Unit 1

Electric Charge & Electric Fields

Every circuit, chip, motor and antenna you will ever design runs on one fact: charges push and pull on each other. This unit builds the language for that push and pull. We start with charge itself, write the force between two charges (Coulomb's law), and then make the big conceptual move of the course: instead of asking “what force does charge A put on charge B?” we ask “what does charge A do to the space around it?” The answer is the electric field, and from here to Maxwell's equations in Unit 9 the field is the main character. The lab lets you chop a charged rod, ring or disk into little pieces and watch their fields add up, arrow by arrow, at a point you drag around.

1 · Electric charge

Rub a balloon on your hair and it sticks to the wall. Something has been moved from one object to the other, and that something is electric charge. Benjamin Franklin named the two kinds positive and negative; the names are arbitrary, but the behaviour is not:

Like charges repel, unlike charges attract.
Ordinary matter is made of positive protons (in nuclei) and negative electrons (around them). In solids it is almost always electrons that move; the nuclei stay put. “Positively charged” usually means “missing some electrons”.
The SI unit of charge is the coulomb (C). One coulomb is a lot: a static shock moves around \(10^{-8}\) to \(10^{-6}\) C, so you will see nC (\(10^{-9}\)) and \(\mu\)C (\(10^{-6}\)) constantly.

Charge is quantized

\[ q = n e,\qquad n = 0, \pm1, \pm2, \dots,\qquad e = 1.60\times10^{-19}\ \text{C} \] \(e\) is the elementary charge: the magnitude of the charge on one proton or one electron (the electron has \(-e\)). Every free charge is a whole number of \(e\). (Quarks carry \(\pm\tfrac13 e\) and \(\pm\tfrac23 e\), but they are never found alone.) Because \(e\) is so tiny, 1 nC is \(10^{-9}/(1.60\times10^{-19}) = 6.25\times10^{9}\) electrons — so for everyday amounts we treat charge as a smooth, continuous quantity.

Charge is conserved

The total charge of an isolated system never changes. Rubbing does not create charge; it moves electrons, so if the balloon gains \(-5\) nC your hair gains \(+5\) nC. Even in particle physics, where particles are created and destroyed, the net charge before equals the net charge after.

EE connection

Kirchhoff's current law (Unit 5) — “the current into a node equals the current out” — is nothing but charge conservation. A circuit node cannot store charge, so whatever flows in per second must flow out per second. You will use this law thousands of times as an engineer; this is where it comes from.

Conductors and insulators

In a conductor (copper, aluminium, salt water) some electrons are free to wander through the whole material. In an insulator (glass, plastic, dry air) every electron is tied to its own atom. Semiconductors (silicon, germanium) sit in between, and — the basis of all modern electronics — their number of free carriers can be controlled by doping, light, temperature and electric fields.

Two ways to charge something follow from this. By contact: touch a charged object to a conductor and charge flows across; both end with the same sign. By induction: bring a charged rod near (not touching) a conductor, so its free electrons shift toward or away from the rod; connect the far side to ground, then break the ground connection before removing the rod. The conductor is left with the opposite sign to the rod. Even a neutral object is attracted to a charged one: the nearby side becomes oppositely charged (it is polarized), and the nearer, opposite charge wins because force grows as distance shrinks. That is why the balloon sticks to a neutral wall.

EE connection

Electrostatic discharge (ESD). Walking across a carpet can charge you to thousands of volts. Touch a circuit board and that charge rushes through whatever path it finds — and the gate insulator of a MOSFET is only a few nanometres thick. This is why chips ship in conductive bags, why you wear a grounded wrist strap at the bench, and why every chip has ESD protection diodes on its pins.

2 · Coulomb's law

In 1785 Charles-Augustin de Coulomb measured the force between two small charged spheres with a torsion balance. The force is proportional to each charge, falls off as the square of the distance, and acts along the line joining them. Start with the magnitude:

\[ F = k\,\frac{|q_1|\,|q_2|}{r^2},\qquad k = \frac{1}{4\pi\varepsilon_0} = 8.99\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2 \] \(q_1, q_2\) are the charges (C), \(r\) the distance between them (m), and \(F\) the force each exerts on the other (N). \(k\) is the Coulomb constant; \(\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)\) is the permittivity of free space. They carry the same information; \(k\) is handy for point charges, \(\varepsilon_0\) makes the Gauss's-law formulas of Unit 2 cleaner.

Force is a vector, so we need a direction too. Let \(\hat{\mathbf r}\) (“r-hat”) be the unit vector pointing from the source charge to the charge feeling the force. Then one formula handles attraction and repulsion automatically:

\[ \mathbf{F}_{\text{on }2\text{ by }1} = k\,\frac{q_1 q_2}{r^2}\,\hat{\mathbf r}_{12}, \qquad \hat{\mathbf r}_{12} = \frac{\mathbf r_2 - \mathbf r_1}{|\mathbf r_2 - \mathbf r_1|} \] Here the charges keep their signs. If \(q_1q_2 > 0\) (like charges) the force on 2 points along \(\hat{\mathbf r}_{12}\), away from 1: repulsion. If \(q_1q_2 < 0\) it points back toward 1: attraction. Newton's third law holds: \(\mathbf F_{\text{on }1\text{ by }2} = -\mathbf F_{\text{on }2\text{ by }1}\).

The geometry of Coulomb's law. \(\mathbf r\) runs from the source charge \(q_1\) to the charge feeling the force, \(q_2\); \(\hat{\mathbf r}\) is that direction with length 1. The sign of \(q_1q_2\) decides whether the force points along \(\hat{\mathbf r}\) or against it.

Common mistakes

Double-counting signs. Either use magnitudes in \(k|q_1||q_2|/r^2\) and pick the direction from “like repel, unlike attract”, or keep signs and use \(\hat{\mathbf r}\) carefully. Mixing the two methods is the single most common way to get a force backwards.
Forgetting to square \(r\), or squaring it in centimetres. Convert to metres first: \(3\) cm is \(0.03\) m and \(r^2 = 9\times10^{-4}\ \text{m}^2\).
Adding magnitudes of forces from different charges. Forces are vectors; add components (Section 3).

3 · Superposition

What if there are many charges? Experiment says the force from each one is completely unaffected by the presence of the others. So you compute each pairwise force as if the others were not there, and add the vectors:

\[ \mathbf F_{\text{on }q} = \mathbf F_{1} + \mathbf F_{2} + \cdots + \mathbf F_{N} = \sum_{i=1}^{N} k\,\frac{q_i\,q}{r_i^2}\,\hat{\mathbf r}_i \] \(r_i\) is the distance from charge \(q_i\) to \(q\), and \(\hat{\mathbf r}_i\) the unit vector from \(q_i\) toward \(q\). This is the principle of superposition. It is why electrostatics is “linear” — double all the sources and every force doubles. (It is also why the superposition theorem works in linear circuits.)

Recipe for vector sums

Draw the charges and the point of interest. Draw each force arrow, with the right direction, before you calculate anything.
Find each magnitude with \(k|q_i||q|/r_i^2\).
Split each arrow into \(x\) and \(y\) components using the geometry (often a 3-4-5 triangle or a symmetric angle).
Add all \(x\) components, add all \(y\) components. Then \(|\mathbf F| = \sqrt{F_x^2 + F_y^2}\), direction \(\tan\theta = F_y/F_x\).
Look for symmetry first: components that cancel by symmetry need not be computed.

4 · The electric field

Coulomb's law looks like “action at a distance”: charge 1 somehow reaches across empty space to push charge 2. The field picture says something different and more useful. Charge 1 changes the space around it, setting up an electric field at every point; charge 2 then responds to the field where it is. To measure the field, place a tiny positive test charge \(q_0\) at the point and see what force it feels:

\[ \mathbf E = \frac{\mathbf F_{\text{on }q_0}}{q_0}\qquad\Longleftrightarrow\qquad \mathbf F = q\,\mathbf E \] \(\mathbf E\) is the force per unit charge, unit N/C (in Unit 3 you will see N/C is the same as V/m, the unit engineers use). The test charge is “tiny” so it does not disturb the charges making the field. Once you know \(\mathbf E\), the force on any charge \(q\) placed there is \(q\mathbf E\): along \(\mathbf E\) for positive \(q\), opposite to \(\mathbf E\) for negative \(q\) (electrons!).

Dividing Coulomb's law by \(q_0\) gives the field of a point charge:

\[ \mathbf E = k\,\frac{q}{r^2}\,\hat{\mathbf r},\qquad E = k\,\frac{|q|}{r^2} \] \(\hat{\mathbf r}\) points from the source charge \(q\) to the field point. For \(q > 0\) the field points away from the charge; for \(q < 0\) it points toward it. Fields from several sources superpose: \(\mathbf E = \sum_i k q_i\hat{\mathbf r}_i/r_i^2\).

Common mistake

A charge does not feel its own field. When you compute the force on \(q\), the field you use is the field made by all the other charges at \(q\)'s location.

EE connection

In a MOSFET, a voltage of about 1 V sits across a gate insulator only about 2 nm thick. That is a field of roughly \(1/(2\times10^{-9}) = 5\times10^{8}\) V/m — a hundred times what makes air spark. This field reaches through the insulator and pulls electrons into a thin channel in the silicon, turning the transistor on. “Field-effect transistor” means exactly what it says.

5 · Field lines

A field is an arrow at every point in space — impossible to draw completely. Michael Faraday's trick is to draw field lines: curves whose tangent at each point is the direction of \(\mathbf E\). The rules:

Lines start on positive charges and end on negative charges (or run off to infinity).
The number of lines leaving or entering a charge is proportional to its magnitude: draw twice as many from \(2q\) as from \(q\).
Where lines are close together the field is strong; where they spread apart it is weak.
Lines never cross: the field has only one direction at each point.

Field lines of a dipole (equal and opposite charges, Section 7). The field is strongest between and next to the charges, where the lines crowd together. At any point, \(\mathbf E\) is tangent to the line through it.

Common mistake

Field lines are not trajectories. A charge's acceleration is along the line, but its velocity need not be, so a moving charge generally cuts across curved field lines. Only a charge starting from rest in a field with straight lines moves exactly along one.

6 · Continuous charge distributions

A charged rod, a charged ring, the charge on a metal plate: these contain \(10^{10}\) or more electrons, so we treat the charge as smeared out smoothly. The idea is still superposition — just with infinitely many infinitely small pieces, so the sum becomes an integral.

\[ \mathbf E = \int k\,\frac{dq}{r^2}\,\hat{\mathbf r} \] Chop the object into tiny pieces of charge \(dq\). Each acts like a point charge and makes a field \(d\mathbf E = k\,dq\,\hat{\mathbf r}/r^2\) at the field point, where \(r\) is the distance from that piece to the point. Add (integrate) all the \(d\mathbf E\)'s — as vectors.

Charge densities

Object	Density	Unit	Piece of charge	Uniform case
Line, rod, wire, ring	\(\lambda\) (lambda)	C/m	\(dq = \lambda\,dx\) or \(\lambda\,ds\)	\(\lambda = Q/L\)
Surface, sheet, disk	\(\sigma\) (sigma)	C/m²	\(dq = \sigma\,dA\)	\(\sigma = Q/A\)
Volume, solid ball	\(\rho\) (rho)	C/m³	\(dq = \rho\,dV\)	\(\rho = Q/V\)

The set-up recipe

Draw the object, the field point P, and one typical piece \(dq\) at a general position (not at a special spot like the end or centre).
Write \(dq\) in terms of a coordinate: \(\lambda\,dx\), \(\lambda\,a\,d\theta\), \(\sigma\,2\pi s\,ds\), …
Write \(r\), the distance from \(dq\) to P, in terms of the same coordinate (Pythagoras, usually).
Use symmetry to decide which components of \(d\mathbf E\) cancel. Keep only the surviving component — typically by multiplying \(dE\) by a \(\cos\theta\) that you also write in terms of the coordinate.
Integrate over the object with the right limits. Constants (\(k\), \(\lambda\), the distance of P) come outside.
Check: right units? Far away, does it look like \(kQ/r^2\)?

Common mistake

Integrating the magnitudes \(dE\) without components. The pieces of a ring push P in different directions, and their sideways parts cancel. If you integrate \(k\,dq/r^2\) directly you get \(kQ/r^2\) — the wrong answer for every point except very far away. Always ask “which component survives?” before integrating.

A uniformly charged rod, point on its axis

Rod of length \(L\) and charge \(Q\) along the \(x\)-axis from \(x = 0\) to \(x = L\); P on the axis at distance \(d\) beyond the end, i.e. at \(x = L + d\). Every \(d\mathbf E\) points along \(+x\) (no cancellation needed), \(dq = \lambda\,dx\), \(r = L + d - x\):

\[ E = \int_0^L \frac{k\lambda\,dx}{(L+d-x)^2} = k\lambda\left[\frac{1}{L+d-x}\right]_0^L = k\lambda\left(\frac1d - \frac{1}{L+d}\right) = \frac{kQ}{d\,(d+L)} \] Check: when \(d \gg L\), \(d(d+L)\approx d^2\) and \(E \approx kQ/d^2\) — a point charge, as it must be from far away.

A uniformly charged rod, point on its perpendicular bisector

Now the rod runs from \(x = -L/2\) to \(L/2\) and P is at height \(y\) above its centre. Pieces at \(+x\) and \(-x\) have equal and opposite \(x\)-components, which cancel; only \(E_y\) survives, with \(\cos\theta = y/r\) and \(r = \sqrt{x^2+y^2}\):

\[ E_y = \int_{-L/2}^{L/2} \frac{k\lambda\,dx}{x^2+y^2}\cdot\frac{y}{\sqrt{x^2+y^2}} = \frac{kQ}{y\sqrt{y^2 + L^2/4}} \] The integral \(\int dx/(x^2+y^2)^{3/2} = x/(y^2\sqrt{x^2+y^2})\) is a standard trig-substitution result (\(x = y\tan\theta\)). Far away (\(y\gg L\)) this is \(kQ/y^2\) again.

The infinite line of charge (a limit)

Let the rod become very long, \(L\to\infty\), with \(\lambda = Q/L\) fixed. Then \(\sqrt{y^2+L^2/4}\to L/2\) and \(Q/(L/2) = 2\lambda\):

\[ E = \frac{2k\lambda}{r} = \frac{\lambda}{2\pi\varepsilon_0\, r}\qquad(\text{infinite line, distance } r) \] Pointing radially away from the line (for \(\lambda > 0\)). Note \(1/r\), not \(1/r^2\): a line spreads its field over a cylinder, not a sphere. Unit 2 gets this in two lines with Gauss's law. It is a good model whenever you are much closer to a wire than to its ends.

A uniformly charged ring, point on its axis

A ring of radius \(a\) and total charge \(Q\), P on the axis at distance \(z\) from the centre. Every piece \(dq\) is the same distance \(r = \sqrt{z^2+a^2}\) from P. The piece directly opposite, \(dq'\), has a mirror-image field: the components perpendicular to the axis cancel in pairs, and only the axial components (fraction \(\cos\theta = z/r\)) survive.

\[ E_z = \int \frac{k\,dq}{z^2+a^2}\cdot\frac{z}{\sqrt{z^2+a^2}} = \frac{k z}{(z^2+a^2)^{3/2}}\int dq = \frac{kQz}{(z^2+a^2)^{3/2}} \] Everything except \(dq\) is the same for every piece, so it comes out of the integral, and \(\int dq = Q\). Checks: at the centre (\(z = 0\)) \(E = 0\) — every piece is cancelled by the one opposite. For \(z\gg a\), \(E\approx kQ/z^2\). In between, \(E\) peaks at \(z = a/\sqrt2\) (set \(dE_z/dz = 0\)).

A uniformly charged disk, point on its axis

Build a disk of radius \(R\) out of thin rings. The ring of radius \(s\) and width \(ds\) has area \(2\pi s\,ds\) and charge \(dq = \sigma\,2\pi s\,ds\). Use the ring result for each one and add the rings:

\[ E_z = \int_0^R \frac{k z\,\sigma\,2\pi s\,ds}{(z^2+s^2)^{3/2}} = 2\pi k\sigma z\left[\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}\right] = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{z}{\sqrt{z^2+R^2}}\right) \] (For \(z > 0\); we used \(2\pi k = 1/(2\varepsilon_0)\).) This is the “build big things from things you already solved” strategy: a disk is rings, a ring is points.

The infinite sheet (a limit)

\[ R\to\infty:\qquad E = \frac{\sigma}{2\varepsilon_0}\qquad(\text{infinite sheet, either side}) \] Independent of distance! Moving away from an infinite sheet does not weaken its field, because you see ever more of the sheet at a shallow angle. Real plates behave like this when you are much closer to them than to their edges — the key to the parallel-plate capacitor of Unit 4.

Source	Field	How it falls off
Point charge	\(kQ/r^2\)	\(1/r^2\)
Ring, on axis	\(kQz/(z^2+a^2)^{3/2}\)	0 at centre, \(1/z^2\) far
Infinite line	\(\lambda/(2\pi\varepsilon_0 r)\)	\(1/r\)
Infinite sheet	\(\sigma/(2\varepsilon_0)\)	constant
Dipole, far away	\(\sim kp/r^3\)	\(1/r^3\)

Interactive lab Fields of continuous charge Open it in Interactive Laboratories →

EE connection

A long straight trace on a circuit board, a power line and the centre conductor of a coaxial cable are all “lines of charge” to a good approximation, and the \(1/r\) field around them sets how strongly neighbouring traces couple (crosstalk) and how much insulation a high-voltage line needs. The sheet result is the starting point for every capacitor calculation.

7 · Electric dipoles

An electric dipole is a pair of equal and opposite charges \(+q\) and \(-q\) separated by a small distance \(d\). Dipoles are everywhere: a water molecule is one (its oxygen end is slightly negative), and any neutral atom becomes one when a field tugs its electrons one way and its nucleus the other. The whole dipole is described by one vector:

\[ \mathbf p = q\,\mathbf d,\qquad p = qd\quad(\text{unit: C}\cdot\text{m}) \] The dipole moment. \(\mathbf d\) points from \(-q\) to \(+q\). (Chemists sometimes draw the opposite arrow; physics and engineering use this one.)

The field far from a dipole

\[ \text{On the axis: } E \approx \frac{2kp}{r^3}\ (\text{along } \mathbf p),\qquad \text{on the perpendicular bisector: } E \approx \frac{kp}{r^3}\ (\text{opposite } \mathbf p) \] Valid for \(r \gg d\). On the axis, for example, \(E = kq\left[\frac{1}{(r-d/2)^2} - \frac{1}{(r+d/2)^2}\right] = \frac{2kqdr}{(r^2-d^2/4)^2}\approx\frac{2kp}{r^3}\). The fields of \(+q\) and \(-q\) almost cancel, leaving a remainder that falls off one power faster than a single charge: \(1/r^3\).

A dipole in a uniform field

In a uniform field the forces on the two ends are equal and opposite, so the net force is zero — but they do not act along the same line, so they twist the dipole toward alignment with \(\mathbf E\).

\[ \boldsymbol\tau = \mathbf p\times\mathbf E,\qquad \tau = pE\sin\theta \] \[ U = -\mathbf p\cdot\mathbf E = -pE\cos\theta \] \(\theta\) is the angle from \(\mathbf p\) to \(\mathbf E\). The torque \(\boldsymbol\tau\) (N·m) rotates \(\mathbf p\) toward \(\mathbf E\). The potential energy \(U\) (J) is lowest, \(-pE\), when \(\mathbf p\) is aligned with \(\mathbf E\) (stable equilibrium), and highest, \(+pE\), when anti-aligned (unstable). Turning a dipole from aligned to anti-aligned takes work \(2pE\). (Zero of \(U\) is chosen at \(\theta = 90^\circ\).)

Common mistake

“No net force, so nothing happens.” Wrong: zero net force and zero net torque are separate conditions. In a non-uniform field there is also a net force (the end in the stronger field wins) — which is how a charged comb picks up neutral bits of paper: it polarizes them, then pulls the induced dipoles into the stronger field near the comb.

EE connection

A microwave oven makes the field oscillate \(2.45\times10^{9}\) times per second; water dipoles keep trying to follow, and the jostling heats the food. Dipoles aligning in a field are also what makes a dielectric increase a capacitor's capacitance (Unit 4). And the simplest radio antenna is literally called a dipole antenna: charge sloshing between its two halves is an oscillating \(\mathbf p\), which radiates (Unit 9).

8 · Charges moving in a uniform field

Between two large, parallel, oppositely charged plates the field is (nearly) uniform: same magnitude and direction everywhere. A charge \(q\) of mass \(m\) there feels a constant force, so it has a constant acceleration — exactly like a ball under gravity, with \(g\) replaced by \(qE/m\):

\[ \mathbf a = \frac{q\,\mathbf E}{m} \] Newton's second law with \(\mathbf F = q\mathbf E\). For a positive charge \(\mathbf a\) is along \(\mathbf E\); for an electron it is opposite. All the constant-acceleration kinematics from Physics I apply unchanged: \(v = v_0 + at\), \(x = x_0 + v_0t + \tfrac12at^2\), \(v^2 = v_0^2 + 2a\Delta x\).

An electron fired at speed \(v_0\) between deflection plates. Horizontally nothing acts, so it crosses the plates in time \(t = L/v_0\). Vertically it accelerates at \(eE/m\) — against \(\mathbf E\), because its charge is negative. The result is a parabola, exactly like a horizontally thrown ball.

\[ y = \frac12\,\frac{eE}{m}\left(\frac{L}{v_0}\right)^2 = \frac{eEL^2}{2mv_0^2},\qquad \tan\phi = \frac{v_y}{v_0} = \frac{eEL}{mv_0^2} \] \(y\) is the sideways deflection on leaving the plates and \(\phi\) the exit angle. Deflection is proportional to \(E\) — so by putting the signal you want to display on the plates, the spot on the screen traces the signal.

Ignore gravity (for particles)

For an electron in even a weak field of 100 N/C, \(eE = 1.6\times10^{-17}\) N while \(m_eg \approx 8.9\times10^{-30}\) N — gravity is about \(10^{12}\) times smaller. For electrons, protons and ions, drop gravity. For charged droplets, compare the two first.

EE connection

This is exactly how an analog cathode-ray oscilloscope drew its trace: one pair of plates deflected the beam horizontally with a sweep voltage, the other vertically with your signal. The same physics steers ink in continuous inkjet printers and accelerates electrons in X-ray tubes.

9 · Worked examples

1. Field of two point charges (superposition)

\(q_1 = +2.0\ \mu\text{C}\) sits at the origin and \(q_2 = -3.0\ \mu\text{C}\) at \((0.30\text{ m}, 0)\). Find \(\mathbf E\) at P \(= (0, 0.40\text{ m})\), and the force on a \(+1.0\) nC charge placed there.

From \(q_1\): distance 0.40 m, positive charge, so \(\mathbf E_1\) points away from it: straight up.

\[ E_1 = \frac{(8.99\times10^{9})(2.0\times10^{-6})}{(0.40)^2} = 1.124\times10^{5}\ \text{N/C},\qquad \mathbf E_1 = (0,\ 1.124\times10^{5})\ \text{N/C} \]

From \(q_2\): distance \(\sqrt{0.30^2+0.40^2} = 0.50\) m (a 3-4-5 triangle). Negative charge, so \(\mathbf E_2\) points from P toward \(q_2\): direction \((0.30, -0.40)/0.50 = (0.6, -0.8)\).

\[ E_2 = \frac{(8.99\times10^{9})(3.0\times10^{-6})}{(0.50)^2} = 1.079\times10^{5}\ \text{N/C},\qquad \mathbf E_2 = (6.47\times10^{4},\ -8.63\times10^{4})\ \text{N/C} \]

Add components:

\[ E_x = 6.47\times10^{4},\quad E_y = 1.124\times10^{5} - 8.63\times10^{4} = 2.61\times10^{4}\ \text{N/C} \] \[ |\mathbf E| = \sqrt{(6.47\times10^4)^2 + (2.61\times10^4)^2} = 6.98\times10^{4}\ \text{N/C},\qquad \theta = \tan^{-1}\frac{2.61}{6.47} = 21.9^\circ \text{ above } +x \] \[ F = qE = (1.0\times10^{-9})(6.98\times10^{4}) = 6.98\times10^{-5}\ \text{N, in the same direction} \]

The drawing-first step pays off here: \(\mathbf E_2\) has a large downward component that partly cancels \(\mathbf E_1\). Adding magnitudes (\(2.2\times10^5\)) would be wrong by a factor of three.

2. A charged rod versus a point charge

A thin rod 0.20 m long carries 4.0 nC spread uniformly. Find \(E\) on its axis 0.10 m beyond one end. Compare with treating the rod as a point charge at its centre.

\[ E = \frac{kQ}{d(d+L)} = \frac{(8.99\times10^{9})(4.0\times10^{-9})}{(0.10)(0.30)} = \frac{35.96}{0.030} = 1.20\times10^{3}\ \text{N/C} \]

Directed along the axis, away from the rod. Point-charge estimate (distance 0.20 m to the centre):

\[ E_{\text{pt}} = \frac{35.96}{(0.20)^2} = 899\ \text{N/C} \]

The real field is 33% larger: the near half of the rod is much closer than the centre and, because of \(1/r^2\), contributes disproportionately. The point-charge model is only good when you are far away compared with the object's size.

3. Field on the axis of a ring

A ring of radius \(a = 5.0\) cm carries \(Q = 10\) nC. Find \(E\) on the axis 12 cm from the centre, and where on the axis the field is largest.

\[ \sqrt{z^2+a^2} = \sqrt{0.12^2 + 0.050^2} = \sqrt{0.0169} = 0.130\ \text{m} \] \[ E = \frac{kQz}{(z^2+a^2)^{3/2}} = \frac{(8.99\times10^{9})(10\times10^{-9})(0.12)}{(0.130)^3} = \frac{10.79}{2.197\times10^{-3}} = 4.91\times10^{3}\ \text{N/C} \]

Directed along the axis away from the ring. Maximum: differentiate \(z(z^2+a^2)^{-3/2}\):

\[ \frac{d}{dz}\left[\frac{z}{(z^2+a^2)^{3/2}}\right] = \frac{(z^2+a^2) - 3z^2}{(z^2+a^2)^{5/2}} = 0\ \Rightarrow\ z = \frac{a}{\sqrt2} = 3.54\ \text{cm} \]

There \(E_{\max} = \dfrac{2}{3\sqrt3}\dfrac{kQ}{a^2} = 0.385\times\dfrac{89.9}{0.0025} = 1.38\times10^{4}\) N/C. The field is zero at the centre, rises to this peak, then falls off like \(1/z^2\) — try it in the lab.

4. A finite disk versus an infinite sheet

A plastic disk of radius 10 cm has \(\sigma = 5.0\ \mu\text{C/m}^2\). Find \(E\) on the axis 2.0 cm from the disk, and the error made by treating it as an infinite sheet.

\[ \frac{\sigma}{2\varepsilon_0} = \frac{5.0\times10^{-6}}{2(8.85\times10^{-12})} = 2.82\times10^{5}\ \text{N/C} \] \[ 1 - \frac{z}{\sqrt{z^2+R^2}} = 1 - \frac{0.020}{\sqrt{0.0004 + 0.0100}} = 1 - \frac{0.020}{0.1020} = 0.804 \] \[ E = (2.82\times10^{5})(0.804) = 2.27\times10^{5}\ \text{N/C} \]

The infinite-sheet value \(2.82\times10^{5}\) is about 24% too high at this distance (\(z/R = 0.2\)). Closer in, say \(z/R = 0.01\), the error is only 1%. Rule of thumb: “infinite” means the distance is small compared with the size.

5. A water molecule in a field

A water molecule has \(p = 6.2\times10^{-30}\) C·m. It sits in a uniform field \(E = 2.0\times10^{5}\) N/C with \(\mathbf p\) at \(30^\circ\) to \(\mathbf E\). Find the torque, the potential energy, and the work needed to turn it from aligned to anti-aligned.

\[ \tau = pE\sin\theta = (6.2\times10^{-30})(2.0\times10^{5})(0.500) = 6.2\times10^{-25}\ \text{N}\cdot\text{m} \] \[ U = -pE\cos\theta = -(1.24\times10^{-24})(0.866) = -1.07\times10^{-24}\ \text{J} \] \[ W = U(180^\circ) - U(0^\circ) = pE - (-pE) = 2pE = 2.48\times10^{-24}\ \text{J} \]

Perspective: the typical thermal energy at room temperature is \(k_BT \approx 4.1\times10^{-21}\) J, over a thousand times larger. So thermal jostling keeps the molecules almost randomly oriented, and a field only produces a slight average alignment. That small bias is still enough to make water's dielectric constant about 80.

6. Electron through deflection plates

An electron enters the space between parallel plates moving horizontally at \(v_0 = 2.0\times10^{7}\) m/s. The plates are 4.0 cm long and the field between them is \(5.0\times10^{3}\) N/C, pointing up. Find the deflection and direction on leaving.

\[ a = \frac{eE}{m_e} = \frac{(1.60\times10^{-19})(5.0\times10^{3})}{9.11\times10^{-31}} = 8.78\times10^{14}\ \text{m/s}^2\ (\text{downward, opposite } \mathbf E) \] \[ t = \frac{L}{v_0} = \frac{0.040}{2.0\times10^{7}} = 2.0\times10^{-9}\ \text{s} \] \[ y = \tfrac12 at^2 = \tfrac12(8.78\times10^{14})(2.0\times10^{-9})^2 = 1.76\times10^{-3}\ \text{m} = 1.76\ \text{mm (down)} \] \[ v_y = at = 1.76\times10^{6}\ \text{m/s},\qquad \phi = \tan^{-1}\frac{1.76\times10^{6}}{2.0\times10^{7}} = 5.0^\circ \text{ below horizontal} \]

The acceleration is \(10^{14}\) times \(g\) — another reason gravity is ignored.

Checklist

I know charge is quantized in units of \(e\) and conserved, and that KCL is charge conservation.
I can explain charging by contact and by induction, and why a neutral object is attracted to a charged one.
I can apply Coulomb's law in vector form, with \(\hat{\mathbf r}\) pointing from source to target, without double-counting signs.
I add fields and forces from several charges by components, and use symmetry to skip cancelling ones.
I know \(\mathbf F = q\mathbf E\), and that an electron is pushed opposite to \(\mathbf E\).
I can sketch field lines and read field strength from their spacing.
I can set up \(\int k\,dq\,\hat{\mathbf r}/r^2\) for a rod, ring or disk, keeping only the surviving component, and check the far-field limit.
I remember: line \(\lambda/(2\pi\varepsilon_0r)\), sheet \(\sigma/(2\varepsilon_0)\), dipole \(\sim 1/r^3\).
I can use \(\boldsymbol\tau = \mathbf p\times\mathbf E\) and \(U = -\mathbf p\cdot\mathbf E\).
I treat a charge in a uniform field as projectile motion with \(a = qE/m\).