Electric Fields

Coulomb's law, electric field strength, and electric potential

SPEC 3.7.3 • FIELDS AND THEIR CONSEQUENCES

Coulomb's Law

The force between two point charges is given by Coulomb's law:

Coulomb's Law

F = \frac{1}{4 π ε _{0}} \frac{Q _{1} Q _{2}}{r ^{2}} = \frac{k Q _{1} Q _{2}}{r ^{2}}

Key Points

Like charges repel (F is positive → force outward)

Unlike charges attract (F is negative → force inward)

Force is an inverse square law: double r → quarter F

Permittivity & Coulomb Constant

Permittivity of Free Space (ε₀)

ε_{0} = 8.85 \times 1 0^{- 12} F m^{- 1}

Permittivity measures how easily an electric field can pass through a medium. A vacuum (free space) has the lowest permittivity — electric fields pass through most easily.

Coulomb Constant (k)

k = \frac{1}{4 π ε _{0}} = 8.99 \times 1 0^{9} N m^{2} C^{- 2}

The constant k is simply a convenient way to combine 1/(4πε₀) into a single number. It tells us how strong the electric force is for given charges and separation.

Exam Note

Air ≈ vacuum: The permittivity of air is so close to ε₀ that we treat them as equal in A-level calculations.

Charged Sphere Rule

For a uniformly charged conducting sphere, the charge acts as if concentrated at the centre (for external points).

Worked Example01

Two point charges of +3.0 μC and −5.0 μC are separated by 0.20 m in air. Calculate the magnitude and direction of the electrostatic force between them.

1

Write down known values: Q₁ = 3.0 × 10⁻⁶ C, Q₂ = 5.0 × 10⁻⁶ C, r = 0.20 m

2

Apply Coulomb's law:

F = \frac{k Q _{1} Q _{2}}{r ^{2}}

3

Substitute:

F = \frac{8.99 \times 1 0 ^{9} \times 3.0 \times 1 0 ^{- 6} \times 5.0 \times 1 0 ^{- 6}}{( 0.20 ) ^{2}}

4

Calculate:

F = \frac{8.99 \times 1 0 ^{9} \times 1.5 \times 1 0 ^{- 11}}{0.04} = \frac{0.135}{0.04} = 3.4 N

AnswerF = 3.4 N, attractive (charges are opposite signs)

Electric vs Gravitational Fields

Gravitational

F = \frac{G m _{1} m _{2}}{r ^{2}}

Acts on mass

Always attractive

Very weak at atomic scale

G = 6.67 × 10⁻¹¹ N m² kg⁻²

Electric

F = \frac{k Q _{1} Q _{2}}{r ^{2}}

Acts on charge

Attractive OR repulsive

Very strong at atomic scale

k = 8.99 × 10⁹ N m² C⁻²

Key Difference

Electric forces between subatomic particles are ~10³⁶ times stronger than gravitational forces! This is why gravity is negligible at the atomic scale, but dominates at planetary scales where objects are electrically neutral.

Exam Insight

Since the electrostatic repulsion between protons is so much stronger than gravitational attraction, there must be another attractive force holding nuclei together — this implies the existence of the strong nuclear force.

Similarities

Both are inverse square laws, act at a distance (non-contact forces), and can be described using field theory.

Electric Field Strength

Electric field strength E at a point is defined as the force per unit positive charge placed at that point:

Definition

E = \frac{F}{Q}

Units: N C⁻¹ or equivalently V m⁻¹

Important

E is defined using a positive test charge. The field exists whether or not a test charge is present — it tells you what force a charge would experience.

Vector Quantity

Electric field strength is a vector. Direction of E = direction of force on a positive charge:

Points away from positive charges

Points towards negative charges

Exam Tip

Unlike magnetic fields, an electric field exerts force on stationary charged particles.

Radial Electric Fields

Around a point charge or charged sphere, the field is radial:

Radial Field Formula

E = \frac{1}{4 π ε _{0}} \frac{Q}{r ^{2}} = \frac{k Q}{r ^{2}}

Lines point OUTWARD

Lines point INWARD

Field Line Rules

Field lines radiate outward like spokes on a wheel from a positive charge. Lines are closer together where the field is stronger. They never cross.

Worked Example02

A metal sphere carries a charge of +2.0 nC. Calculate the electric field strength at a distance of 0.15 m from the centre of the sphere.

1

Identify values: Q = 2.0 × 10⁻⁹ C, r = 0.15 m

2

Apply radial field formula:

E = \frac{k Q}{r ^{2}}

3

Substitute:

E = \frac{8.99 \times 1 0 ^{9} \times 2.0 \times 1 0 ^{- 9}}{( 0.15 ) ^{2}}

4

Calculate:

E = \frac{18.0}{0.0225} = 800 N C^{- 1}

AnswerE = 800 N C⁻¹ (directed radially outward from sphere)

Uniform Electric Fields

Between two parallel charged plates, the electric field is uniform (constant magnitude and direction):

Uniform Field Formula

E = \frac{V}{d}

where V = potential difference between plates, d = separation

Derivation

Step 1: Work done moving charge: W = F × d

Step 2: Work done also equals: W = Q × ΔV

Step 3: Equating: F × d = Q × ΔV

Step 4: Rearranging: F/Q = ΔV/d

Step 5: Since E = F/Q: E = ΔV/d

Worked Example03

Two parallel plates are separated by 5.0 mm and have a potential difference of 200 V between them. Calculate (a) the electric field strength between the plates, and (b) the force on an electron placed between them.

1

Part (a): V = 200 V, d = 5.0 × 10⁻³ m

2

Apply:

E = \frac{V}{d} = \frac{200}{5.0 \times 1 0 ^{- 3}} = 40000 V m^{- 1}

3

Part (b): Use F = QE with Q = e = 1.60 × 10⁻¹⁹ C

4

F = 1.60 \times 1 0^{- 19} \times 40000 = 6.4 \times 1 0^{- 15} N

Answer(a) E = 4.0 × 10⁴ V m⁻¹; (b) F = 6.4 × 10⁻¹⁵ N (towards positive plate)

Electric Potential

The absolute electric potential V at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point:

Definition

V = \frac{W}{Q}

Units: Volts (V) = J C⁻¹

Zero Reference

Electric potential is defined as zero at infinity

The potential is largest right next to the charge

V falls off as 1/r (slower than E which falls off as 1/r²)

For a point charge or charged sphere:

Radial Potential Formula

V = \frac{1}{4 π ε _{0}} \frac{Q}{r} = \frac{k Q}{r}

Scalar Quantity

Unlike E, potential V is a scalar — potentials from multiple charges add algebraically (not as vectors).

Worked Example04

Calculate the electric potential at a distance of 0.30 m from a point charge of +4.0 μC.

1

Identify values: Q = 4.0 × 10⁻⁶ C, r = 0.30 m

2

Apply:

V = \frac{k Q}{r}

3

Substitute:

V = \frac{8.99 \times 1 0 ^{9} \times 4.0 \times 1 0 ^{- 6}}{0.30}

4

Calculate:

V = \frac{3.60 \times 1 0 ^{4}}{0.30} = 1.2 \times 1 0^{5} V

AnswerV = 1.2 × 10⁵ V (or 120 kV)

Work Done Moving Charge

The work done in moving a charge Q through a potential difference ΔV is:

Work Done Formula

Δ W = Q Δ V

Sign Convention

Positive work: Energy must be supplied to move +Q from low V to high V (against the field)

Negative work: Energy is released when charge moves "naturally" (with the field)

Energy-Potential Link

This formula is key for understanding particle accelerators — charged particles gain KE as they move through potential differences.

Important

ΔV = V₂ − V₁ where V₂ is the final potential and V₁ is the initial potential. The sign of W tells you whether energy was supplied or released.

Worked Example05

An electron is accelerated from rest through a potential difference of 500 V. Calculate (a) the work done on the electron, and (b) its final speed.

1

Part (a): W = QΔV = eV (for an electron accelerated through V)

2

W = 1.60 × 10⁻¹⁹ × 500 = 8.0 × 10⁻¹⁷ J

3

Part (b): Work done = kinetic energy gained: ½mv² = eV

4

v = \frac{2 e V}{m} = \frac{2 \times 8.0 \times 1 0 ^{- 17}}{9.11 \times 1 0 ^{- 31}} = 1.76 \times 1 0^{14}

5

v = 1.33 × 10⁷ m s⁻¹

Answer(a) W = 8.0 × 10⁻¹⁷ J (or 500 eV); (b) v = 1.3 × 10⁷ m s⁻¹

Equipotential Surfaces

Equipotential surfaces are surfaces where every point has the same potential. No work is done moving a charge along an equipotential.

Key Properties

Equipotentials are perpendicular to field lines

No work is done moving charge along an equipotential

Closer spacing = stronger field (steeper potential gradient)

Relationship Between E and V

Electric field strength and potential are related by:

E-V Relationship

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

E equals the negative potential gradient

Graphical Interpretation

E = gradient of V-r graph (with negative sign)

ΔV = area under E-r graph

Uniform field: V varies linearly with distance

Radial field: V ∝ 1/r (curve gets shallower)

Worked Example06

The electric field strength between two points varies linearly from 2000 V m⁻¹ to 500 V m⁻¹ over a distance of 0.10 m. Calculate the potential difference between these points.

1

ΔV = area under E-r graph

2

The graph is a trapezium with parallel sides 2000 and 500, height 0.10

3

Area = ½ × (2000 + 500) × 0.10 = ½ × 2500 × 0.10

4

Area = 125 V

AnswerΔV = 125 V

Motion in Uniform Electric Fields

A charged particle entering a uniform field experiences a constant force, similar to a projectile in a gravitational field:

Both paths are parabolic — positive charges deflect toward negative plate, electrons toward positive plate

Analysis Method

Horizontal: No force → constant velocity (vₓ = constant)

Vertical: Constant force F = QE → constant acceleration a = QE/m

Path is parabolic (same maths as projectile motion)

Worked Example07

An electron enters a uniform electric field of 5000 V m⁻¹ at right angles with a horizontal velocity of 2.0 × 10⁷ m s⁻¹. The plates are 4.0 cm long. Calculate the vertical deflection of the electron as it leaves the field.

1

Time in field:

t = \frac{length}{v _{x}} = \frac{0.040}{2.0 \times 1 0 ^{7}} = 2.0 \times 1 0^{- 9} s

2

Vertical acceleration:

a = \frac{e E}{m} = \frac{1.60 \times 1 0 ^{- 19} \times 5000}{9.11 \times 1 0 ^{- 31}} = 8.78 \times 1 0^{14} m s^{- 2}

3

Vertical deflection:

y = \frac{1}{2} a t^{2} = \frac{1}{2} \times 8.78 \times 1 0^{14} \times (2.0 \times 1 0^{- 9})^{2}

4

y = \frac{1}{2} \times 8.78 \times 1 0^{14} \times 4.0 \times 1 0^{- 18} = 1.76 \times 1 0^{- 3} m

Answery = 1.8 mm (towards positive plate)

Practice

Question 1: Two protons are separated by a distance of 5.0 × 10⁻¹⁵ m (about the diameter of a nucleus). Calculate the electrostatic force between them, and compare this to the gravitational force between them.

1

Electric force:

F_{E} = \frac{k q ^{2}}{r ^{2}} = \frac{8.99 \times 1 0 ^{9} \times ( 1.60 \times 1 0 ^{- 19} ) ^{2}}{( 5.0 \times 1 0 ^{- 15} ) ^{2}}

2

F_{E} = \frac{8.99 \times 1 0 ^{9} \times 2.56 \times 1 0 ^{- 38}}{2.5 \times 1 0 ^{- 29}} = 9.2 N

(repulsive)

3

Gravitational force:

F_{G} = \frac{G m ^{2}}{r ^{2}} = \frac{6.67 \times 1 0 ^{- 11} \times ( 1.67 \times 1 0 ^{- 27} ) ^{2}}{( 5.0 \times 1 0 ^{- 15} ) ^{2}}

4

F_{G} = 7.4 \times 1 0^{- 36} N

AnswerF_E = 9.2 N, F_G = 7.4 × 10⁻³⁶ N. Ratio: F_E/F_G ≈ 10³⁶

Practice

Question 2: A Van de Graaff generator has a dome of radius 0.15 m carrying a charge of 2.5 μC. Calculate (a) the electric field strength and (b) the electric potential at the surface of the dome.

1

(a)

E = \frac{k Q}{r ^{2}} = \frac{8.99 \times 1 0 ^{9} \times 2.5 \times 1 0 ^{- 6}}{( 0.15 ) ^{2}}

2

E = \frac{2.25 \times 1 0 ^{4}}{0.0225} = 1.0 \times 1 0^{6} V m^{- 1}

3

(b)

V = \frac{k Q}{r} = \frac{8.99 \times 1 0 ^{9} \times 2.5 \times 1 0 ^{- 6}}{0.15}

4

V = 1.5 \times 1 0^{5} V

Answer(a) E = 1.0 × 10⁶ V m⁻¹; (b) V = 150 kV

Practice

Question 3: An alpha particle (charge +2e, mass 6.64 × 10⁻²⁷ kg) is accelerated from rest through a potential difference of 50 kV. Calculate its final speed.

1

Work done = kinetic energy gained: QV = ½mv²

2

Q = 2e = 2 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹ C

3

v = \frac{2 Q V}{m} = \frac{2 \times 3.20 \times 1 0 ^{- 19} \times 50000}{6.64 \times 1 0 ^{- 27}}

4

v = \frac{3.20 \times 1 0 ^{- 14}}{6.64 \times 1 0 ^{- 27}} = 4.82 \times 1 0^{12} = 2.2 \times 1 0^{6} m s^{- 1}

Answerv = 2.2 × 10⁶ m s⁻¹

Summary

Key Formulas

QuantityRadial FieldUniform Field

Force

F = \frac{k Qq}{r ^{2}}

F = QE

Field Strength

E = \frac{k Q}{r ^{2}}

E = \frac{V}{d}

Potential

V = \frac{k Q}{r}

V varies linearly

Work Done

W = Q Δ V

Key Exam Points

E = force per unit positive charge (direction matters)

V is a scalar; E is a vector

Equipotentials are perpendicular to field lines

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

— field strength is negative potential gradient

Charged particles follow parabolic paths in uniform fields