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

A system of point charges

Coulomb's assumption contains, besides the inverse-square law for two charges, a second statement: when several charges act on one charge, the forces simply add. The presence of the other charges does not modify the force that any one of them exerts. This is the superposition principle. It is an experimental fact, and it is what makes electrostatics a linear theory: double all the sources and every force doubles, and adding a second set of sources simply adds its forces to the old ones.

A numerical illustration

Before writing the general formula, let us add two forces by hand. A charge $q_{1} = + 4 nC$ sits at the origin and a charge $q_{2} = - 5 nC$ at $(3, 0, 0) m$ . What force acts on a small charge $q = + 1 nC$ at $P (0, 4, 0) m$ ? We use $1/ (4 π ε_{0}) = 9 \times 1 0^{9} m/F$ and treat the two sources one at a time (Figure 3.1.13).

Force of $q_{1}$ . $R_{1} = (0 - 0) a_{x} + (4 - 0) a_{y} = 4 a_{y} m$ , $R_{1} = 4 m$ , so

$F_{1} = \frac{q q _{1}}{4 π ε _{0}} \frac{R _{1}}{R _{1}^{3}} = 9 \times 1 0^{9} \times 1 0^{- 9} \times 4 \times 1 0^{- 9} \times \frac{4 a _{y}}{64} = 2.25 a_{y} nN .$

It points straight up, away from $q_{1}$ (both charges are positive: repulsion).
Force of $q_{2}$ . $R_{2} = (0 - 3) a_{x} + (4 - 0) a_{y} = - 3 a_{x} + 4 a_{y} m$ , $R_{2} = 5 m$ , so

$F_{2} = 9 \times 1 0^{9} \times 1 0^{- 9} \times (- 5 \times 1 0^{- 9}) \times \frac{- 3 a _{x} + 4 a _{y}}{125} = (1.08 a_{x} - 1.44 a_{y}) nN .$

It points from P towards $q_{2}$ (unlike charges: attraction). Its magnitude is $1.8 nN$ .
Total. Adding component by component,

$F ∣ F ∣ = F_{1} + F_{2} = (0 + 1.08) a_{x} + (2.25 - 1.44) a_{y} = (1.08 a_{x} + 0.81 a_{y}) nN, = 1.35 nN .$

Notice that $∣ F ∣ = 1.35 nN$ is much less than $2.25 + 1.8 = 4.05 nN$ : the two forces partly cancel because they point in different directions.

Superposition as vector addition. (a) Each source charge acts on q at P as if the other were absent: \vec F_1 (repulsion, away from q_1) and \vec F_2 (attraction, towards q_2). (b) Placing the tail of \vec F_2 at the tip of \vec F_1 gives the total force \vec F=(1.08\,\vec a_x+0.81\,\vec a_y)\ \mathrm{nN}, of magnitude 1.35\ \mathrm{nN}. The force arrows in both panels are drawn to the same scale (bar in (b)). — **Figure 3.1.13.** Superposition as vector addition. (a) Each source charge acts on $q$ at P as if the other were absent: $F_{1}$ (repulsion, away from $q_{1}$ ) and $F_{2}$ (attraction, towards $q_{2}$ ). (b) Placing the tail of $F_{2}$ at the tip of $F_{1}$ gives the total force $F = (1.08 a_{x} + 0.81 a_{y}) nN$ , of magnitude $1.35 nN$ . The force arrows in both panels are drawn to the same scale (bar in (b)).

The general formula

Now consider $N$ point charges $q_{1}, q_{2}, \dots, q_{N}$ at the source points $r_{1}, \dots, r_{N}$ , and a charge $q$ at the field point $r$ (Figure 3.1.14). Let $R_{k} = r - r_{k}$ be the displacement vector from $q_{k}$ to $q$ , $R_{k} = ∣ R_{k} ∣$ its length and $a_{R_{k}} = R_{k} / R_{k}$ its unit vector. By Coulomb's law (3.1.7) the charge $q_{k}$ alone would exert on $q$ the force $q q_{k} a_{R_{k}} / (4 π ε_{0} R_{k}^{2})$ . Superposition says that the total force is the sum of these $N$ contributions:

Superposition of Coulomb forces

F = \frac{q}{4 π ε _{0}} k = 1 \sum N \frac{q _{k}}{( R _{k} ) ^{2}} a_{R_{k}} = \frac{q}{4 π ε _{0}} k = 1 \sum N \frac{q _{k}}{( R _{k} ) ^{3}} R_{k}, a_{R_{k}} = \frac{R _{k}}{R _{k}} . (3.1.15)

Superposition. The charge q at the field point feels the vector sum of the forces exerted by q_1,q_2,\dots,q_N. Each displacement vector \vec R_k points from its own source q_k to q. — **Figure 3.1.14.** Superposition. The charge $q$ at the field point feels the vector sum of the forces exerted by $q_{1}, q_{2}, \dots, q_{N}$ . Each displacement vector $R_{k}$ points from its own source $q_{k}$ to $q$ .

The sum is a vector sum: each term has its own direction, and in practice we add the $x$ -, $y$ - and $z$ -components separately, exactly as in the numerical illustration. The second form, with $R_{k} / R_{k}^{3}$ , is again the most practical one because each $R_{k}$ is written directly in Cartesian components.

In plain words

Each charge pushes or pulls on $q$ as if it were alone. To get the total, draw all these arrows and add them tip to tail, or, more reliably, add their $x$ -, $y$ - and $z$ -components.

Move the three charges and watch the force on each one: it is always the tip-to-tail sum of the two pair forces acting on it, never the sum of their magnitudes.

InteractiveLoading simulation…

Common mistake · Adding magnitudes

Forces are vectors. Adding the magnitudes, $∣ F_{1} ∣ + ∣ F_{2} ∣$ , is wrong unless the forces happen to be parallel. In the illustration above it would give $4.05 nN$ instead of $1.35 nN$ . Add components.

Quick check

Three equal charges $q$ sit at the corners of an equilateral triangle. What is the total force on a fourth charge placed at the centre of the triangle?

Zero. The three forces have equal magnitudes (equal charges, equal distances) and directions $12 0^{\circ}$ apart, so placed tip to tail they close a triangle.

Continuous distributions: line, surface and volume sources

What if the source charge is not a set of points but a continuous distribution? We use the same idea three times over:

Chop the distribution into small elements, each so small that it can be treated as a point charge $d q^{'}$ located at its source point $Q (x^{'}, y^{'}, z^{'})$ .
Apply Coulomb's law to each element: it contributes a force given by (3.1.7) with $q^{'} \to d q^{'}$ .
Add up: the sum over all elements becomes, in the limit of infinitely many infinitely small elements, an integral.

The displacement vector from the element to the field point $P (x, y, z)$ is, as in (3.1.6),

R R = (x - x^{'}) a_{x} + (y - y^{'}) a_{y} + (z - z^{'}) a_{z}, = (x - x^{'})^{2} + (y - y^{'})^{2} + (z - z^{'})^{2}, (3.1.16)

and depends on both sets of coordinates. The geometry is that of Figure 3.1.3.

Line source

A charge spread along a curve $C$ with density $ρ_{L} (x^{'}, y^{'}, z^{'})$ : the element $d l^{'}$ carries $d q^{'} = ρ_{L} d l^{'}$ , and the force on a point charge $q$ at P is

F = \frac{q}{4 π ε _{0}} \int_{C} \frac{ρ _{L} ( x ^{'} , y ^{'} , z ^{'} )}{R ^{2}} a_{R} d l^{'} = \frac{q}{4 π ε _{0}} \int_{C} ρ_{L} (x^{'}, y^{'}, z^{'}) \frac{R}{R ^{3}} d l^{'} . (3.1.17)

Surface charge

A charge spread over a surface $S$ with density $ρ_{S} (x^{'}, y^{'}, z^{'})$ : the element $d S^{'}$ carries $d q^{'} = ρ_{S} d S^{'}$ , and

F = \frac{q}{4 π ε _{0}} \iint_{S} \frac{ρ _{S} ( x ^{'} , y ^{'} , z ^{'} )}{R ^{2}} a_{R} d S^{'} = \frac{q}{4 π ε _{0}} \iint_{S} ρ_{S} (x^{'}, y^{'}, z^{'}) \frac{R}{R ^{3}} d S^{'} . (3.1.18)

Volume charge

A charge filling a region $V$ with density $ρ_{v} (x^{'}, y^{'}, z^{'})$ : the element $d V^{'}$ carries $d q^{'} = ρ_{v} d V^{'}$ , and

F = \frac{q}{4 π ε _{0}} ∭_{V} \frac{ρ _{v} ( x ^{'} , y ^{'} , z ^{'} )}{R ^{2}} a_{R} d V^{'} = \frac{q}{4 π ε _{0}} ∭_{V} ρ_{v} (x^{'}, y^{'}, z^{'}) \frac{R}{R ^{3}} d V^{'} . (3.1.19)

The three formulas look heavy, but each is just "Coulomb's law for one element, added over the source". Several points deserve emphasis, because they are the source of most mistakes in practice.

Which coordinates are integrated? The integration runs over the source coordinates $(x^{'}, y^{'}, z^{'})$ only. The field point P is held fixed during the integration, and the result is a function of $(x, y, z)$ .
The integrals are vector integrals. The integrand $R / R^{3}$ changes direction from element to element. We therefore always expand $R$ in the constant Cartesian unit vectors $a_{x}, a_{y}, a_{z}$ , take them outside the integral and integrate three scalar functions (Figure 3.1.16).
Use the right element. On a straight line $d l^{'} = d z^{'}$ , but on a circle of radius $b$ the length of a small arc is $d l^{'} = b d ϕ^{'}$ , and on a disk the area of a small patch is $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ (Figure 3.1.15, see also Section 2.3).
Symmetry first. Before integrating it pays to ask which components must vanish by symmetry. Problems 3.1.1 to 3.1.3 show how much work this saves.

(a) Straight line: dl'=dz' — (a) Straight line: $d l^{'} = d z^{'}$

(b) Circle of radius b: dl'=b\,d\phi' — (b) Circle of radius $b$ : $d l^{'} = b d ϕ^{'}$

(c) Disk, polar coordinates: dS'=\rho'\,d\rho'\,d\phi'. The elements that carry the charge dq'. (a) On a straight line along z, dl'=dz'. (b) On a circle of radius b an increase d\phi' of the angle sweeps the arc length b\,d\phi', not d\phi'. (c) On a disk the small patch has sides d\rho' and \rho'\,d\phi', so dS'=\rho'\,d\rho'\,d\phi'. The factor \rho' is why patches far from the centre are larger. — **Figure 3.1.15.** (c) Disk, polar coordinates: $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ . The elements that carry the charge $d q^{'}$ . (a) On a straight line along $z$ , $d l^{'} = d z^{'}$ . (b) On a circle of radius $b$ an increase $d ϕ^{'}$ of the angle sweeps the arc length $b d ϕ^{'}$ , not $d ϕ^{'}$ . (c) On a disk the small patch has sides $d ρ^{'}$ and $ρ^{'} d ϕ^{'}$ , so $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ . The factor $ρ^{'}$ is why patches far from the centre are larger.

Why we integrate in Cartesian unit vectors. On a ring, the radial unit vector \vec a_\rho of the source point (red) turns with \phi', whereas \vec a_x (blue) is the same everywhere. A constant vector may be taken out of an integral, but \vec a_\rho may not. Adding the four red arrows gives zero, adding the four blue ones gives 4\vec a_x. — **Figure 3.1.16.** Why we integrate in Cartesian unit vectors. On a ring, the radial unit vector $a_{ρ}$ of the source point (red) turns with $ϕ^{'}$ , whereas $a_{x}$ (blue) is the same everywhere. A constant vector may be taken out of an integral, but $a_{ρ}$ may not. Adding the four red arrows gives zero, adding the four blue ones gives $4 a_{x}$ .

Common mistake · Unit vectors inside the integral

Writing $\int a_{ρ} d ϕ^{'} = a_{ρ} \int d ϕ^{'} = 2 π a_{ρ}$ for a full ring is wrong: $a_{ρ} = cos ϕ^{'} a_{x} + sin ϕ^{'} a_{y}$ depends on $ϕ^{'}$ , and the correct result is $\int_{0}^{2 π} a_{ρ} d ϕ^{'} = 0$ . Expand every unit vector that belongs to the source point in $a_{x}, a_{y}, a_{z}$ before integrating. (A unit vector of the field point, which is fixed, may stay outside.)

Common mistake ·

d l^{'} = d ϕ^{'}

On a circle of radius $b$ the length element is $d l^{'} = b d ϕ^{'}$ , and on a disk the area element is $d S^{'} = ρ^{'} d ρ^{'} d ϕ^{'}$ . Forgetting the factor $b$ or $ρ^{'}$ gives a result with the wrong units: a quick unit check catches it.

3.1.5 The superposition principle