1. Electric charge

Every atom contains charged particles: positive protons and negative electrons. Normally an object has equal numbers of each, so it has no overall charge — we say it is neutral. If an object gains or loses electrons, it becomes charged.

Key word — charge: a property of matter that causes it to experience a force in an electric field. Charge is measured in coulombs (C).

Two types of charge

There are only two types of electric charge: positive and negative. The rule for the direction of the force between them is simple.

Like charges repel. Unlike charges attract. Two positive charges push each other apart; two negative charges push each other apart; a positive and a negative charge pull towards each other.

Fig. 1.1 Like charges repel; unlike charges attract.

Conservation of charge

Charge cannot be created or destroyed. In any closed system, the total charge stays the same, even when charge moves from one object to another. If one object loses a certain amount of negative charge, another object must gain exactly that amount.

Conservation of charge: the total electric charge of an isolated system is always constant.

Quantisation of charge

Charge does not come in just any amount — it always comes in whole-number multiples of a smallest possible charge, called the elementary charge, e. Every proton carries charge +e and every electron carries charge −e.

e = 1.60 × 10⁻¹⁹ C

This means any charge you measure — however it was produced — will always be a whole-number multiple of 1.60 × 10⁻¹⁹ C. This surprising fact was confirmed experimentally by Robert Millikan in 1909. Workbook 2 in this pair is entirely devoted to that experiment, including a simulation you can run yourself, so we won't go into it further here.

Worked example 1.1

A charged sphere has a charge of −6.4 × 10⁻¹⁹ C. How many extra electrons does it carry?

Answer: number of electrons = charge ÷ e = (6.4 × 10⁻¹⁹) ÷ (1.60 × 10⁻¹⁹) = 4 electrons.

1A metal sphere P, carrying charge +3e, is held near an identical neutral sphere Q. State whether P and Q would attract, repel, or feel no force at this moment, and explain why.

No overall force yet, because Q is neutral — it has no net charge for P's field to act on (although P's field can still induce a slight charge separation inside Q; that is covered in Section 2). Once Q becomes charged, the direction of the force depends on the sign it ends up with.

2Two identical conducting spheres carry charges of +9.0 nC and −3.0 nC. They are touched together and then separated. Use conservation of charge to determine the charge on each sphere afterwards. (Hint: identical conducting spheres share any charge equally.)

Hint: add the two charges together to find the total charge on the pair, then split that total equally between the two identical spheres.

3Explain, in your own words, why the statement "electrons can be created by rubbing two objects together" is incorrect.

Rubbing does not create charge — it only transfers existing electrons from one object's surface to the other. The total charge of the two objects together is unchanged (conservation of charge); one object ends up with extra electrons (negative) and the other with a deficit of electrons (positive).

2. Charging and discharging

There are three ways that charge can be transferred between objects: friction, contact and induction. You need to be able to explain each one in terms of the movement of electrons.

Charging by friction

When two insulators are rubbed together, electrons are transferred from one surface to the other. The material that gains electrons becomes negatively charged; the material that loses electrons becomes positively charged (because it is now left with more protons than electrons).

Fig. 2.1 Friction transfers electrons from one insulator to another.

When friction occurs between two different insulating materials, electrons are transferred from the surface of one to the surface of the other.

Charging by contact

If a charged object physically touches a neutral (or oppositely-charged) conductor, some charge flows between them until it is shared. The object that was charged keeps some of its original charge; the object it touched now carries charge of the same sign.

Charging by induction

Induction charges an object without contact. A charged object is brought close to a neutral conductor. Electrons in the conductor are attracted or repelled, so they redistribute — one side becomes negative, the other positive — even though the conductor's overall charge is still zero. If the conductor is then earthed while the charged object is still nearby, electrons can flow to or from the ground, leaving the conductor with a net charge once the earth connection and the charged object are both removed.

Key word — earthing (grounding): connecting an object to the ground with a conductor so that charge can flow freely between the object and the Earth, leaving the object at 0 V.

4A negatively charged rod is brought close to (but does not touch) a neutral metal sphere on an insulating stand.

a) Use the simulator below to predict how charge redistributes on the sphere. Click on each side of the sphere until it shows the charge you think is correct.

Click the near half and the far half of the sphere to set a charge on each.

Metals contain free electrons that can move within the sphere. Think about which type of charge would be attracted towards a nearby negative rod, and which type would be pushed away from it.

b) The sphere is then earthed briefly while the rod stays in place. Explain what happens.

c) The earth connection is removed, then the rod is taken away. What is the final charge on the sphere?

b) While earthed, the repelled electrons can escape to the ground, since they are being pushed away from the rod and towards the earth wire. c) Once the earth wire is removed (rod still present) the sphere has a deficit of electrons; removing the rod afterwards leaves the sphere with an overall positive charge, spread evenly.

5Aircraft are earthed with a bonding cable before refuelling. Explain, using ideas about friction and charge build-up, why this matters and what could happen if it were skipped.

As fuel and air flow past the aircraft's surfaces, friction transfers charge, so the aircraft can build up a large static charge. Without earthing, a spark could jump between the aircraft and the fuel truck as they reach different potentials, and this spark could ignite fuel vapour. The bonding cable keeps both at the same potential (0 V) so charge can drain away safely instead of sparking.

6A student rubs a balloon on a wool jumper. The balloon becomes negatively charged. State the sign of charge left on the jumper, and explain how you know, without measuring anything.

The jumper must be left positively charged. Friction only transfers electrons — it cannot create charge — so whatever negative charge the balloon has gained must be exactly balanced by a positive charge left behind on the jumper (conservation of charge).

3. Coulomb's law

Coulomb's law tells us how strong the electric force is between two point charges. It was published by Charles-Augustin de Coulomb in 1783.

F = k q₁q₂ / r²

Coulomb's law: the force, F, between two point charges q₁ and q₂, separated by a distance r, is given by F = kq₁q₂/r², where k is the Coulomb constant, k = 8.99 × 10⁹ N m² C⁻².

Notice the shape of this equation — it is an inverse square law, just like Newton's law of gravitation. If you double the separation, the force becomes four times weaker; if you triple it, the force becomes nine times weaker.

All boards · Light

The Inverse Square
Law

Spread the same energy over a sphere and its area grows as r². So whatever you measure at a point — brightness, field strength, force — falls as 1/r². Pick a quantity, then move the detector.

Light intensity I ∝ 1/r²

Where the inverse square law turns up

Any influence that streams outward from a point and isn't absorbed obeys it — across mechanics, fields, waves and nuclear physics, and far beyond the exam spec.

?The law fails when the spreading isn't over a full sphere — a laser beam stays roughly parallel, and a long wire or charged plate spreads over a cylinder or plane, giving 1/r or a constant field instead.

Interactive simulation by Dr Dan Jones — drjonesphysics.com/general-apps. Used with permission.

A positive answer for F means the force is repulsive (both charges have the same sign); a negative answer means the force is attractive (the charges have opposite signs).

Key word — point charge: a charge treated as though it exists at a single point in space. A uniformly charged sphere behaves, from the outside, exactly like a point charge at its centre.

Worked example 3.1

Calculate the force between two point charges of +2.0 × 10⁻⁸ C and −5.0 × 10⁻⁸ C, separated by 4.0 cm in air.

Answer:
F = kq₁q₂/r² = (8.99 × 10⁹) × (2.0 × 10⁻⁸) × (−5.0 × 10⁻⁸) / (0.040)²
F = −5.6 × 10⁻³ N (the negative sign shows the force is attractive)

7Two point charges of +4.0 μC and +6.0 μC are separated by 25 cm in air. Calculate the force between them, and state whether it is attractive or repulsive.

Hint: substitute directly into F = kq₁q₂/r², using k = 8.99×10⁹ N m² C⁻². Convert the separation to metres before squaring it. The sign of both charges tells you whether the force is attractive or repulsive.

8The force between two identical point charges is 8.0 × 10⁻³ N when they are 12 cm apart. Determine the magnitude of each charge.

Hint: since the two charges are identical, Coulomb's law becomes F = kq²/r². Rearrange to make q² the subject, substitute in SI units, then take the square root.

9The force between two point charges separated by 20 cm is 6.0 × 10⁻⁵ N. Predict the new force if the separation is reduced to 5.0 cm, without recalculating the charges.

Hint: you don't need to find the actual charges. Work out the factor by which the separation has changed, then use F ∝ 1/r² to see how the force scales by the square of that factor.

4. Electric fields

A region of space where a charge would experience an electric force is called an electric field. We describe electric fields using two related ideas: field lines (a picture) and field strength (a number).

Electric field strength

E = F / q

Electric field strength, E, is the force per unit charge that a small positive test charge would feel at that point. SI unit: N C⁻¹ (equivalent to V m⁻¹).

Field lines

Field lines are a way of drawing an electric field. Three rules always apply:

Field lines point in the direction of the force on a positive test charge.
Field lines never cross.
The closer together the lines are, the stronger the field — field line density represents field strength.

Fig. 4.1 Radial field lines around an isolated positive charge (left) and negative charge (right). Lines point away from positive charge, towards negative charge.

Worked example 4.1

A charge of +4.0 nC experiences a force of 2.4 × 10⁻⁵ N at a certain point. Calculate the electric field strength at that point.

Answer: E = F/q = (2.4 × 10⁻⁵) / (4.0 × 10⁻⁹) = 6.0 × 10³ N C⁻¹

Drag charges onto the canvas below (or use a preset), then tick Show Field Lines to see the pattern build up. Try recreating the two-equal-positive-charges setup described above and look for the point where the lines cancel.

Interactive simulation by Dr Nigel Hanbury — drhanburyphysics.com/labs/electric-fields.html. Used with permission.

10A point charge of 5.0 nC feels a field strength of 3.0 × 10⁴ N C⁻¹. Calculate the force on the charge.

Hint: rearrange E = F/q to make F the subject, then substitute the given field strength and charge directly.

11Two diagrams show field lines around a charged object. In diagram A the lines are close together; in diagram B (same type of charge, but drawn further from the object) the lines are much more spread out. Explain what this tells you about how field strength changes with distance from a point charge.

Because field line density represents field strength, lines spreading out further from the charge shows that the field gets weaker with increasing distance — consistent with the inverse-square relationship E = kq/r².

5. The uniform field between parallel plates

A very useful electric field can be made by connecting a potential difference across two parallel metal plates. Between the plates (away from the edges), the field is uniform — the same strength and direction everywhere.

Fig. 5.1 The uniform field between two oppositely charged parallel plates, separated by distance d.

E = V / d

Uniform field strength between parallel plates: E = V/d, where V is the potential difference across the plates and d is the separation between them.

This equation shows that N C⁻¹ and V m⁻¹ are equivalent units for electric field strength.

Worked example 5.1

Two parallel plates are separated by 1.2 cm and connected to a 600 V supply. Calculate the electric field strength between them.

Answer: E = V/d = 600 / 0.012 = 5.0 × 10⁴ V m⁻¹

12Parallel plates 8.0 mm apart produce a field of 2.5 × 10⁵ V m⁻¹. Calculate the potential difference across them.

Hint: rearrange E = V/d to make V the subject, and convert the plate separation to metres before substituting.

13A student halves the plate separation but keeps the same potential difference across a parallel-plate arrangement. Explain what happens to the field strength between the plates, and to the force on a charge placed there.

Since E = V/d and V is unchanged while d halves, E doubles. Because F = Eq, the force on any given charge placed between the plates also doubles.

6. Magnetic field lines

A magnetic field exists anywhere a magnetic force acts. Just like electric fields, we represent magnetic fields on paper using field lines — but the rules for drawing them are slightly different.

Magnetic field lines always form closed loops — they never simply start or stop.
Outside a magnet, field lines point from the north pole to the south pole.
Like electric field lines, they never cross, and are closest together where the field is strongest.

Fig. 6.1 Magnetic field lines around a bar magnet

Permanent magnets and poles

A bar magnet has two poles: a north pole and a south pole. Like poles repel; unlike poles attract — exactly like electric charges. But there is one big difference: an isolated single magnetic pole has never been found. If you cut a bar magnet in half, you don't get a separate N piece and a separate S piece — you get two smaller magnets, each with its own N and S pole.

The Earth's magnetic field

The Earth behaves like a giant, weak bar magnet, which is why a compass needle (a small permanent magnet, free to rotate) lines up with it. The end of a compass needle that points towards geographic North is called its "north pole" — which means it is actually attracted towards a magnetic south pole located near the Earth's geographic North.

14Two bar magnets are placed end to end, north pole facing north pole, a small gap between them. Sketch the magnetic field lines you would expect to see in the gap between the two north poles.

Because like poles repel, the field lines from each north pole curve away from the gap rather than crossing it; there is a point on the line joining the poles, midway between them, where the resultant field is zero.

15Explain why cutting a bar magnet in half does not produce an isolated north pole and an isolated south pole.

Magnetic poles always occur in pairs (dipoles) — a single isolated pole ("monopole") has never been observed. Cutting the magnet simply creates two smaller magnets, each with its own complete N and S pole.

7. Electric potential energy

Just as work must be done to lift a mass in a gravitational field, work must be done to move charges closer together or further apart in an electric field. This work is stored as electric potential energy.

Electric potential energy, E_p: the work done to assemble a system of charges, bringing each charge in from infinite separation to its present position. We define E_p = 0 when all charges are infinitely far apart.

For a system of just two point charges, this works out to a very similar equation to Coulomb's law:

E_p = k q₁q₂ / r

Notice the sign carefully:

If the charges have the same sign (both positive or both negative), E_p is positive. Energy would be released as kinetic energy if the charges were let go — they repel.
If the charges have opposite signs, E_p is negative. Energy must be supplied to pull them apart — they attract.

Worked example 7.1

Calculate the electric potential energy of a system consisting of a +3.0 × 10⁻⁸ C charge and a −5.0 × 10⁻⁸ C charge, separated by 6.0 cm.

Answer:
E_p = kq₁q₂/r = (8.99×10⁹)(3.0×10⁻⁸)(−5.0×10⁻⁸)/(0.060) = −2.2 × 10⁻⁴ J
(negative, because the charges attract — energy would need to be supplied to separate them fully)

16Two point charges of +2.0 nC are separated by 15 cm. Calculate the electric potential energy of the system, and explain the sign of your answer.

Hint: substitute directly into E_p = kq₁q₂/r. Before you calculate, decide the sign of your answer from the types of charge involved — this tells you whether energy would be released or would need to be supplied.

17A proton and an electron are separated by 1.0 × 10⁻¹⁰ m (roughly the size of an atom). Calculate the electric potential energy of this pair. (Charge of proton = +1.60×10⁻¹⁹ C, charge of electron = −1.60×10⁻¹⁹ C.)

Hint: substitute the charges of the proton and electron directly into E_p = kq₁q₂/r, keeping the negative sign on the electron's charge.

8. Electric potential

Electric potential energy depends on both charges involved. It is often more useful to describe a point in space on its own — regardless of what charge might be placed there. That's what electric potential does.

Electric potential, V_e, at a point is the work done per unit charge to bring a small positive test charge from infinity to that point.

V_e = kQ / r

Some important features of electric potential:

Electric potential is a scalar quantity — it has size but no direction.
Ve = 0 is defined to be at infinite distance from all charges.
The SI unit is the volt (V), equivalent to J C⁻¹.
Around a positive charge, potential is positive; around a negative charge, potential is negative.
Because potential is a scalar, the combined potential from several charges is found by simple addition (not vector addition, unlike field strength).

Worked example 8.1

Calculate the electric potential at a distance of 3.0 cm from a point charge of −4.0 × 10⁻⁹ C.

Answer: Ve = kQ/r = (8.99×10⁹)(−4.0×10⁻⁹)/(0.030) = −1.2 × 10³ V

18Calculate the combined electric potential at a point which is 20 cm from a charge of +5.0 nC, and 30 cm from a charge of −8.0 nC.

Hint: calculate the potential due to each charge separately using V_e = kQ/r, then add the two values as ordinary signed numbers — potential is a scalar.

19Explain why electric potential can be added directly as numbers, while electric field strength from two charges must be combined using vector addition.

Electric potential is a scalar quantity — it has no direction, so contributions from different charges just add up like ordinary numbers. Electric field strength is a vector — it has both size and direction, so contributions from different charges must be added "tip-to-tail" or using components/Pythagoras.

9. Potential gradient and work done

Electric field strength and electric potential are closely linked. The field strength tells you how quickly the potential is changing as you move through space — this is called the potential gradient.

E = − ΔV_e / Δr

The electric field strength equals the negative of the potential gradient. The minus sign shows that the field points in the direction of decreasing potential, even though we usually just quote the size of E.

This tells us something important: where the potential is constant, the field strength is zero. Fields only exist where potential is changing with position.

Fig. 9.1 Electric field strength, E, and electric potential, V_e, plotted against distance r from the centre of a uniformly charged conducting sphere of radius R. Inside the sphere, E = 0 and V_e is constant — outside, both fall off with distance exactly as they would for a point charge at the centre.

Work done moving a charge

W = q ΔV_e

The work done in moving a charge q between two points in an electric field equals the charge multiplied by the potential difference between those points.

Worked example 9.1

A charge of +2.0 × 10⁻⁶ C is moved from a point at 40 V to a point at 15 V. Calculate the work done on the charge, and state whether the charge gained or lost kinetic energy.

Answer:
ΔV_e = 15 − 40 = −25 V
W = qΔV_e = (2.0×10⁻⁶)(−25) = −5.0 × 10⁻⁵ J
The negative sign shows that the electric field did negative work on the charge as it moved this way — so if nothing else acted on it, the charge would have to be pushed there by an external force; it would not accelerate there on its own (a positive charge moves naturally from high to low potential only if that direction also matches the field's force direction — check the full context of the question before concluding).

20A potential-distance graph for a point charge shows that potential falls from +80 V at r = 2.0 cm to +20 V at r = 8.0 cm. Use the gradient of a straight line joining these two points to estimate the electric field strength over this interval.

Hint: use E = −ΔV/Δr. Work out ΔV (final potential minus initial potential) and Δr (final distance minus initial distance) between the two points, then substitute — don't drop the negative sign in the formula.

21A charge of −3.0 μC moves from a point at −10 V to a point at −55 V. Calculate the work done on the charge by the field.

Hint: find ΔV between the two points first (final potential minus initial potential), then substitute into W = qΔV_e, keeping careful track of both negative signs.

10. Equipotential surfaces

Equipotential surface (or line): a surface connecting all the points that have the same electric potential.

Equipotentials are a second, very useful way of picturing an electric field, alongside field lines.

Equipotential surfaces are always perpendicular to electric field lines. No work is done moving a charge between two points on the same equipotential surface, because there is no potential difference between them.

Fig. 10.1 Equipotential circles (dashed) are always at right angles to the radial field lines (solid) around a point charge.

22A charge of +3.0 C is moved from one point to another point on the same equipotential surface, 12 cm away. Calculate the work done. Explain your answer without doing any further calculation.

Hint: think about what ΔV_e equals when both points lie on the same equipotential surface, then use W = qΔV_e — you shouldn't need a calculator for this one.

23Sketch the field lines and equipotential surfaces around an isolated negative point charge. Label which set of lines you drew as solid and which as dashed.

Field lines are radial straight lines pointing inward, towards the negative charge; equipotentials are concentric circles (spheres in 3D) centred on the charge, each circle crossing every field line at 90°.

11. Equation summary

Quantity	Equation	Unit
Coulomb's law	F = kq₁q₂/r²	N
Electric field strength	E = F/q	N C⁻¹ (= V m⁻¹)
Field strength, point charge	E = kQ/r²	N C⁻¹
Field strength, parallel plates	E = V/d	V m⁻¹
Electric potential energy	E_p = kq₁q₂/r	J
Electric potential	V_e = kQ/r	V (= J C⁻¹)
Field as potential gradient	E = −ΔV_e/Δr	V m⁻¹
Work done moving a charge	W = qΔV_e	J
Coulomb constant	k = 1/(4πε₀) = 8.99 × 10⁹	N m² C⁻²
Elementary charge	e	1.60 × 10⁻¹⁹ C

12. Mixed practice

These questions draw on several sections at once, the way an exam question might.

24Two point charges, +6.0 nC and −6.0 nC, are fixed 40 cm apart.

a) Calculate the force between them.
b) Calculate the electric potential energy of the system.
c) Calculate the electric potential at the midpoint between the two charges.
d) State the electric field strength at the midpoint, giving a reason without calculating.

Hint: a) use Coulomb's law directly. b) use E_p = kq₁q₂/r. c) calculate the potential from each charge separately using V_e = kQ/r and add them, keeping their signs. d) think about direction, not just size — at the midpoint, do the two field contributions point the same way, or opposite ways?

25A student sets up two parallel plates 5.0 cm apart with a p.d. of 2500 V across them, in order to hold a charged dust particle stationary between the plates.

a) Calculate the electric field strength between the plates.
b) The particle has a mass of 8.0 × 10⁻¹⁵ kg. Calculate the charge on the particle needed for it to be held stationary. (g = 9.81 m s⁻²)
c) State how many excess or deficit electrons this corresponds to.

Hint: a) use E = V/d. b) at balance, the electric force equals the weight — set Eq = mg and rearrange to make q the subject. c) once you have q in coulombs, divide by e and round to the nearest whole number.

26A hollow charged conducting sphere of radius 5.0 cm carries a charge of −2.0 nC.

a) State the electric field strength inside the sphere, with a reason.
b) Calculate the electric potential at the surface of the sphere.
c) Calculate the electric potential at the centre of the sphere.
d) Sketch the field lines and two equipotential surfaces, one inside and one outside the sphere.

Hint: a) think about where excess charge sits on a conductor. b) use V_e = kQ/r with the sphere's radius as r. c) think about how potential can change (or not) if the field strength is zero all the way to the centre. d) sketch field lines only outside the sphere, and remember equipotentials are always perpendicular to them.

13. Glossary

Charge (electric): A property of matter, measured in coulombs, that causes objects to experience forces in electric fields.
Coulomb's law: F = kq₁q₂/r² — the equation for the force between two point charges.
Earthing (grounding): Connecting an object to the ground so charge can flow freely, bringing the object to 0 V.
Elementary charge, e: The smallest possible unit of charge, 1.60 × 10⁻¹⁹ C; every charge is a whole-number multiple of e.
Electric field: A region of space in which a charge experiences an electric force.
Electric field strength, E: Force per unit charge, E = F/q, measured in N C⁻¹.
Electric potential, V_e: Work done per unit charge to bring a positive test charge from infinity to a point; a scalar, zero at infinity.
Electric potential energy, E_p: Work done to assemble a system of charges from infinite separation.
Equipotential surface: A surface joining points of equal electric potential; always perpendicular to field lines.
Induction (electrostatic): Charge separation caused in an object by a nearby charge, without contact.
Point charge: A charge treated as existing at a single point in space.
Potential difference: The work done per unit charge moving between two points, ΔV_e = W/q.
Quantisation of charge: The fact that charge only ever occurs in whole-number multiples of e.