The electric field

Coulomb's law says what two charges do to each other and leaves an embarrassment behind it, which is that neither charge has any way of knowing the other is there.

The previous lesson ended on that point. This one takes Michael Faraday's way out, which was to insist that the space between the charges is not empty but is in a state, and that the state is the real object of study. The move looks at first like bookkeeping, and for static charges it is. It stops being bookkeeping in the ninth lesson, where a field turns up with no charge responsible for it, and it stops being anything but the whole subject in the twelfth.

Dividing out the test charge

Put a charge $q_{0}$ at some point near a fixed charge $Q$ and it feels a force $F = k Q q_{0} / r^{2}$ . Double $q_{0}$ and the force doubles. Halve it and the force halves. The force therefore contains a factor that has nothing to do with $Q$ or with the geometry, and dividing it out leaves something that depends only on the source and the point:

E = \frac{F}{q_{0}}

That is the electric field, and its units are newtons per coulomb, later shown to be the same thing as volts per metre. It is a vector at every point of space, which is what makes it a field rather than a number: a temperature field assigns a scalar to every point, and this assigns an arrow.

Two cautions come with the definition. The first is that $q_{0}$ has to be small. A real test charge repels the charges on the conductors producing the field, redistributing them and changing the very thing being measured, so the definition is properly $E = lim_{q_{0} \to 0} F / q_{0}$ . The limit is a fiction, since charge is quantised and cannot go below $e$ , but it is a harmless one: laboratory charges are $10^{12}$ elementary charges or more, so there is plenty of room to be small without being zero.

The second is that the field is defined by the force on a positive charge. Place a negative charge in a field and the force is opposite to $E$ , which is a permanent minor nuisance and the source of a great many sign errors.

For a point charge $Q$ the field follows at once by dividing Coulomb's law by $q_{0}$ :

E = \frac{k Q}{r^{2}}

pointing away from $Q$ if $Q$ is positive and towards it if negative. A charge of 1 nC produces 8.99 N/C at a metre. The proton in a hydrogen atom produces $k e / r^{2} = 5.15 \times 10^{11}$ N/C at the Bohr radius of 52.9 pm, which is why atomic physics needs no external fields to be interesting: nothing built in a laboratory comes within five orders of magnitude of that.

What the field is for

At this stage the field is a repackaging, and it is worth being honest that nothing has been explained yet. $E = F / q_{0}$ gives the same answers as Coulomb's law because it was derived from Coulomb's law by dividing.

The reason to prefer it is that the two-step story it tells, source makes field, field pushes charge, survives when the one-step story does not. If the source charge is jiggled, the force on a distant charge does not change instantly; the change propagates outward at a finite speed, and during the transit the field near the distant charge is what it was, still pushing. Something has to hold the information in the interval, and that something is the field. By the end of this course the field will be carrying energy and momentum in its own right, and a light wave will be a piece of field that has detached from its source entirely and no longer cares where it came from.

Superposition, and the picture

Fields add, because forces add. The field of several charges is the vector sum of the fields each would produce alone, and the fact that the sum is unaffected by the presence of the other charges is the linearity noted in the previous lesson.

Faraday's own way of seeing a field was lines of force: curves drawn everywhere tangent to $E$ , starting on positive charges and ending on negative ones. The convention that makes them quantitative is that the number of lines through a small area perpendicular to them is proportional to the field strength there, so lines crowd where the field is strong. Around an isolated point charge the lines are radial spokes, and the number crossing a sphere of radius $r$ is fixed while the sphere's area grows as $r^{2}$ , so the density falls as $1 / r^{2}$ , which is the field. The picture has the inverse square built into the geometry of three dimensions.

Two things the lines cannot do are worth stating, because both are commonly assumed. They never cross: the field at a point has one direction, so two tangents cannot pass through the same place. And they are not trajectories. A charge released in a field accelerates along the field, but it then has velocity, and its subsequent path curves away from the line, exactly as a projectile does not follow the direction of gravity.

Example. A charge $q_{1} = + 4.0$ nC sits at $x = 0$ and $q_{2} = - 6.0$ nC at $x = 8.0$ cm. What is the field at the midpoint, $x = 4.0$ cm?

Each is 4.0 cm from the point. The positive charge pushes a test charge away from itself, that is in the $+ x$ direction, with $E_{1} = (8.9876 \times 10^{9}) (4.0 \times 10^{- 9}) / (0.040)^{2} = 2.25 \times 10^{4}$ N/C. The negative charge pulls a test charge towards itself, which from the midpoint is also the $+ x$ direction, with $E_{2} = (8.9876 \times 10^{9}) (6.0 \times 10^{- 9}) / (0.040)^{2} = 3.37 \times 10^{4}$ N/C. They point the same way, so $E = 5.62 \times 10^{4}$ N/C in the $+ x$ direction.

Now you. A charge $q_{1} = + 3.0$ nC sits at $x = 0$ and $q_{2} = + 5.0$ nC at $x = 10.0$ cm. What is the field at the midpoint, $x = 5.0$ cm?

Answer

Both are positive, so their fields at the midpoint oppose. From $q_{1}$ : $(8.9876 \times 10^{9}) (3.0 \times 10^{- 9}) / (0.050)^{2} = 1.08 \times 10^{4}$ N/C in $+ x$ . From $q_{2}$ : $(8.9876 \times 10^{9}) (5.0 \times 10^{- 9}) / (0.050)^{2} = 1.80 \times 10^{4}$ N/C in $- x$ . The net field is $7.19 \times 10^{3}$ N/C in the $- x$ direction, towards the smaller charge.

Continuous charge, and the integral it forces

Real charged objects are not points. A charged rod holds something like $10^{13}$ elementary charges spread along it, and adding $10^{13}$ vectors is not a plan. The standard move is to go the other way and treat the charge as a continuum, cut it into pieces $d q$ small enough to count as points, and integrate.

Three densities cover the cases. A line carries $λ$ coulombs per metre, so $d q = λ d s$ . A surface carries $σ$ coulombs per square metre, so $d q = σ d A$ . A volume carries $ρ$ coulombs per cubic metre, so $d q = ρ d V$ . In each case the contribution of one piece to the field at the point of interest is $d E = k d q / r^{2}$ directed from the piece to the point, and the total is the vector sum, which means integrating the components separately.

The component bookkeeping is the part that makes these integrals painful, and the part that symmetry usually rescues. The uniformly charged ring is the cleanest illustration.

Take a ring of radius $a$ carrying total charge $Q$ spread evenly, and ask for the field at a point on its axis a distance $x$ from the centre. Every piece $d q$ of the ring is the same distance $\sqrt{x^{2} + a^{2}}$ from that point, so each contributes $k d q / (x^{2} + a^{2})$ . The directions differ, but for each piece there is a piece diametrically opposite whose contribution has the same axial component and the opposite transverse one, so the transverse components cancel in pairs and only the axial survives. The axial fraction of each contribution is $\cos θ = x / \sqrt{x^{2} + a^{2}}$ , so

E = \int \frac{k d q}{x^{2} + a^{2}} \cdot \frac{x}{\sqrt{x^{2} + a^{2}}} = \frac{k Q x}{(x^{2} + a^{2})^{3 / 2}}

where everything except $d q$ came out of the integral because it is the same for every piece, and $\int d q = Q$ .

Read the answer at its two extremes, which is always worth doing. At the centre, $x = 0$ , the field is zero, as symmetry demands: every piece is opposed by the piece across the ring. Far away, $x ≫ a$ , the $a^{2}$ is negligible and $E \to k Q / x^{2}$ , which is a point charge, as it must be. In between the field rises, peaks, and falls, and differentiating shows the peak sits at $x = a / \sqrt{2}$ .

Example. A ring of radius 5.0 cm carries 20 nC spread uniformly. What is the field on its axis 12 cm from the centre?

$x^{2} + a^{2} = (0.12)^{2} + (0.05)^{2} = 0.0144 + 0.0025 = 0.0169$ m², whose square root is exactly 0.13 m, so $(x^{2} + a^{2})^{3 / 2} = {0.13}^{3} = 2.197 \times 10^{- 3}$ . Then $E = (8.9876 \times 10^{9}) (20 \times 10^{- 9}) (0.12) / (2.197 \times 10^{- 3}) = 9.82 \times 10^{3}$ N/C, directed along the axis away from the ring.

Now you. A ring of radius 4.0 cm carries 15 nC. What is the field on its axis 3.0 cm from the centre?

Answer

$x^{2} + a^{2} = 0.0009 + 0.0016 = 0.0025$ , so the square root is 0.050 and the cube of that is $1.25 \times 10^{- 4}$ . Then $E = (8.9876 \times 10^{9}) (15 \times 10^{- 9}) (0.030) / (1.25 \times 10^{- 4}) = 3.24 \times 10^{4}$ N/C.

The line of charge, and a first infinity

Now a straight rod of length $2 L$ carrying $λ$ per metre, with the field wanted at a perpendicular distance $y$ from its midpoint. Put the rod along the $z$ axis from $- L$ to $+ L$ . A piece at height $z$ is $\sqrt{y^{2} + z^{2}}$ away and contributes $k λ d z / (y^{2} + z^{2})$ , of which the fraction $y / \sqrt{y^{2} + z^{2}}$ points along the perpendicular. Components along the rod cancel between $+ z$ and $- z$ , so

E = \int_{- L}^{L} \frac{k λ y d z}{(y^{2} + z^{2})^{3 / 2}} = \frac{2 k λ L}{y \sqrt{y^{2} + L^{2}}}

using the standard integral $\int d z / (y^{2} + z^{2})^{3 / 2} = z / (y^{2} \sqrt{y^{2} + z^{2}})$ , which can be checked by differentiating the right side.

Two limits again. For $y ≫ L$ the rod looks like a point of charge $2 λ L$ and the expression collapses to $k (2 λ L) / y^{2}$ . For $L ≫ y$ , meaning very close to a long rod, $\sqrt{y^{2} + L^{2}} \to L$ and

E = \frac{2 k λ}{y} = \frac{λ}{2 π ε_{0} y}

which falls off as $1 / y$ rather than $1 / y^{2}$ . That is not a failure of Coulomb's law. It is what happens when moving away from a line brings more of the line into view at a shallow angle, partly compensating the loss. The same logic will shortly give a field that does not fall off at all.

An infinite rod does not exist, and the result is still the one to use in practice, because "infinite" here means only $L ≫ y$ . At 1 cm from a 1 m rod the finite formula and the infinite one differ by about 0.02 per cent.

The dipole

Two equal and opposite charges $+ q$ and $- q$ separated by a small distance $d$ form an electric dipole, the most important arrangement in the subject after the point charge, because a neutral molecule with its charge slightly off-centre is one. Water is the standard example, with a dipole moment of $6.2 \times 10^{- 30}$ coulomb metres, and essentially all of the chemistry of water follows from it.

Define the dipole moment $p = q d$ , treated as a vector pointing from the negative charge to the positive one. On the axis, at a distance $r$ from the centre with $r ≫ d$ , the two fields nearly cancel. Writing them out, $E = k q / (r - d / 2)^{2} - k q / (r + d / 2)^{2}$ , and expanding each denominator to first order in $d / r$ gives

E = \frac{2 k p}{r^{3}}

The cancellation costs a power of $r$ : a dipole's field dies as the inverse cube, faster than a point charge's, because at a distance the two charges are increasingly well disguised as nothing. That single fact explains why neutral matter can be handled at all. If molecules produced $1 / r^{2}$ fields, every object in a room would be electrically screaming at every other.

Put a dipole in a uniform external field $E$ and the two forces, $+ q E$ and $- q E$ , cancel exactly, so there is no net force. There is a net torque, since the forces act at different places, of magnitude

τ = p E \sin θ

with $θ$ the angle between $p$ and $E$ . The torque turns the dipole into alignment with the field, which is why a charged rod attracts an uncharged scrap of paper: the field polarises the paper, aligning and slightly separating its molecular charges, and then, because the field is not uniform near the rod, the induced near end feels a stronger pull than the far end repels. The attraction of neutral matter to charge is a second-order effect requiring a non-uniform field, and it is the reason the very first observation in the subject, amber picking up chaff, took two thousand years to explain.

Motion in a field

Once $E$ is known, mechanics takes over: the force on a charge is $q E$ , and the rest is Newton's second law. In a uniform field the acceleration is constant, so the constant-acceleration results from mechanics apply unchanged, and a charged particle fired across a uniform field follows a parabola for exactly the reason a projectile does.

Example. An electron starts from rest and is accelerated through 1.0 cm by a uniform field of $1.50 \times 10^{4}$ N/C. How fast is it going?

The force is $e E = (1.602 \times 10^{- 19}) (1.50 \times 10^{4}) = 2.40 \times 10^{- 15}$ N, so the acceleration is $a = e E / m_{e} = 2.40 \times 10^{- 15} / 9.109 \times 10^{- 31} = 2.64 \times 10^{15}$ m/s². With $v^{2} = 2 a d$ and $d = 0.010$ m, $v^{2} = 5.28 \times 10^{13}$ and $v = 7.26 \times 10^{6}$ m/s. That is 2.4 per cent of the speed of light, so treating it non-relativistically is good to about a tenth of a per cent, and a slightly larger field would make that excuse fail.

Now you. An electron starts from rest and is accelerated through 2.5 cm by a uniform field of $8.0 \times 10^{3}$ N/C. How fast is it going?

Answer

$a = e E / m_{e} = (1.602 \times 10^{- 19}) (8.0 \times 10^{3}) / (9.109 \times 10^{- 31}) = 1.41 \times 10^{15}$ m/s². Then $v^{2} = 2 a d = 2 (1.41 \times 10^{15}) (0.025) = 7.04 \times 10^{13}$ , so $v = 8.39 \times 10^{6}$ m/s.

Gravity does not appear in that calculation, and the omission is not laziness. The field needed to hold an electron up against its own weight is $m_{e} g / e = 5.6 \times 10^{- 11}$ N/C, which is eleven orders of magnitude below any field a laboratory can avoid producing by accident. In charged-particle problems, gravity is noise.

Where the integrals run out

The method of this lesson is complete in principle: any static charge distribution, cut into pieces, integrated. In practice it collapses almost immediately. The ring was easy because every piece sat at the same distance. Move off the axis and the same ring gives an integral with no expression in elementary functions. A uniformly charged sphere, the single most important case in the subject, can be done this way, but it takes a page of unpleasant work with the cosine rule.

What rescues it is that the difficulty is in the wrong place. The integral is hard because it tracks where the field points, and yet the ring, the sphere and the infinite line all have enough symmetry that the direction of the field is obvious before any calculation begins. A method that used the symmetry first and asked only for the magnitude would be doing a fraction of the work.

Such a method exists, it takes a page to derive, it converts Coulomb's inverse square into a statement about closed surfaces, and it solves the sphere in three lines. It is Gauss's law, and it is next.