Coulomb’s law and electric fields

AshPhys · Revision cheat sheet · Sheet 1 of 6

9702 A Level Ch.22Chapter 22: Coulomb’s lawLesson 22.1Sections 22.1–22.5A Level only (not AS)

Goal: Use F = Qq/4πε₀r², E = Q/4πε₀r² and V = Q/4πε₀r for point charges, combine fields and potentials, and compare with gravity.

force

01The big picture

Every charge sets up an electric field around it. The field strength at a point is the force per unit positive charge on a small test charge placed there, $E = F/q$ , in N C⁻¹ (the same as V m⁻¹). Like charges repel; unlike charges attract.

Two point charges exert a force $F = \dfrac{Qq}{4\pi\varepsilon_0 r^2}$ on each other: Coulomb’s law, another inverse-square law. Outside a charged conducting sphere, the sphere acts as if its charge were a point charge at its centre, so $r$ is measured from the centre.

Dividing by $q$ gives the field of a point charge, $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$ . It is radial: outwards from a positive charge, inwards towards a negative one. Fields are vectors, so fields from several charges add with their directions.

Electric potential $V = \dfrac{Q}{4\pi\varepsilon_0 r}$ is zero at infinity, positive near a positive charge and negative near a negative one. It is a scalar, so potentials simply add with their signs. Moving a charge $q$ through a potential difference $\Delta V$ takes energy $q\Delta V$ , and the field is the negative potential gradient, $E = -\Delta V/\Delta r$ .

The maths matches gravity ( $F = GMm/r^2$ ) with one big difference: gravity only attracts, while electric forces can repel. Between two protons the electric force is about $10^{36}$ times the gravitational one.

Fig. 1 · Radial fields of point charges

Field lines point away from a positive charge and towards a negative one: the direction a positive test charge would be pushed.

Fig. 2 · Fields add as vectors

Midway between +Q and −Q both fields point the same way (away from +, towards −), so they add; the potentials (+ and −) cancel.

Electric field strength (syllabus) [syllabus wording]: Force per unit positive charge.
Electric potential (syllabus) [syllabus wording]: The work done per unit positive charge in bringing a small test charge from infinity to the point.
Coulomb’s law [paraphrase]: Two point charges exert a force on each other proportional to the product of the charges and inversely proportional to the square of their separation.
Field lines [paraphrase]: Show the direction of the force on a positive charge. They start on positive charges, end on negative charges, and are closer together where the field is stronger.

Coulomb’s law and electric fields · 9702 A Level Ch.22Sheet 2 of 6

02Deriving the equations

D1 Field strength of a point charge

Put a small positive test charge $q$ at distance $r$ from a point charge $Q$ , and use the definition $E = F/q$ .

Coulomb’s law gives the force on the test charge:
$F = \frac{Qq}{4\pi\varepsilon_{0}r^{2}}$
Field strength is force per unit positive charge, so divide by $q$ ; the test charge cancels:
$E = \frac{F}{q} = \frac{Q}{4\pi\varepsilon_{0}r^{2}}$
Inverse square: doubling $r$ quarters $E$ . For a negative $Q$ the field points inwards, towards the charge.

Result

E = \frac{Q}{4\pi\varepsilon_{0}r^{2}}

D2 Potential, potential energy and the potential gradient

Potential is zero at infinity. Bringing a positive test charge towards a positive $Q$ means pushing against the repulsion, so the work done, and the potential, are positive.

Fig. 3 · Potential against distance

Around +Q the potential is positive and falls towards zero; around −Q it is negative and rises towards zero. The field strength is minus the gradient.

The potential of a point charge (negative for a negative $Q$ ):
$V = \frac{Q}{4\pi\varepsilon_{0}r}$
A charge $q$ at that point has potential energy $qV$ . The sign tells you the force: positive for repulsion, negative for attraction.
$E_\mathrm{p} = qV = \frac{Qq}{4\pi\varepsilon_{0}r}$
Moving $q$ through a potential difference $\Delta V$ :
$\Delta E_\mathrm{p} = q\,\Delta V$
The field strength is the negative of the potential gradient: on a $V$ – $r$ graph, $E$ is minus the gradient.
$E = -\frac{\Delta V}{\Delta r}$

Result

V = \frac{Q}{4\pi\varepsilon_{0}r} \qquad E_\mathrm{p} = \frac{Qq}{4\pi\varepsilon_{0}r}

Coulomb’s law and electric fields · 9702 A Level Ch.22Sheet 3 of 6

02Deriving the equations (continued)

D3 Combining two charges

With several charges, find each one’s field and potential at the point, then combine them.

Fig. 4 · The neutral point

At N the field from each charge has the same size and points the opposite way, so the resultant field is zero.

Fields are vectors: draw each one (away from +, towards −) and add them with their directions.
$\vec{E} = \vec{E}_{1} + \vec{E}_{2}$
Potentials are scalars: add them with their signs.
$V = \frac{Q_{1}}{4\pi\varepsilon_{0}r_{1}} + \frac{Q_{2}}{4\pi\varepsilon_{0}r_{2}}$
Between two like charges there is a neutral point where the fields cancel. Distance $x$ from $Q_1$ , with $d$ between the charges:
$\frac{Q_{1}}{x^{2}} = \frac{Q_{2}}{(d-x)^{2}} \;\Longrightarrow\; \frac{d-x}{x} = \sqrt{\frac{Q_{2}}{Q_{1}}}$
At the neutral point $E = 0$ , but $V$ is not zero: both potentials there are positive.

Result

\text{add } E \text{ as vectors,} \quad \text{add } V \text{ as scalars}

D4 Electric and gravitational fields compared

Both are inverse-square fields, so every equation has a partner: replace $GM$ by $Q/4\pi\varepsilon_0$ .

Force, field and potential side by side:
$\frac{GMm}{r^{2}} \leftrightarrow \frac{Qq}{4\pi\varepsilon_{0}r^{2}}, \qquad \frac{GM}{r^{2}} \leftrightarrow \frac{Q}{4\pi\varepsilon_{0}r^{2}}, \qquad -\frac{GM}{r} \leftrightarrow \frac{Q}{4\pi\varepsilon_{0}r}$
Differences: gravity only attracts and its potential is always negative; electric forces attract or repel and $V$ can have either sign.
Both forces fall as $1/r^2$ , so their ratio for two particles does not depend on $r$ . For a proton and an electron:
$\frac{F_\mathrm{E}}{F_\mathrm{G}} = \frac{e^{2}}{4\pi\varepsilon_{0}Gm_\mathrm{p}m_\mathrm{e}} = \frac{8.99\times10^{9} \times (1.60\times10^{-19})^{2}}{6.67\times10^{-11} \times 1.67\times10^{-27} \times 9.11\times10^{-31}} = 2.3\times10^{39}$

Result

\text{same } 1/r^{2} \text{ form;} \quad F_\mathrm{E} \gg F_\mathrm{G} \text{ for particles}

Coulomb’s law and electric fields · 9702 A Level Ch.22Sheet 4 of 6

03Key equations

Field strength (definition)

E = \frac{F}{q}

Coulomb’s law

F = \frac{Qq}{4\pi\varepsilon_{0}r^{2}}

Field of a point charge

E = \frac{Q}{4\pi\varepsilon_{0}r^{2}}

r from the centre

Electric potential

V = \frac{Q}{4\pi\varepsilon_{0}r}

zero at infinity; sign of Q

Potential energy of two charges

E_\mathrm{p} = \frac{Qq}{4\pi\varepsilon_{0}r}

Work done / energy change

W = q\,\Delta V

Field and potential gradient

E = -\frac{\Delta V}{\Delta r}

Uniform field (parallel plates)

E = \frac{\Delta V}{\Delta d}

The Coulomb constant

\frac{1}{4\pi\varepsilon_{0}} = 8.99\times10^{9}

N m² C⁻²

Symbols and units

$Q,\ q$ charges $\mathrm{C}$
$r$ distance between centres $\mathrm{m}$
$\varepsilon_{0}$ permittivity of free space, 8.85 × 10⁻¹² $\mathrm{F\,m^{-1}}$
$F$ electric force $\mathrm{N}$
$E$ electric field strength $\mathrm{N\,C^{-1}}$
$V$ electric potential $\mathrm{V}$
$E_\mathrm{p}$ electric potential energy $\mathrm{J}$
$W$ work done moving a charge $\mathrm{J}$
$d$ plate separation $\mathrm{m}$
$e$ elementary charge, 1.60 × 10⁻¹⁹ $\mathrm{C}$
$m_\mathrm{p}$ proton mass, 1.67 × 10⁻²⁷ $\mathrm{kg}$
$m_\mathrm{e}$ electron mass, 9.11 × 10⁻³¹ $\mathrm{kg}$

04Sample problems with solutions

Sample 1 · Force between two charged spheres
Two small spheres carry charges of +4.0 nC and −6.0 nC, with their centres 5.0 cm apart. Calculate the force between them and state its direction.

In SI units (nC = 10⁻⁹ C, 5.0 cm = 0.050 m), Coulomb’s law with the size of each charge. Opposite signs, so the force is attractive.
$F = \frac{8.99\times10^{9} \times 4.0\times10^{-9} \times 6.0\times10^{-9}}{0.050^{2}} = 8.63\times10^{-5}\ \mathrm{N}$

Answer:

F = 8.6\times10^{-5}

N, attractive

Unit check:

\frac{\mathrm{N\,m^{2}\,C^{-2}} \times \mathrm{C} \times \mathrm{C}}{\mathrm{m^{2}}} = \mathrm{N}\ \checkmark

Sample 2 · The neutral point between two charges
Charges of +2.0 μC and +8.0 μC are 30 cm apart. Find the point between them where the electric field is zero, and calculate the potential there.

At distance $x$ from the 2.0 μC charge the two fields are equal and opposite:
$\frac{2.0}{x^{2}} = \frac{8.0}{(0.30-x)^{2}} \;\Longrightarrow\; \frac{0.30-x}{x} = \sqrt{4} = 2 \;\Longrightarrow\; x = 0.10\ \mathrm{m}$
Potentials add as scalars:
$V = 8.99\times10^{9}\left(\frac{2.0\times10^{-6}}{0.10} + \frac{8.0\times10^{-6}}{0.20}\right) = 5.39\times10^{5}\ \mathrm{V}$

Answer: 10 cm from the +2.0 μC charge;

V = 5.4\times10^{5}

V there

Unit check:

\mathrm{N\,m^{2}\,C^{-2}} \times \frac{\mathrm{C}}{\mathrm{m}} = \mathrm{N\,m\,C^{-1}} = \mathrm{J\,C^{-1}} = \mathrm{V}\ \checkmark

Coulomb’s law and electric fields · 9702 A Level Ch.22Sheet 5 of 6

04Sample problems with solutions (continued)

Sample 3 · Closest approach of an α-particle
An α-particle (charge +2e) with 7.7 MeV of kinetic energy is fired straight at a gold nucleus (charge +79e). Calculate how close it gets to the centre of the nucleus.

Kinetic energy in joules:
$E_\mathrm{k} = 7.7\times10^{6} \times 1.60\times10^{-19} = 1.23\times10^{-12}\ \mathrm{J}$
At closest approach it stops for an instant: all the kinetic energy has become electric potential energy.
$E_\mathrm{k} = \frac{Qq}{4\pi\varepsilon_{0}r}$
$r$ is in the denominator and $E_\mathrm{k}$ is the lone term on the other side: swap them in one step (cross-multiplication).
$r = \frac{Qq}{4\pi\varepsilon_{0}E_\mathrm{k}} = \frac{8.99\times10^{9} \times (2 \times 1.60\times10^{-19})(79 \times 1.60\times10^{-19})}{1.23\times10^{-12}} = 2.95\times10^{-14}\ \mathrm{m}$

Answer:

r = 3.0\times10^{-14}

m

Unit check:

\frac{\mathrm{N\,m^{2}\,C^{-2}} \times \mathrm{C^{2}}}{\mathrm{J}} = \frac{\mathrm{N\,m^{2}}}{\mathrm{N\,m}} = \mathrm{m}\ \checkmark

Sample 4 · Two charged spheres hanging from one point
Two identical small spheres, each of mass 0.20 g and charge Q, hang from the same point on insulating threads 0.50 m long. They repel and come to rest with each thread at 15° to the vertical. Calculate Q.

Each sphere is in equilibrium. Resolving, $F_\mathrm{T}\cos\theta = mg$ and $F_\mathrm{T}\sin\theta = F$ ; dividing the two:
$F = mg\tan\theta = 0.20\times10^{-3} \times 9.81 \times \tan 15^\circ = 5.26\times10^{-4}\ \mathrm{N}$
Separation of the centres:
$d = 2L\sin\theta = 2 \times 0.50 \times \sin 15^\circ = 0.259\ \mathrm{m}$
Coulomb’s law with equal charges, $F = Q^2/(4\pi\varepsilon_0 d^2)$ :
$Q = d\sqrt{4\pi\varepsilon_{0}F} = 0.259 \times \sqrt{\frac{5.26\times10^{-4}}{8.99\times10^{9}}} = 6.26\times10^{-8}\ \mathrm{C}$

Fig. 5 · Two charged spheres in equilibrium

Tension, weight and the electric repulsion balance on each sphere: F = mg tan θ.

Answer:

Q = 6.3\times10^{-8}

C (63 nC)

Unit check:

\mathrm{m} \times \sqrt{\frac{\mathrm{N}}{\mathrm{N\,m^{2}\,C^{-2}}}} = \mathrm{m} \times \frac{\mathrm{C}}{\mathrm{m}} = \mathrm{C}\ \checkmark

Coulomb’s law and electric fields · 9702 A Level Ch.22Sheet 6 of 6

05Practice problems

P1
Calculate the force between two protons 1.0 × 10⁻¹⁵ m apart.
Final answer
2.3 × 10² N, repulsive
Answer: 2.3 × 10² N, repulsive
P2
Calculate the electric field strength 0.20 m from a +5.0 μC point charge.
Final answer
1.1 × 10⁶ N C⁻¹, away from the charge
Answer: 1.1 × 10⁶ N C⁻¹, away from the charge
P3
Calculate the force on an electron in a field of strength 3.0 × 10⁴ N C⁻¹.
Final answer
4.8 × 10⁻¹⁵ N, opposite to the field
Answer: 4.8 × 10⁻¹⁵ N, opposite to the field
P4
Calculate the potential 0.10 m from a −3.0 nC point charge.
Final answer
−2.7 × 10² V
Answer: −2.7 × 10² V
P5
Calculate the work done moving a +2.0 μC charge from a point at 50 V to a point at 250 V.
Final answer
4.0 × 10⁻⁴ J
Answer: 4.0 × 10⁻⁴ J
P6
The distance from a point charge is doubled. By what factor do F, E and V change?
Final answer
F and E to ¼; V to ½
Answer: F and E to ¼; V to ½
P7
An isolated sphere of radius 0.15 m is charged to a potential of 3.0 × 10⁵ V. Calculate its charge.
Final answer
5.0 × 10⁻⁶ C
Answer: 5.0 × 10⁻⁶ C
P8
Along a field line the potential falls by 60 V over 2.0 mm. Calculate the field strength there.
Final answer
3.0 × 10⁴ V m⁻¹
Answer: 3.0 × 10⁴ V m⁻¹
P9
Calculate the ratio of the electric force to the gravitational force between two protons.
Final answer
1.2 × 10³⁶
Answer: 1.2 × 10³⁶
P10
Calculate the electric potential energy of an electron and a proton 5.3 × 10⁻¹¹ m apart.
Final answer
−4.3 × 10⁻¹⁸ J
Answer: −4.3 × 10⁻¹⁸ J

06Other useful information

Common mistakes

✗ Wrong: Add the fields of two charges as plain numbers.
✓ Right: E is a vector: draw each field’s direction first. V is a scalar: add with signs.
✗ Wrong: Where E = 0, V must be 0 too.
✓ Right: No: at the neutral point between like charges E = 0 but V is not.
✗ Wrong: Use the distance from the surface of a charged sphere.
✓ Right: r is measured from the centre.
✗ Wrong: Forget to convert nC and μC.
✓ Right: nC = 10⁻⁹ C, μC = 10⁻⁶ C. Square r in metres.
✗ Wrong: Electric potential is always negative, like gravitational potential.
✓ Right: V is positive near a positive charge, negative near a negative one.

Exam tips

Sketch the charges and draw each field arrow (away from +, towards −) before calculating.
Neutral point between like charges: Q₁/x² = Q₂/(d − x)², then take square roots.
Closest approach and similar energy questions: kinetic energy lost = potential energy gained.
Hanging charged spheres: resolve forces, F = mg tan θ, then Coulomb’s law.

Explore

What the syllabus asks

9702 topic 18, Electric fields: E = F/q, field lines, Coulomb’s law, E = Q/4πε₀r², V = Q/4πε₀r, E as the negative potential gradient, Ep = Qq/4πε₀r. Coursebook Chapter 22.
A Level only: not part of AS.
For a point outside a charged conducting sphere, treat its charge as a point charge at its centre.
The matching gravity results are in Chapter 17.