Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Field problems involve two kinds of points (Figure 2.3.1):

the field point (observation point) $P (x, y, z)$ , where we want to know the field, and
the source point $Q (x^{'}, y^{'}, z^{'})$ , where a source (a charge, a current element) is located.

Coordinates of source points are always written with a prime, while unprimed coordinates belong to the field point. This convention removes any doubt about which variables are held fixed when we integrate over a source distribution (the primed ones run) or differentiate a field (the unprimed ones vary).

Position vector

The position vector of $P$ is the vector from the origin $O$ to $P$ :

r ∣ r ∣ a_{r} = x a_{x} + y a_{y} + z a_{z}, = r = x^{2} + y^{2} + z^{2}, = \frac{r}{r} . (2.3.1)

Its components are simply the coordinates of $P$ . Its magnitude $r$ is the distance of $P$ from the origin and $a_{r}$ is the unit vector pointing from the origin towards $P$ (it will become the radial unit vector of spherical coordinates in Section 2.3.6). Likewise the source point has

r^{'} = x^{'} a_{x} + y^{'} a_{y} + z^{'} a_{z} . (2.3.2)

Displacement vector

The vector from the source point $Q$ to the field point $P$ is the displacement vector. Since $r^{'} + R = r$ (head to tail, Figure 2.3.1),

Displacement vector

R = r - r^{'} = (x - x^{'}) a_{x} + (y - y^{'}) a_{y} + (z - z^{'}) a_{z}, (2.3.3)

∣ R ∣ = R a_{R} = [(x - x^{'})^{2} + (y - y^{'})^{2} + (z - z^{'})^{2}]^{1/2}, = \frac{R}{R} . (2.3.4)

$R$ is the distance between source and field point and $a_{R}$ is the direction “away from the source”. Coulomb's law (Section 3.1) and every field formula derived from it are written with these quantities. The order “field minus source” in $R$ is essential, because reversing it reverses the direction of the field.

In plain words

$R$ is the arrow you would draw from the charge to the place where you measure: “where I am, minus where the source is”. Its length is the distance between them, and $a_{R}$ is the direction a positive charge pushes.

Common mistake

Writing $R = r^{'} - r$ (source minus field point). This vector points towards the source, and a field computed with it comes out with the wrong sign: a positive charge would seem to attract. Remember “field minus source”, and check that $Q + R$ lands on $P$ .

Quick check

A source sits at $Q (1, 0, 0)$ and we observe at the origin $P (0, 0, 0)$ . Find $R$ and $a_{R}$ .

$R = r - r^{'} = (0 - 1) a_{x} = - a_{x}$ , $R = 1$ , $a_{R} = - a_{x}$ : at the origin “away from the source” means towards $- x$ .

Example 2.3.1Displacement vector Example 2.3.2Unit vector along the displacement Example 2.3.3Gradient of

1/ R

The function 1/R around a source point \mathrm Q, in a plane through \mathrm Q (computed). The blue circles are level curves 1/R=0.6,\,0.9,\,1.4,\,2.2, and the arrows show \nabla(1/R)=-\vec a_R/R^2 on three rings, R=0.75, 1.25 and 1.85. They are perpendicular to the level curves, point uphill (towards \mathrm Q, where 1/R grows without bound), and their length falls as 1/R^2: on the outer ring they are (1.85/0.75)^2\approx6.1 times shorter than on the inner ring. — **Figure 2.3.2.** The function $1/ R$ around a source point $Q$ , in a plane through $Q$ (computed). The blue circles are level curves $1/ R = 0.6, 0.9, 1.4, 2.2$ , and the arrows show $\nabla (1/ R) = - a_{R} / R^{2}$ on three rings, $R = 0.75$ , $1.25$ and $1.85$ . They are perpendicular to the level curves, point uphill (towards $Q$ , where $1/ R$ grows without bound), and their length falls as $1/ R^{2}$ : on the outer ring they are $(1.85/0.75)^{2} \approx 6.1$ times shorter than on the inner ring.

2.3.1 Position and displacement vectors

Position vector

Displacement vector

In this part

Examples · 3