Electrostatic Fields

How to use this guide

Every section ends with a short recall check. Answer it with the guide closed before you move on: retrieving an idea fixes it far better than rereading it.

1. Day 0 (two 45-minute blocks): do the six prompts above, then work through 1.1–1.14. After each section, close the page and answer its recall check. Mark every miss.
2. Day 1: rebuild the derivations in Derive it cold on blank paper, every step. Compare, then redo only the ones that broke.
3. Day 3: mixed practice 1–18 without notes. Mark misses, read the section named in each answer, take a break, then redo only the marked problems. Stop when a round adds no new marks.
4. Day 6 or 7: mixed practice 19–38, then answer the Explain it questions aloud in full sentences. Cover the chapter map and say what each idea rests on.
5. About two weeks later and before the exam: do the two exam-style problems against the clock, mixed with problems from later chapters. Tick a self-check line only when you have done it cold on a later day.

When stuck: after about ten minutes, reveal only the next step of the answer, not the whole working. Later, redo the full problem from a blank page. Looking at a solution and thinking “I could have done that” is not the same as doing it.

How the chapter fits together

Every idea in the chapter rests on others. Choose one to see what it is built on and why; the three on the left are measured, everything else follows. To practise, cover the map and say what each idea rests on.

Which tool for which question

1.1 Electric charge

The observation. Rub two glass rods with wool and they repel; two plastic rods repel too; a glass rod and a plastic rod attract. Mechanics cannot explain this, so a new property is introduced: electric charge. Because there is both attraction and repulsion, there must be two kinds. By convention glass gets +, plastic gets −.

In plain words: charge is a property of matter, like mass, except that it comes in two kinds, + and −. Like charges push each other apart and unlike charges pull together. Charge is never created or destroyed, only moved, and it always comes in whole multiples of the smallest amount, the charge e of one proton.

• Charge is separated, not created. Rubbing glass moves electrons onto the wool, leaving the glass with an electron deficit (+). Plastic takes electrons from the wool and ends up with an excess (−).
• Conservation: the total charge of a closed system is constant. Charges only appear or disappear in equal and opposite pairs.

Near miss: a rubbed glass rod seems to have gained charge, but none was created: electrons moved from the glass to the wool, leaving the rod positive and the wool equally negative. The total charge is still zero.

1.2 Coulomb’s law

For two point charges (bodies much smaller than their distance), the force grows with each charge and falls with the square of the distance. The same 1/r² appears in Newton’s gravitation; the 4πr² in the denominator is the area of a sphere of radius r.

In plain words: Coulomb’s law says how hard two point charges push or pull: the force grows with each charge and falls with the square of the distance between them. Like charges repel and unlike charges attract, along the line joining them.

• Signs are built in. If Q₁Q₂ > 0 the force points along e_r (repulsion); if < 0 it points against it (attraction). Put the signed charges in and the direction comes out right.
• Action = reaction: the force on charge 1 has the same size and the opposite direction.

Trap: r is the distance between the charges, not a position coordinate. Point charges are an idealisation; Coulomb’s law in this form needs bodies much smaller than their distance.

Near miss: doubling the distance between two charges looks as if it should halve the force, but the force falls to a quarter: it depends on 1/r², not on 1/r.

1.3 Electric field strength

Coulomb’s law describes a pair of charges but says nothing about the space around them. The field idea fixes that: a charge changes the surrounding space, and that change is felt by any probe charge placed there. A field assigns a physical quantity to every point in space; temperature is a scalar field, gravity and E are vector fields.

In plain words: instead of asking how two charges act on each other, describe what one charge does to the space around it. The field E says, at every point, the force a charge of one coulomb would feel there; any charge Q placed at that point then feels the force Q E.

• E does not depend on the probe: a bigger probe charge feels a bigger force, but F/Q stays the same.
• The field exists in vacuum; no matter is needed.
• A positive probe is pushed along E, a negative probe against it.
• The vector r always runs from the source point Q to the observation point P.

Near miss: a bigger probe charge feels a bigger force, so the field might seem to depend on the probe, but E = F/Q_probe stays the same: double the probe and the force doubles with it.

1.4 Superposition of fields

Experiments show that the forces of several charges on one probe add as vectors. In a homogeneous medium the total field is therefore the vector sum of the individual fields.

In plain words: with several charges, each one makes its own field as if the others were not there, and the total field is their vector sum. Add the fields as arrows, component by component, not as plain numbers.

1. Sketch the charges and P with coordinates.
2. For each charge write r_i = r_P − r_Qi and its length.
3. Write each field with the signed charge.
4. Add x, y and z components separately; look for components that cancel by symmetry.

Trap: adding magnitudes. Two 7.19 kV/m fields gave 11.5 kV/m here, not 14.4 kV/m, because they point in different directions.

Near miss: +Q and −Q at equal distances on either side of a point P make fields of the same size there, yet they do not cancel: both point the same way, from the + charge towards the − charge, so at P they add.

Try it in the lab: Point charges: field, potential and work · predict where the fields of +Q and +2Q cancel before you watch it.

1.5 Forces between charge distributions

The notes only pose two questions here, using two charged bodies V₁ and V₂. Both follow from what you already know:

In plain words: for charged bodies rather than points, cut each body into tiny pieces, treat every piece as a point charge, and add up the forces. The adding up becomes an integral over the volume; the hard part is knowing the field that the other body produces.

1. Forces inside one body. Every pair of charges inside the body pushes or pulls on each other with equal and opposite forces, so these internal forces add to zero for the body as a whole. They still matter: they decide where charge settles. In a conductor, for example, excess charge spreads out to the surface.
2. Forces between two bodies. Cut V₂ into small elements dQ = ϱ dV, find the field E₁ that V₁ alone produces at each element, and add up dF = E₁ dQ. The force on V₁ is equal and opposite.

Near miss: treating two charged metal spheres as point charges at their centres works while they are far apart, but close together their charges shift towards or away from each other on the metal, and the point-charge force is no longer accurate.

1.6 Charge densities

A point charge is an idealised charge squeezed into a single point, so its density would be infinite. Real charge is spread out, and you describe it with a density defined as a limit: take a small piece of line, area or volume around a point, divide its charge by its size, and let the piece shrink onto the point.

In plain words: real charge is spread out, so you describe how much of it sits on each unit of length, area or volume: the charge densities λ, σ_A and ϱ_V. The total charge is the density added up over the line, surface or body.

A surface charge is idealised as infinitely thin. Pick the volume element that matches the shape (appendix 7.3 of the notes): dV = dx dy dz (Cartesian), ρ dρ dφ dz (cylinder), r² sin ϑ dr dϑ dφ (sphere). If ϱ depends only on r, use thin shells: dV = 4πr² dr.

Near miss: stretching a uniformly charged rod to twice its length while keeping the same total charge halves its line density λ: a density is charge per length, not the total charge.

1.7 Field lines

A field line is a curve whose tangent at every point has the direction of E. Arrows on the line show the direction. In vector plots the arrow length shows the field strength; in line plots the density of lines does.

In plain words: field lines are a picture of E: at every point the line runs in the direction of the field, starting on positive charges and ending on negative ones. Where the lines crowd together the field is strong; where they spread out it is weak.

1. Construction: find E at a starting point.
2. Step a short distance in that direction and find E again.
3. Repeat. The field line is the curve that is tangent to all these vectors.

• Lines start on positive charges and end on negative charges (or at infinity).
• Lines meet conductor surfaces at right angles (shown in 1.8 and 1.11).
• Lines never cross where E ≠ 0: a point cannot have two field directions. They can only meet at field-free points.

Patterns to sketch from memory: a positive point charge (radial, outward), a negative one (radial, inward), the ideal plate capacitor (parallel lines from + to −), a + and − pair (lines bend from one charge into the other), and two equal + charges (lines push apart, with a field-free point exactly between them).

Uniqueness theorem: a field that satisfies Maxwell’s equations and the boundary conditions is the only possible solution. Two arrangements with the same boundary conditions therefore have the same field. This is the deeper reason field lines cannot cross.

Near miss: field lines look like the paths charges would follow, but they are not: they show the direction of the force at each point, and a moving charge carried by its momentum can cut across them. Field lines also never cross, because E has one direction at each point.

1.8 Electrostatic potential

Work. Moving a charge Q from P₀ to P₁ against the field costs energy; moving it with the field releases energy. The notes count the work supplied from outside, hence the minus sign.

In plain words: moving a charge against the field takes work, like carrying a ball uphill. The potential φ_e is that work per coulomb, measured from a chosen reference point. Because the electrostatic field is conservative, the work depends only on where you start and finish, not on the path between.

Path independence. Carry the charge from P₀ along one path to P₁ and back along another. Afterwards the field is exactly as before, so the system’s energy is unchanged and the total work is zero. Since the return path was arbitrary, the work from P₀ to P₁ cannot depend on the path.

Potential. Because the work depends only on the end points, every point P can be given a number: the work per unit charge needed to bring a charge there from a reference point P₀, where φ_e(P₀) = 0.

Equipotential surfaces

An equipotential surface collects all points of equal potential; in a drawing its cuts appear as equipotential lines. Moving along one costs no work, so E · ds = 0 there: equipotentials cross field lines at right angles. Drawn at equal potential steps, closely spaced equipotentials mean a strong field.

Conductors. A tangential field at a metal surface would push the free charges along it, and they would keep moving until that component vanished. In the static state the whole conductor therefore sits at one potential, its surface is an equipotential, field lines meet it at right angles, and the interior is field-free.

Near miss: the potential of a point charge falls as 1/r and its field as 1/r², so at twice the distance the potential is half as large but the field only a quarter. Mixing up the two powers is the most common slip in this section.

Try it in the lab: Point charges: field, potential and work · predict the work along an equipotential, and its sign before you watch it.

1.9 Voltage

Write φ_e(P₁) − φ_e(P₂) with the definition of potential. The two integrals from the reference point P₀ join into one integral from P₁ to P₂, and P₀ drops out: a potential difference does not depend on the reference point.

In plain words: voltage is a difference of potentials: U₁₂ = φ₁ − φ₂, the work per coulomb the field does carrying a charge from P₁ to P₂. The reference point drops out, so a voltage, unlike a potential, needs no agreed zero.

• U₁₂ > 0 means P₁ is at the higher potential. E points from high to low potential.
• Link to work: moving Q from P₁ to P₂ needs W_e = Q(φ₂ − φ₁) = −Q U₁₂.

Sign check: U₁₂ is “first point minus second point”: φ₁ − φ₂. Read the subscripts in that order every time.

Near miss: a bird on a high-voltage line sits at a very high potential, yet it is safe, because the voltage between its two feet is almost zero. What matters is the difference, not the potential itself.

1.10 Electric flux density and Gauss’s law

Two field quantities. E describes the effect (the force on charges). The flux density D describes the cause and makes the influence of matter explicit. For the linear materials in this course they are proportional.

In plain words: the flux density D describes the field from the side of its sources, the charges, while E describes its effect on charges. Gauss’s law says that the total flux of D out of any closed surface equals the charge inside it, whatever shape the surface has.

Language trap: “flux” and “flux density” are historical names. Nothing flows in an electrostatic field; do not confuse flux with current, where charges really move.

Gauss’s law

Around a point charge in vacuum, D = Q/(4πr²) e_r. On a concentric sphere D is parallel to dA and has constant size, so the flux is D × 4πr² = Q, whatever the radius. The same holds for any closed surface and any position of the charge inside it, and by superposition for any charge distribution:

Gauss’s law always holds, but it only gives you the field when symmetry tells you the direction of D and that its size is constant on the surface. In practice that means three shapes: sphere (point charge, charged shell or ball), infinitely long cylinder (line charge, coax) and infinite plane (plate capacitor).

1. From the symmetry, state the direction of D and which coordinate it can depend on.
2. Choose a closed surface on which D is either perpendicular and constant, or parallel to the surface (zero flux).
3. Find the enclosed charge as a function of the surface size.
4. Set flux = enclosed charge, solve for D, then E = D/ε.

Near miss: a closed surface with zero total flux might seem to contain no charge at all, but it can hold +Q and −Q together: Gauss’s law gives only the net charge inside.

Try it in the lab: Gauss’s law: a charged cylinder · predict how D(ρ) rises, peaks and falls before you watch it.

1.11 Interface conditions

At a boundary, split E and D into a normal part (along the unit normal n) and a tangential part (along a unit tangent t). Each part is fixed by one law:

In plain words: where two materials meet, the field changes, but not arbitrarily. Split E and D at the boundary into the part across it and the part along it: the part of E along the boundary is the same on both sides, and so is the part of D across it, as long as no free charge sits on the boundary.

1. Normal part from Gauss’s law. Take a flat pillbox that straddles the boundary and shrink its height h → 0. Only the two faces contribute: (D_n2 − D_n1)A = enclosed charge = σ_AA.
2. Tangential part from ∮ E · ds = 0. Take a thin rectangular loop across the boundary and shrink its width h → 0. Only the two long sides of length l contribute: (E_t2 − E_t1)l = 0.
3. Get the remaining parts from D = εE on each side.

Refraction of field lines. The two parts change differently, so a field line bends at the boundary, much like light. Measuring the angle α from the normal, tan α = E_t/E_n, and the two conditions give tan α₁/tan α₂ = ε₁/ε₂: in the medium with the larger ε the line leans further from the normal. (This rule follows directly from the table; the notes only mention the analogy with light.)

Ideal plate capacitor from two charged sheets

1. One infinite sheet with +σ_A: by symmetry D has the same size on both sides and points away from the sheet. The jump condition gives D − (−D) = σ_A, so D = σ_A/2 on each side.
2. The −σ_A sheet gives σ_A/2 pointing toward it on both sides.
3. Superpose: outside the plates the two contributions cancel; between them they add to D = σ_A, directed from + to −.

Near miss: at a boundary between air and glass, E might seem to keep its size across it, but only its part along the boundary does. Its part across the boundary drops by the factor ε_r of the glass, because it is the part of D across the boundary that stays equal.

Try it in the lab: Interface conditions: field refraction · predict which way a field line bends at a boundary before you watch it.

1.12 Matter in an electrostatic field

The notes compare three versions of the same ideal plate capacitor. The plates carry a fixed charge ±Q (the capacitor is isolated), so D = σ_A stays the same throughout. Only E, and therefore U, changes.

In plain words: in an insulator the charges cannot flow away, but each atom or molecule stretches slightly in the field and becomes a tiny dipole. Their combined effect, the polarisation P, weakens the field inside the material by the factor ε_r; a metal goes further and cancels the field inside it completely.

Memory hook: empty gap > dielectric > metal. The notes name these voltages U_b and U_c differently on two pages (see Slips in the notes), so learn the order by material, not by letter.

Conductors: electrostatic induction

Put an uncharged metal body into a field. Its free electrons drift toward the positive plate, leaving a deficit on the far side. The body as a whole stays neutral, but its surfaces now carry opposite charges. If the body is made of two touching sheets that you pull apart while they are still in the field, each sheet keeps its charge. The charges stop moving once the whole body is one equipotential and the inside is field-free. Because these induced charges depend on the field they create, general field problems with conductors are hard to compute.

Hollow conductor. Put a charge Q at the centre of a metal shell with inner radius a and outer radius b. The inner surface collects −Q, the outer surface +Q. D = Q/(4πr²) in the cavity, it drops to zero inside the metal, and outside it looks like the field of the point charge again. At each surface the induced σ equals the D that arrives there, so D jumps by exactly σ, as the interface condition demands.

Dielectrics: polarisation

An insulator has no free charge carriers, but its bound charges can shift or turn:

• Displacement polarisation: the field pulls an atom’s electron cloud and nucleus slightly apart (electronic), or shifts ions in a crystal (ionic).
• Orientation polarisation: molecules with a built-in dipole, such as water, turn toward the field.

The aligned dipoles leave a thin layer of bound charge on each face of the slab, opposite in sign to the neighbouring plate. These polarisation charges weaken the field inside. Read ε_r two ways: at the same D, E in the dielectric is ε_r times smaller; at the same E, D is ε_r times larger.

Near miss: sliding a dielectric slab into an isolated charged capacitor lowers its voltage, which can seem odd because no charge was removed: Q and D stay the same, but E = D/ε falls, and with it U = E d.

Try it in the lab: Plate capacitor: gap, dielectric and energy · predict empty, dielectric and metal-filled gaps before you watch it.

1.13 Capacitors and capacitance

A capacitor is two conductors carrying equal and opposite charge, so that all the flux leaving the positive one ends on the negative one. Scale the charge by any factor and D, E and U scale by the same factor, so Q ∝ U.

In plain words: a capacitor stores charge on two conductors held apart. Put +Q on one and −Q on the other and a voltage U appears between them; the ratio C = Q/U is the capacitance, the charge it holds per volt. It depends only on shape, size and material, not on how much charge is there.

Recipe for any geometry: assume ±Q → find D with Gauss → E = D/ε → U = ∫ E · ds → C = Q/U. The Q must cancel; if it does not, there is a mistake.

• Spherical: U = Q/(4π ε₀) (1/a − 1/b) = Q(b − a)/(4π ε₀a b). Limit check: for a thin gap d = b − a ≪ a, C ≈ ε₀4πa²/d, the plate formula with the sphere’s area.
• Coaxial: U = Q/(2π ε l) ∫_a^b dρ/ρ = Q ln(b/a)/(2π ε l).
• Large C in practice: small gap (but the risk of breakdown grows), large area (metal foil on both sides of a thin film, rolled up: a wound capacitor), or high ε_r (electrolytic capacitor).

Capacitor networks

Quick checks: a parallel total is larger than the largest capacitor; a series total is smaller than the smallest. In series, the smallest capacitor takes the largest voltage. These are the reverse of the resistor rules.

Built-up capacitors

• Multilayer: n interleaved plates per terminal form 2n − 1 equal capacitors in parallel: C = (2n − 1) εA/d.
• Variable (rotary) capacitor: the overlap of rotor and stator is a ring sector of angle α, area α(r_a² − r_i²)/2, so C = (2n − 1) ε₀α(r_a² − r_i²)/(2d). Turning the rotor changes α and hence C.
• Stray capacitance: any two conductors form a capacitor. A three-phase overhead line has partial capacitances between every pair of wires and from each wire to earth.

Near miss: doubling the charge on a capacitor does not double its capacitance: U doubles too, so C = Q/U stays the same.

Try it in the lab: Capacitor networks: series and parallel · predict series and parallel capacitors on one source before you watch it.

1.14 Energy stored in an electric field

Charge a capacitor step by step. When it already holds ±q, its voltage is u = q/C. Carrying a further dq from the negative to the positive plate, against the field, costs dW_e = u dq = (q/C) dq. (Lower-case letters mark quantities that change with time.)

In plain words: charging a capacitor takes work, because each extra bit of charge is pushed against the voltage already there. That work is stored in the field between the plates and comes back out when the capacitor discharges.

Where is the energy? In the field. For a plate capacitor, ½C U² = ½(εA/d)(E d)² = ½εE² × A d, and A d is the volume of the field region. Dividing by the volume gives an energy density that also holds for non-uniform fields:

Spring analogy: a spring with constant c_s stores ½c_sX² = ½F_X²/c_s. Matching it to ½Q²/C, capacitance plays the part of 1/c_s (a soft spring) and voltage the part of force.

Near miss: doubling the voltage on a capacitor doubles its charge, but it quadruples the stored energy, because W_e = ½ C U² grows with the square of U.

Try it in the lab: Plate capacitor: gap, dielectric and energy · predict where the stored energy goes before you watch it.

Formula map and units

Close the page and rebuild this table from memory before checking it.

Check every answer: the units must match the quantity. Signs must agree with a sketch: attraction or repulsion, field away from + and toward −, work positive when you push against the field. Limits must make sense: fields fall off far away, piecewise formulas agree at the boundary, and a thin spherical gap behaves like plates.

Derive it cold

Rebuild each derivation on blank paper, every step, without looking. The section number tells you where to check.

1. From Coulomb’s law to the field of a point charge, and to the superposition sum. §1.3–1.4
2. Path independence from a round trip, and ∮ E · ds = 0. §1.8
3. The potential of a point charge with the reference at infinity, including the sign discussion. §1.8
4. Why U₁₂ does not depend on the reference point. §1.9
5. Gauss’s law for a point charge on a sphere, and D_ρ(ρ) for the charged cylinder. §1.10
6. Both interface conditions with a pillbox and a thin loop. §1.11
7. The ideal plate capacitor field from two charged sheets. §1.11
8. C of the plate, spherical and coaxial capacitor with the five-step recipe. §1.13
9. The series and parallel rules. §1.13
10. W_e = ½C U² by charging step by step, and w_e = ½E D from the plate capacitor. §1.14

Explain it

Answer aloud in full sentences, as if teaching someone. If you stall, that is the gap to fix.

1. Why is the electrostatic field conservative, and what does that let you define?
2. Why must E be zero inside a conductor and perpendicular to its surface?
3. Why does a dielectric lower the voltage of an isolated charged capacitor, and why does a metal slab lower it even more?
4. When does Gauss’s law give you the field, and when is it true but useless?
5. At an uncharged interface, why is the normal part of D continuous while the normal part of E jumps?
6. Why is a potential tied to a reference point, while a voltage is not?
7. Why do capacitors in series share the same charge?
8. Why is the energy of a capacitor ½Q U and not Q U?

Mixed practice

Use ε₀ = 8.854 × 10⁻¹² F/m and 1/(4π ε₀) ≈ 8.988 × 10⁹ V m/(A s). The 38 problems are shuffled across sections, so first decide which idea a problem needs, then solve it. Each answer names the section to revisit if you miss it.

1. Q₁ = +1 μC sits at the origin and Q₂ = +2 μC at (3 cm, 4 cm). Find the force vector on Q₂.
2. A material has ε_r = 4 and E = 250 V/m. Find |D|.
3. A line density varies as λ(x) = 2x μC/m from x = 0 to 3 m. Find the total charge.
4. φ_A = 12 V and φ_B = −3 V. Find U_AB and the work needed to move +2 μC from A to B.
5. A 10 μF capacitor is charged to 12 V. Find its stored energy.
6. At an uncharged interface, ε₁ = 2ε₀, ε₂ = 4ε₀, E_n1 = 100 V/m and E_t1 = 80 V/m. Find E_n2 and E_t2.
7. Find the field magnitude 0.20 m from a −4 nC point charge and give its direction.
8. 2 μF and 8 μF are connected in series across 10 V. Find the charge and the voltage on each.
9. Find the potential 0.30 m from a +3 nC point charge, with zero potential at infinity.
10. An isolated plate capacitor has a 4 mm gap and 1000 V/m between the plates. Find the voltage (a) empty, (b) with a 1 mm uncharged metal slab inserted parallel to the plates, (c) with a 1 mm slab of ε_r = 5 instead.
11. An object carries +4.806 × 10⁻¹⁹ C. How many elementary charges is this, and what does the sign mean for its electrons?
12. Show that the two formulas for the charged cylinder give the same D at ρ = a. Why must they agree?
13. +3 nC sits at the origin and −3 nC at (6 cm, 0). Find E at P = (3 cm, 4 cm).
14. A vacuum region has a uniform field of 1.0 × 10⁵ V/m. Find the energy density.
15. A uniform 500 V/m field points from A to B, which are 0.08 m apart along the field. Find U_AB.
16. A spherical capacitor in vacuum has a = 5 cm and b = 6 cm. Find C and compare it with the thin-gap plate approximation.
17. A conducting surface in vacuum carries σ_A = 4 nC/m². Find the field just outside and inside the conductor.
18. A line carries a uniform 5 μC/m over 0.40 m. Find its charge.
19. −2 nC is moved from 0.20 m to 0.10 m from a fixed +5 nC. Find W_e and say who does the work.
20. Find the equivalent of 6 μF and 3 μF in parallel and in series, and say what is common to the elements in each case.
21. A uniform D = 3 μC/m² is normal to a 0.20 m² surface. Find the flux.
22. Find the force between +2 μC and −3 μC that are 0.20 m apart, and say whether it attracts or repels.
23. A 10 nF capacitor is charged to 100 V. A dielectric with ε_r = 4 then fills the gap. Find U, Q and W_e afterwards if (a) the source was disconnected first, (b) the source stays connected.
24. A long uniformly charged cylinder has ϱ_V = 4 μC/m³ and a = 0.10 m. Find D at ρ = 0.05 m.
25. A field of 200 V/m in a medium with ε_r = 2 meets a medium with ε_r = 6 at 30° to the normal. There is no surface charge. Find the size and angle of the field in medium 2.
26. The distance between two point charges doubles. What happens to the force?
27. A multilayer capacitor has 5 plates per terminal, plate area 1 cm², layer thickness 10 μm and ε_r = 1000. Find C.
28. +2 μC moves 0.10 m along a uniform 200 V/m field. Find the externally supplied energy.
29. A ball of radius R = 0.1 m has ϱ(r) = k r with k = 10⁻⁶ C/m⁴. Find its total charge and D at its surface.
30. +2 nC at x = −0.10 m and −2 nC at x = +0.10 m. Find E at the origin.
31. Plates with ε_r = 2, area 0.020 m², gap 1.0 mm. Find C.
32. A metal sphere of radius 0.10 m carries 2 nC. Find σ_A, the field just outside, and compare with the point-charge field at the same radius.
33. A closed surface encloses −8 nC. Find the outward flux of D.
34. −3 nC is brought from infinity to 0.20 m from a fixed +6 nC. Find W_e.
35. A uniform surface density of 8 nC/m² covers 0.25 m². Find the charge.
36. A coaxial cable has inner radius 1.0 mm, outer radius 4.0 mm and ε_r = 2.5. Find its capacitance per metre.
37. +Q and +2Q sit 30 cm apart. Where on the line between them is the field zero, and what does that point mean for the field-line picture?
38. A thin rod of length 0.5 m carries a uniform 2 nC/m and lies in a uniform external field of 4 kV/m. Find the force on the rod. How would you find it if the external field were not uniform?

Exam-style problems

Allow about 20 minutes each, closed book. Each one chains several sections together.

A. Coaxial cable

Inner conductor radius 0.5 mm, outer conductor radius 2 mm, dielectric ε_r = 2.2, length 10 m, voltage 100 V.

1. Derive C/l with the five-step recipe, then find C.
2. Find the charge on the inner conductor.
3. Find E at the inner and the outer conductor. Where would the dielectric break down first?
4. Find the stored energy.

B. Two-layer plate capacitor

Plate area 100 cm². Layer 1: 1 mm thick, ε_r = 2. Layer 2: 2 mm thick, ε_r = 4. The layers touch with no free charge between them, and 300 V is applied.

1. Which quantity is the same in both layers, and which interface condition says so?
2. Find D, E₁, E₂ and the voltage across each layer.
3. Find C in two ways: from Q/U, and as two capacitors in series.
4. Find the energy stored in each layer.

Slips in the notes

We sent the professor a list of small slips in the WS 2026/27 notes, and the revised script (November 2026) fixes most of them. These three remain; page numbers are the ones printed on the pages.

• p. 47: not an error, but two notation switches to watch. Here Q₁ is the probe and Q the source (pp. 43–46 use Q and Q₀), and U stands for φ(r₂) − φ(r₁), end minus start, the opposite order to U₁₂ = φ₁ − φ₂ on p. 52. With that U, W_e = Q₁U is right; with the p. 52 convention the same work is W_e = −Q U_start,end.
• p. 84 vs p. 90: both figures now call the dielectric case (b) and the metal case (c), but the voltage names are swapped: pp. 84–87 call the metal-slab voltage U_b and the dielectric one U_c; p. 90 does the reverse. Physics: empty gap > dielectric > metal slab.
• p. 92: “ε_r describes the ratio χ” mixes two quantities. ε_r is the ratio of the field without and with dielectric at the same D; χ = ε_r − 1 is the susceptibility. “Dieletric” is also misspelt. The professor agrees and plans to reword it.

Self-check

Tick a skill only when you have done its problems without notes on a later day than you studied it. These are chapter learning goals, not an official exam specification.

• Explain charge sign, quantisation and conservation. (11)
• Use Coulomb’s law as a vector with signed charges. (1, 22, 26)
• Tell field from force and find one from the other. (7)
• Add point-charge fields as vectors, using symmetry, and apply them to extended bodies. (13, 30, 38)
• Choose λ, σ or ϱ and integrate it, including non-uniform cases. (3, 18, 29, 35)
• Sketch and read field-line patterns and explain why lines cannot cross. (37)
• Derive the point-charge potential and get the sign of the work right. (9, 19, 28, 34)
• Use U₁₂ = φ₁ − φ₂ and link it to work. (4, 15)
• Relate E, D, flux and Gauss’s law with correct units. (2, 21, 33)
• Choose a Gaussian surface and solve for D. (12, 24, 29)
• Apply the interface conditions, including refraction and conductors. (6, 17, 25, 32)
• Explain induction and polarisation, and order the voltages for an empty, dielectric and metal-filled gap. (10, 23)
• Find C for plate, spherical, coaxial and multilayer capacitors, and combine them. (8, 16, 20, 27, 31, 36)
• Find capacitor energy and energy density by both routes. (5, 14, 23)
• Rebuild every derivation in Derive it cold on blank paper.
• Solve both exam-style problems in about 20 minutes each.