every \(ds\) is the same \(r\) (but different \(\mathbf{r}\)) from \(P\), consider breaking the electric field strength into two components. There will be one component pointing up, and each of these vertical components from all the \(ds\) will point in the same direction i.e. sum. However, the horizontal components will not sum up. In fact, for each \(ds\) there is another component exactly \(180^\circ\) away which has an equal magnitude and opposite direction horizontal component. In other words, before any kind of integration is done, a symmetry argument is used to eliminate the horizontal components.
So, each component \(ds\) can be thought of as contributing a scalar electric field of strength \(dE \cos(\theta)\) in the z direction. \(\theta\) is the angle made by \(ds P O\) (\(O\) being the centre).
\(\theta\) can be rewritten in terms of symbols we already have though, like so
\[ \cos(\theta) = \frac{z}{r} = \frac{z}{\sqrt{z^2 + R^2}} \]
Now we need to sum up each of these \(ds\) to get the total electric field. We already have an expression for \(dE\), and we have reasoned that the sum of all \(dE \cos(\theta)\) is \(E\). There's nothing left to do now but integrate.
\[ E = \int dE \cos(\theta) = \frac{z \lambda}{4 \pi \epsilon_0 (z^2 + R^2)^{3/2}} \int_0^{2 \pi R} ds = \frac{z \lambda (2 \pi R)}{4 \pi \epsilon_0 (z^2 + R^2)^{3/2}} \]
Optionally, we can replace the charge per unit length with the total charge \(q = \lambda 2 \pi R\)
\[E = \frac{qz}{4 \pi \epsilon_0 (z^2 + R^2)^{3/2}}\]