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D.2

Electric and magnetic fields

Theme D · Fields · SL and HL, with additional HL material marked

🎯What you need to be able to do

  • Apply Coulomb’s law and use electric field strength; sketch field patterns.
  • Use the uniform field between parallel plates, \( E = V/d \).
  • Sketch magnetic field patterns for magnets, straight wires and solenoids.
  • Use \( F = BIL\sin\theta \) for a current in a field.
  • HLUse electric potential and electric potential energy, and compare gravitational with electric fields.

📚The physics

Coulomb’s law

\[ F = \frac{kQ_1Q_2}{r^{2}} \qquad k = 8.99 \times 10^{9}\ \text{N m}^{2}\ \text{C}^{-2} \]

is Newton’s law of gravitation with charge in place of mass — same inverse square, same centre-to-centre distance. The difference that matters is that charge comes in two signs, so the electric force can repel as well as attract, while gravity only ever attracts. Almost every asymmetry between the two field types traces back to that one fact.

The other difference is scale. Between two protons the electric repulsion is about \( 10^{36} \) times the gravitational attraction. Gravity dominates the universe only because large bodies are electrically neutral.

Electric field strength \( E = F/q = kQ/r^{2} \) is force per unit positive charge, so field lines point away from positive charges and towards negative ones. They start on positive, end on negative, never cross, and are always perpendicular to conductor surfaces.

Between parallel plates the field is uniform, \( E = V/d \), with evenly spaced parallel lines except for the fringing at the edges. This is the arrangement used to accelerate and deflect charged particles, and it reappears in D.3.

Magnetic fields are produced by moving charge — there are no magnetic monopoles, so field lines always form closed loops. Three patterns to know cold: around a bar magnet, out of the north pole and round to the south; around a straight wire, concentric circles whose direction is given by the right-hand grip rule with the thumb along the conventional current; inside a solenoid, uniform and parallel, looking like a bar magnet from outside.

Force on a current. \( F = BIL\sin\theta \), where \(\theta\) is the angle between the current and the field. Note that a wire parallel to the field feels no force at all. Direction comes from Fleming’s left-hand rule: first finger field, second finger current, thumb motion. This is the motor effect, and it is the reason a loudspeaker cone moves.

HLElectric potential and potential energy

\( V_e = kQ/r \) and \( E_p = kQ_1Q_2/r \). Unlike the gravitational versions these are not always negative: two like charges have positive potential energy, because they would fly apart if released, while opposite charges have negative potential energy and are bound. Getting the sign right is usually the whole difficulty.

HLGravitational versus electric

The comparison table is worth building yourself, because examiners ask for it directly.

Force\( \dfrac{GMm}{r^{2}} \) vs \( \dfrac{kQq}{r^{2}} \)
Field\( \dfrac{GM}{r^{2}} \) vs \( \dfrac{kQ}{r^{2}} \)
Potential\( -\dfrac{GM}{r} \) vs \( \dfrac{kQ}{r} \)

Both are inverse-square forces with inverse-\(r\) potentials; gravity is always attractive and always negative in potential, while electricity does both. And the constants differ by forty orders of magnitude.

✏️Worked example

Two small spheres carry charges of +3.0 µC and −5.0 µC and are held 0.20 m apart. Take \( k = 8.99 \times 10^{9} \) N m\(^{2}\) C\(^{-2}\).

(a) Force between them. \( F = kQ_1Q_2/r^{2} = (8.99 \times 10^{9} \times 3.0 \times 10^{-6} \times 5.0 \times 10^{-6})/(0.20^{2}) = 0.1348/0.04 = 3.4 \) N, attractive. Use the magnitudes for the size and the signs to decide the direction — carrying the minus sign through the arithmetic only invites confusion.

(b) Electric field midway between them. Each charge is 0.10 m from the midpoint. From the positive charge: \( E_1 = (8.99 \times 10^{9} \times 3.0 \times 10^{-6})/0.01 = 2.70 \times 10^{6} \) N C\(^{-1}\), pointing away from it. From the negative charge: \( E_2 = (8.99 \times 10^{9} \times 5.0 \times 10^{-6})/0.01 = 4.50 \times 10^{6} \) N C\(^{-1}\), pointing towards it. Both point the same way — from the positive charge towards the negative one — so they add: \( E = 7.2 \times 10^{6} \) N C\(^{-1}\).

(c) HLElectric potential at the midpoint. Potential is a scalar, so no directions are involved, but the signs are real: \( V = k(+3.0 \times 10^{-6})/0.10 + k(-5.0 \times 10^{-6})/0.10 = 2.70 \times 10^{5} - 4.50 \times 10^{5} = -1.8 \times 10^{5} \) V.

The lesson in (b) and (c) together. The fields added while the potentials subtracted, from the same pair of charges. Field is a vector and potential is a scalar, and treating either like the other is the most common error in this topic.

🔭See it happen

PhET, Charges and Fields. Drop a positive and a negative charge, switch on the field vectors and the equipotential tool, and click around the midpoint. You can watch the arrows reinforce while the potential reading passes through zero somewhere off-centre.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Two charges of +2.0 µC are held 0.15 m apart. Find the force between them. Take \( k = 8.99 \times 10^{9} \) N m\(^{2}\) C\(^{-2}\).
\( F = \dfrac{kQ_1Q_2}{r^{2}} = \dfrac{8.99 \times 10^{9} \times (2.0 \times 10^{-6})^{2}}{0.15^{2}} = \dfrac{3.60 \times 10^{-2}}{0.0225} = 1.6 \) N, repulsive since both are positive.
2. Find the electric field strength 0.20 m from a point charge of 5.0 nC.
\( E = \dfrac{kQ}{r^{2}} = \dfrac{8.99 \times 10^{9} \times 5.0 \times 10^{-9}}{0.040} = 1.1 \times 10^{3} \) N C\(^{-1}\), pointing away from the charge.
3. Two parallel plates 5.0 mm apart have a potential difference of 200 V across them. Find the field strength between them.
\( E = V/d = 200/(5.0 \times 10^{-3}) = 4.0 \times 10^{4} \) V m\(^{-1}\), uniform except for fringing at the edges.
4. A wire of length 0.40 m carrying 3.0 A lies perpendicular to a magnetic field of 0.25 T. Find the force on it.
\( F = BIL\sin\theta = 0.25 \times 3.0 \times 0.40 \times \sin 90^\circ = 0.30 \) N.
5. The same wire is now rotated so it lies at 30° to the field. Find the new force, and state the force when it lies parallel to the field.
\( F = 0.30 \times \sin 30^\circ = 0.15 \) N. When parallel, \( \sin 0^\circ = 0 \), so the force is zero.
6. HLFind the electric potential 0.20 m from a point charge of 5.0 nC, and explain why the sign convention differs from the gravitational case.
\( V_e = \dfrac{kQ}{r} = \dfrac{8.99 \times 10^{9} \times 5.0 \times 10^{-9}}{0.20} = 225 \) V. It is positive here because the charge is positive; gravitational potential is always negative because mass produces only attraction, whereas charge comes in two signs and so electric potential can be either.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Classroom — Static Electricity and Magnetism
  • HyperPhysics — electric field, potential and magnetic field concepts