D.2 Electric and magnetic fields
Charge and its conservation, charging by friction, contact and induction, Coulomb's law, electric field strength and field lines, the uniform field between plates, Millikan's experiment, magnetic field patterns, and (HL) electric potential and equipotentials.
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In 30 seconds
- Like charges repel, unlike charges attract. Coulomb's law: F = kq1q2/r2, with k = 1/4πε0. Charge is conserved and quantised in units of e.
- Electric field strength E = F/q, in N C−1. Between parallel plates the field is uniform: E = V/d.
- Magnetic fields come from moving charges: around a wire, inside a solenoid, between magnet poles. Field lines show the direction a north pole would move.
- HL: electric potential Ve = kQ/r, potential energy Ep = kq1q2/r, E = −ΔV/Δr and W = qΔV.
By the end you can
- Describe charging by friction, contact and induction, and use conservation of charge.
- Use Coulomb's law and E = F/q, and draw electric field lines.
- Use E = V/d for parallel plates and explain Millikan's experiment.
- Draw magnetic field patterns for magnets, wires and solenoids.
- Use electric potential, potential energy and the potential gradient, and sketch equipotentials. HL
1Electric charge
- Charge is measured in coulombs. The elementary charge e = 1.60 × 10−19 C; every free charge is a whole-number multiple of it (charge is quantised).
- Conservation of charge: the total charge of an isolated system never changes. Charging moves electrons; it does not make charge.
| Method | What happens |
|---|---|
| friction | rubbing transfers electrons from one material to the other: they gain opposite charges |
| contact | touching a charged conductor shares its charge; equal conductors share it equally |
| induction | a charged rod nearby pushes or pulls electrons in a conductor; earthing it and then removing the earth leaves an opposite charge behind |
Millikan's oil-drop experiment
An oil drop of mass 3.0 × 10−15 kg is held stationary between plates where E = 1.84 × 105 N C−1. Find its charge in units of e.
- qE = mg ⇒ q = 3.0 × 10−15 × 9.8/1.84 × 105 M1
- = 1.6 × 10−19 C = 1e (one extra electron) A1
2Coulomb's law
Find the force between +2.0 μC and −3.0 μC charges 10 cm apart.
- F = 8.99 × 109 × 2.0 × 10−6 × 3.0 × 10−6/0.102 M1
- = 5.4 N, attractive A1
- Permittivity ε0 (8.85 × 10−12 C2 N−1 m−2) describes the vacuum. In a material the force is weaker.
3Electric field
- For a point charge, E = kQ/r2. Field lines start on positive charges and end on negative ones; their density shows the field strength; they never cross.
Find the field strength 20 cm from a +1.0 nC point charge.
- E = kQ/r2 = 8.99 × 109 × 1.0 × 10−9/0.202 M1
- = 225 N C−1, pointing away from the charge A1
Plates 2.0 cm apart are connected to 500 V. Find E and the force on an electron between them.
- E = 500/0.020 = 2.5 × 104 V m−1 A1
- F = eE = 1.6 × 10−19 × 2.5 × 104 = 4.0 × 10−15 N towards the + plate A1
4Magnetic fields
- Right-hand grip rule: thumb along the conventional current, fingers curl in the direction of the field. For a solenoid, fingers along the current in the turns and the thumb points to the N end.
- Symbols: × means field into the page, • means out of the page.
5Electric potential HL
Electric potential Ve — the work done per unit positive charge in bringing a small test charge from infinity to that point. A scalar, zero at infinity.
Find the potential 30 cm from a +5.0 nC charge, and the work needed to bring a +2.0 μC charge there from infinity.
- V = kQ/r = 8.99 × 109 × 5.0 × 10−9/0.30 = 150 V A1
- W = qV = 2.0 × 10−6 × 150 = 3.0 × 10−4 J A1
The potential changes by 200 V over 5.0 cm in a region. Find the field strength.
- E = ΔV/Δr = 200/0.050 M1
- = 4.0 × 103 V m−1, pointing towards lower potential A1
- Equipotentials around a point charge are concentric spheres; between parallel plates they are planes parallel to the plates. They are always perpendicular to the field lines.
Test yourself — cover the answers first
Quick check
- What is the elementary charge?
- Define electric field strength.
- If the separation of two charges triples, what happens to the force?
- What is the field like between parallel plates?
- Which way does the field point around a wire carrying current out of the page?
- HL Is electric potential a vector?
Exam-style
- Two charges of +4.0 nC sit 0.20 m apart. Find the field strength at the midpoint and at a point 0.10 m beyond one charge on the line.
- HL An electron is accelerated from rest through 2000 V. Find the kinetic energy gained in J and eV, and its final speed.
Show answers
- e = 1.60 × 10−19 C
- force per unit positive test charge
- it falls to 1/9
- uniform (except near the edges)
- anticlockwise
- no, a scalar
- midpoint: equal and opposite fields, so E = 0
beyond: E = k(4.0 × 10−9)/0.102 + k(4.0 × 10−9)/0.302 = 3596 + 400
= 4.0 × 103 N C−1 - Ek = eV = 1.6 × 10−19 × 2000 = 3.2 × 10−16 J = 2000 eV
v = √(2Ek/m) = √(2 × 3.2 × 10−16/9.11 × 10−31)
= 2.7 × 107 m s−1
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