Electric Fields

An electric field is a region of space in which a charged particle experiences a force. This concept is central to understanding how charges interact at a distance, and it underpins everything from simple circuits to particle accelerators. Electric fields are invisible, but their effects are measurable and predictable using well-defined mathematical relationships.

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Electric Field Strength

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

E = F / Q

where F is the force in newtons (N) and Q is the charge in coulombs (C). The unit of electric field strength is N C⁻¹ (newtons per coulomb), which is equivalent to V m⁻¹ (volts per metre).

Electric field strength is a vector quantity — it has both magnitude and direction. By convention, the direction of the electric field is the direction in which a positive test charge would move if placed in the field. This means field lines point away from positive charges and towards negative charges.

Why We Use a Test Charge

The definition uses a "unit positive charge" as a theoretical probe. In practice, the test charge must be small enough not to disturb the field it is measuring. If you placed a large charge near another charge, the large charge would redistribute the source charge and distort the field. The test charge is an idealisation that lets us map the field without changing it.

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Field Lines

Electric field lines are a visual tool for representing the direction and relative strength of an electric field.

Rules for field lines:

• They start on positive charges and end on negative charges
• They never cross each other
• The density of field lines indicates the field strength — closer lines mean a stronger field
• They meet conducting surfaces at right angles

Radial Fields

A point charge (or a charged sphere, when viewed from outside) produces a radial field. The field lines spread outward in all directions from a positive charge (or converge inward towards a negative charge). The field strength decreases with distance from the charge — the lines spread further apart.

The key feature of a radial field is that it follows an inverse-square law: the field strength is proportional to 1/r². At twice the distance, the field is one quarter as strong. At three times the distance, it is one ninth as strong.

Uniform Fields

Between two parallel plates connected to a potential difference, the electric field is uniform. The field lines are parallel, equally spaced, and perpendicular to the plates. This means the field strength is the same at every point between the plates (ignoring edge effects).

At the edges of the plates, the field lines curve outward — this is called the fringe effect. In exam questions, you are usually told to ignore edge effects and treat the field as perfectly uniform.

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Uniform Field Between Parallel Plates

For a uniform field between parallel plates separated by a distance d with a potential difference V across them:

E = V / d

This is an extremely useful relationship. It tells you that doubling the voltage doubles the field strength, and doubling the plate separation halves it.

Worked Example 1

Two parallel plates are separated by 5.0 cm and have a potential difference of 2000 V across them. Calculate the electric field strength between the plates.

E = V / d = 2000 / 0.050 = 40 000 V m⁻¹ = 4.0 × 10⁴ V m⁻¹

Note the conversion of 5.0 cm to 0.050 m — always work in SI units.

Worked Example 2

A pair of parallel plates separated by 1.2 cm are used in an ink-jet printer to deflect charged droplets. The maximum field strength the air can sustain without breakdown (sparking) is 3.0 × 10⁶ V m⁻¹. What is the maximum voltage that can be applied across the plates?

V = Ed = 3.0 × 10⁶ × 0.012 = 36 000 V = 36 kV

In reality, ink-jet printers use much lower voltages (a few kV) with very small plate separations (a few mm) to achieve the required deflection.

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Direction Conventions

The direction of the electric field is defined as the direction a positive charge would accelerate. This means:

• Between parallel plates, the field points from the positive plate to the negative plate
• Around a positive point charge, the field points radially outward
• Around a negative point charge, the field points radially inward

When solving problems involving negative charges (such as electrons), remember that the force on the charge is in the opposite direction to the field. An electron placed in a uniform field between parallel plates accelerates towards the positive plate — against the field direction.

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Force on a Charge in a Uniform Field

Rearranging E = F / Q gives:

F = EQ

For a uniform field (E = V / d), this becomes:

F = VQ / d

This force is constant throughout the uniform field, meaning a charged particle between parallel plates experiences a constant acceleration — exactly like a mass in a uniform gravitational field near the Earth’s surface.

Worked Example 3

An electron (charge = 1.6 × 10⁻¹⁹ C, mass = 9.11 × 10⁻³¹ kg) enters a uniform electric field of strength 5.0 × 10³ V m⁻¹. Calculate the force on the electron and its acceleration.

Force: F = EQ = 5.0 × 10³ × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹⁶ N

Acceleration: a = F / m = 8.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹ = 8.8 × 10¹⁴ m s⁻²

This is an enormous acceleration — roughly 10¹⁴ times greater than g. This illustrates why electric forces dominate over gravitational forces at the atomic and subatomic scale.

Worked Example 4: Projectile Motion Between Plates

A proton (charge = 1.6 × 10⁻¹⁹ C, mass = 1.67 × 10⁻²⁷ kg) enters horizontally between two parallel plates with a horizontal velocity of 1.0 × 10⁶ m s⁻¹. The plates are 4.0 cm long and separated by 2.0 cm with a potential difference of 200 V. Calculate the vertical deflection as the proton exits the plates.

E = V / d = 200 / 0.020 = 10 000 V m⁻¹

F = EQ = 10 000 × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹⁵ N

a = F / m = 1.6 × 10⁻¹⁵ / 1.67 × 10⁻²⁷ = 9.58 × 10¹¹ m s⁻²

Time to traverse plates: t = L / v = 0.040 / 1.0 × 10⁶ = 4.0 × 10⁻⁸ s

Vertical deflection: s = ½at² = ½ × 9.58 × 10¹¹ × (4.0 × 10⁻⁸)² = ½ × 9.58 × 10¹¹ × 1.6 × 10⁻¹⁵ = 7.7 × 10⁻⁴ m = 0.77 mm

Since this is much less than half the plate separation (10 mm), the proton exits without hitting a plate.

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Comparing Electric and Gravitational Fields

There is a close analogy between electric and gravitational fields:

Property	Electric Field	Gravitational Field
Field strength definition	Force per unit charge (E = F/Q)	Force per unit mass (g = F/m)
Uniform field	Between parallel plates	Near Earth’s surface
Radial field	Around a point charge	Around a point mass
Field lines	Start on +, end on −	Always point towards mass
Force can be	Attractive or repulsive	Attractive only
Source property	Charge (positive or negative)	Mass (always positive)

This analogy is helpful throughout this topic. Many of the mathematical relationships have the same structure, differing only in whether they involve charge or mass. A charged particle moving horizontally between parallel plates follows a parabolic path — exactly like a ball thrown horizontally in a gravitational field.

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Real-World Applications

Cathode ray oscilloscopes (CROs): Electrons are accelerated and then deflected by uniform electric fields between parallel plates. Two sets of plates (horizontal and vertical) allow the electron beam to be steered to any point on the screen. The deflection is proportional to the applied voltage, which is how the CRO displays waveforms.

Electrostatic precipitators: Used in coal-fired power stations to remove particulate pollution from flue gases. Particles pass through a strong electric field between plates, become charged, and are attracted to collecting plates where they accumulate and can be removed. This technology removes over 99% of particulate matter.

Capacitive touchscreens: Your finger changes the local electric field pattern on the screen surface. Sensors detect this change and determine the touch location. The field principles are exactly those covered in this lesson.

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Common Exam Mistakes

• Forgetting unit conversions: Always convert cm to m and kV to V before substituting into equations
• Sign confusion with electrons: The force on an electron is opposite to the field direction because the electron carries negative charge
• Confusing field strength with force: E is the force per unit charge, not the force itself. You need F = EQ to find the actual force
• Edge effects: In calculations, assume the field is uniform unless told otherwise, but be aware that real fields have fringe effects at the edges

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Key Points

• Electric field strength E = F / Q, measured in N C⁻¹ or V m⁻¹
• For a uniform field between parallel plates, E = V / d
• Field lines show direction (the way a positive charge would move) and relative strength
• Radial fields surround point charges; uniform fields exist between parallel plates
• The force on a charge in a uniform field is constant: F = EQ
• A charged particle between parallel plates follows a parabolic path (like a projectile in gravity)
• Electric and gravitational fields share many structural similarities

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A-Level Deep Dive: Electric Fields

Spec mapping

Edexcel 9PH0 specification Topic 7 — Electric and Magnetic Fields covers the concept of an electric field as a region in which a charged object experiences a force; the definition of electric field strength as force per unit positive charge $E = F/Q$E=F/Q; the use of $E = V/d$E=V/d for the uniform field between parallel plates; Coulomb's law $F = kQ_1Q_2/r^2$F=kQ1​Q2​/r2 for the force between point charges; and the resulting radial field $E = kQ/r^2$E=kQ/r2 around a point charge (refer to the official Pearson Edexcel specification document for exact wording). This sub-strand sits inside the broader Topic 7, which also brings in capacitance, magnetic flux density, electromagnetic induction and the motion of charged particles in fields. Electric fields are examined predominantly in 9PH0 Paper 2 (Section A — core physics carried forward and Section B — further mechanics, fields and particles), with synoptic links into Paper 3 (general and practical principles) through deflection-tube practicals. The Edexcel formula booklet does list $F = kQ_1Q_2/r^2$F=kQ1​Q2​/r2, $E = F/Q$E=F/Q, $E = V/d$E=V/d and $E = kQ/r^2$E=kQ/r2, but does not list the value of $k = 1/(4\pi\varepsilon_0) \approx 8.99 \times 10^9$k=1/(4πε0​)≈8.99×109 N m² C⁻² as a labelled formula — candidates are expected to recognise it from the data sheet.

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Worked example with full mark scheme

Question (8 marks):

(a) Two small charged spheres are held $0.15$0.15 m apart in a vacuum. Sphere A carries a charge of $+4.0 \times 10^{-9}$+4.0×10−9 C and sphere B carries a charge of $-6.0 \times 10^{-9}$−6.0×10−9 C. Calculate the magnitude of the electric force between them and state its direction. (3)

(b) The negative sphere is removed and a small particle of charge $+2.0 \times 10^{-12}$+2.0×10−12 C is placed $0.05$0.05 m from sphere A. Calculate the electric field strength at the position of the particle due to sphere A, and hence the force on the particle. (5)

Solution with mark scheme:

(a) Step 1 — apply Coulomb's law.

$F = \dfrac{kQ_1 Q_2}{r^2} = \dfrac{(8.99 \times 10^9)(4.0 \times 10^{-9})(6.0 \times 10^{-9})}{(0.15)^2}$F=r2kQ1​Q2​​=(0.15)2(8.99×109)(4.0×10−9)(6.0×10−9)​

M1 — correct substitution into Coulomb's law with magnitudes only (the negative sign on $Q_2$Q2​ tells us the direction but should not be carried through into the magnitude calculation; carrying it through and quoting a "negative force" loses marks because force is a vector and the algebraic sign is meaningless without a stated reference axis).

Step 2 — evaluate.

$F = \dfrac{(8.99 \times 10^9)(2.4 \times 10^{-17})}{0.0225} = \dfrac{2.158 \times 10^{-7}}{0.0225} \approx 9.6 \times 10^{-6} \text{ N}$F=0.0225(8.99×109)(2.4×10−17)​=0.02252.158×10−7​≈9.6×10−6 N

A1 — magnitude correct to two significant figures.

Step 3 — state direction.

The force is attractive because the two charges have opposite signs; sphere A is pulled towards sphere B and sphere B is pulled towards sphere A.

B1 — correct identification of the attractive nature with reference to the opposite-sign rule. Many candidates leave the direction implicit; the mark requires an explicit statement.

(b) Step 1 — calculate the field due to sphere A at $r = 0.05$r=0.05 m.

$E = \dfrac{kQ}{r^2} = \dfrac{(8.99 \times 10^9)(4.0 \times 10^{-9})}{(0.05)^2}$E=r2kQ​=(0.05)2(8.99×109)(4.0×10−9)​

M1 — correct substitution into the radial-field formula. Critically, the source charge $Q = 4.0 \times 10^{-9}$Q=4.0×10−9 C goes in the numerator; the test charge $2.0 \times 10^{-12}$2.0×10−12 C does not appear here. Conflating the two is the most common error on this style of question.

Step 2 — evaluate.

$E = \dfrac{35.96}{0.0025} \approx 1.44 \times 10^4 \text{ N C}^{-1}$E=0.002535.96​≈1.44×104 N C−1

A1 — field strength correct.

Step 3 — direction of the field.

The field at the particle's position points radially outward from sphere A (because A is positive and $E$E points in the direction a positive test charge would move).

B1 — correct direction stated.

Step 4 — apply $F = EQ$F=EQ to find the force on the test particle.

$F = EQ = (1.44 \times 10^4)(2.0 \times 10^{-12}) = 2.9 \times 10^{-8} \text{ N}$F=EQ=(1.44×104)(2.0×10−12)=2.9×10−8 N

M1 — correct use of $F = EQ$F=EQ with the test charge.

A1 — final force, correctly directed away from sphere A (since the test charge is also positive).

Total: 8 marks (M3 A3 B2).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): An electron of mass $m_e = 9.11 \times 10^{-31}$me​=9.11×10−31 kg and charge $-e = -1.60 \times 10^{-19}$−e=−1.60×10−19 C enters horizontally a uniform electric field between two parallel plates. The plates are $0.020$0.020 m apart with a potential difference of $500$500 V across them, the upper plate being positive. The electron enters along the central axis with horizontal velocity $2.0 \times 10^7$2.0×107 m s⁻¹ and the plates are $0.080$0.080 m long.

(a) Calculate the magnitude of the vertical acceleration of the electron between the plates. (3)

(b) Calculate the vertical displacement of the electron as it leaves the plates and state the direction of deflection. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO2.1) — find the field strength: $E = V/d = 500/0.020 = 2.5 \times 10^4$E=V/d=500/0.020=2.5×104 N C⁻¹.
• M1 (AO2.1) — find the force on the electron: $F = EQ = (2.5 \times 10^4)(1.60 \times 10^{-19}) = 4.0 \times 10^{-15}$F=EQ=(2.5×104)(1.60×10−19)=4.0×10−15 N (using the magnitude of the electron charge).
• A1 (AO1.1b) — apply $a = F/m$a=F/m: $a = 4.0 \times 10^{-15}/9.11 \times 10^{-31} \approx 4.4 \times 10^{15}$a=4.0×10−15/9.11×10−31≈4.4×1015 m s⁻².

(b)

• M1 (AO2.1) — find the time spent between the plates from horizontal motion: $t = L/v = 0.080/(2.0 \times 10^7) = 4.0 \times 10^{-9}$t=L/v=0.080/(2.0×107)=4.0×10−9 s.
• M1 (AO1.1b) — apply $s = \tfrac{1}{2} a t^2$s=21​at2 for vertical displacement: $s = 0.5 \times 4.4 \times 10^{15} \times (4.0 \times 10^{-9})^2 \approx 3.5 \times 10^{-2}$s=0.5×4.4×1015×(4.0×10−9)2≈3.5×10−2 m. (Note: this exceeds the plate-to-axis distance of $0.010$0.010 m, so in practice the electron strikes the upper plate before leaving — credit is awarded for correct calculation; reasoning candidates may add a comment that the electron does not in fact emerge.)
• A1 (AO3.2a) — direction: deflected upward, towards the positive plate, because the field points downward (positive to negative) but the force on the negative electron is opposite to $E$E.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. Paper 2 routinely combines field strength, kinematics and a direction-of-force argument in a single multi-step calculation — exactly the synoptic structure rewarded here.

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Synoptic links

Connects to:

1. Topic 5 — Gravitational fields: the structural analogy is profound. Coulomb's law $F = kQ_1Q_2/r^2$F=kQ1​Q2​/r2 has the identical inverse-square form as Newton's law $F = Gm_1m_2/r^2$F=Gm1​m2​/r2; field strength $E = kQ/r^2$E=kQ/r2 mirrors gravitational field $g = GM/r^2$g=GM/r2; potential $V = kQ/r$V=kQ/r mirrors gravitational potential $V_g = -GM/r$Vg​=−GM/r. The differences (electric force can attract or repel; gravity is always attractive) sharpen the analogy rather than break it. A* candidates can write the parallel formula table from memory.

2. Topic 7 — Capacitance: the parallel-plate capacitor relies entirely on the uniform field between plates. $C = \varepsilon_0 A/d$C=ε0​A/d is derived by combining $E = V/d$E=V/d with $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$E=σ/ε0​=Q/(ε0​A) and $Q = CV$Q=CV. Without secure command of $E = V/d$E=V/d, capacitor questions are ungrounded.

3. Topic 11 — Particle physics: linear accelerators (linacs) use uniform electric fields between drift tubes to accelerate charged particles; cyclotrons combine electric fields (for acceleration in the gap) with magnetic fields (for circular motion). The kinetic energy gained by a charge $Q$Q falling through a potential difference $V$V is $\Delta KE = QV$ΔKE=QV — used continually in particle-physics calculations to convert eV into joules.

4. Topic 12 — Atomic physics and Millikan's experiment: the determination of the elementary charge $e$e rested on balancing the gravitational force on an oil droplet against the electric force from a uniform field between parallel plates: $mg = QE = QV/d$mg=QE=QV/d. Historic experiments echo through the syllabus.

5. Topic 9 (Practical skills) — Cathode-ray deflection: the parabolic-trajectory result for charged particles in uniform fields is exactly the physics of the cathode-ray oscilloscope and the older television tube. Synoptic projectile-motion questions from Topic 2 (mechanics) reappear here with the gravitational acceleration $g$g replaced by $a = QE/m$a=QE/m.

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Mark-scheme literacy

Electric-field questions on 9PH0 split AO marks across all three objectives because the topic combines factual recall, mathematical application and physical reasoning:

AO	Typical share	Earned by
AO1 (knowledge / understanding)	30–45%	Stating definitions, recalling formulae, identifying field directions, distinguishing radial from uniform fields
AO2 (application)	35–50%	Substituting into $E = V/d$E=V/d, $F = kQ_1Q_2/r^2$F=kQ1​Q2​/r2 or $E = kQ/r^2$E=kQ/r2; combining with $F = ma$F=ma or $F = EQ$F=EQ; multi-step kinematics in fields
AO3 (analysis / evaluation)	10–25%	Comparing radial and uniform fields; reasoning about direction conventions; evaluating limitations such as edge effects, non-uniform charge distributions, the assumption of a vacuum

Examiner-rewarded phrasing: "the field points in the direction of the force on a positive test charge"; "since the charges are of opposite sign the force is attractive"; "treating the field between the plates as uniform and ignoring edge (fringe) effects"; "converting the plate separation to metres before substitution". Phrases that lose marks: writing units as "N/m" instead of "N C⁻¹" or "V m⁻¹"; quoting a "negative force" without specifying a positive direction; describing the electron as "moving in the direction of the field" (it moves opposite to $E$E); leaving $r$r in centimetres in Coulomb's law (a factor of $10^4$104 error in the result).

A specific Edexcel pattern: questions asking candidates to "state and explain" the direction of motion of a charged particle expect two discrete sentences — one stating the direction, one giving the physical justification (sign of charge, direction of $E$E). Combining them implicitly forfeits the explanation mark.

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Grade-band model answers

3-mark question

Question: A point charge of $+5.0 \times 10^{-9}$+5.0×10−9 C sits at the origin in a vacuum. Calculate the electric field strength at a distance of $0.20$0.20 m from the charge, and state its direction.

Grade C response (~150 words):

The field around a point charge is given by $E = kQ/r^2$E=kQ/r2.

$E = (8.99 \times 10^9)(5.0 \times 10^{-9})/(0.20)^2 = 44.95/0.04 = 1124$E=(8.99×109)(5.0×10−9)/(0.20)2=44.95/0.04=1124 N C⁻¹.

So $E \approx 1100$E≈1100 N C⁻¹.

The field points away from the charge.

Examiner commentary: Full marks (3/3). Correct formula, correct substitution, sensible significant figures and a direction statement. The candidate avoids the most common error of leaving $r$r in centimetres. The direction statement could be sharper — "radially outward" is the preferred phrasing because it makes clear the field has a particular geometry — but "away from the charge" is acceptable here. A small presentation slip: $1124$1124 rounds to $1100$1100, not $1000$1000, so significant-figure discipline is intact.

Grade A response (~210 words):*

For a point charge $Q$Q in a vacuum, the radial electric field strength at distance $r$r is

$E = \dfrac{kQ}{r^2} = \dfrac{Q}{4\pi\varepsilon_0 r^2}$E=r2kQ​=4πε0​r2Q​

Substituting $k = 8.99 \times 10^9$k=8.99×109 N m² C⁻², $Q = 5.0 \times 10^{-9}$Q=5.0×10−9 C and $r = 0.20$r=0.20 m:

$E = \dfrac{(8.99 \times 10^9)(5.0 \times 10^{-9})}{(0.20)^2} = \dfrac{44.95}{0.0400} \approx 1.1 \times 10^3 \text{ N C}^{-1}$E=(0.20)2(8.99×109)(5.0×10−9)​=0.040044.95​≈1.1×103 N C−1

The field points radially outward from the charge, because the source charge is positive and the field at any point is, by definition, the direction in which a positive test charge placed at that point would accelerate.

Examiner commentary: Full marks (3/3). The candidate gives both forms of Coulomb's constant ($k$k and $1/(4\pi\varepsilon_0)$1/(4πε0​)), which signals familiarity with the fundamental constant; uses display-math notation throughout; gives the answer to two significant figures, matching the precision of the input data; and justifies the direction with reference to the definition of $E$E. The phrase "radially outward" is the precise geometrical descriptor examiners look for. This is examination craft.

6-mark question

Question: Two parallel metal plates are mounted vertically, $8.0$8.0 mm apart, with a potential difference of $320$320 V between them. (a) Calculate the electric field strength between the plates. (b) An electron is released from rest at the negative plate. Calculate its speed as it reaches the positive plate, ignoring any gravitational effects.

Grade B response (~240 words):

(a) Using $E = V/d$E=V/d:

$E = 320/0.008 = 40\,000$E=320/0.008=40000 V m⁻¹.

(b) The work done on the electron is $W = QV = 1.6 \times 10^{-19} \times 320 = 5.12 \times 10^{-17}$W=QV=1.6×10−19×320=5.12×10−17 J.

This equals the kinetic energy: $\tfrac{1}{2}mv^2 = 5.12 \times 10^{-17}$21​mv2=5.12×10−17.

So $v^2 = 2 \times 5.12 \times 10^{-17}/9.11 \times 10^{-31} = 1.12 \times 10^{14}$v2=2×5.12×10−17/9.11×10−31=1.12×1014.

$v = 1.06 \times 10^7$v=1.06×107 m s⁻¹.

Examiner commentary: (a) is correct (2/2 with units stated). (b) is correct in method and correct in arithmetic, scoring 4/4. Total 6/6, but the response is sparse — the candidate has not justified why the work-energy theorem applies (the plate field is uniform, so the electric force is constant and conservative; all the work goes into kinetic energy). On a longer question this terseness would cost the AO3 reasoning mark. Grade B answers tend to be mechanically correct but reasoning-light.

Grade A response (~290 words):*

(a) The field between two parallel plates separated by distance $d$d with potential difference $V$V is uniform and given by

$E = \dfrac{V}{d} = \dfrac{320}{8.0 \times 10^{-3}} = 4.0 \times 10^4 \text{ V m}^{-1}$E=dV​=8.0×10−3320​=4.0×104 V m−1

(equivalently, $4.0 \times 10^4$4.0×104 N C⁻¹). The field is uniform because the plates are large and parallel; we ignore fringe effects at the edges.

(b) The electron, charge magnitude $e$e, is accelerated from rest by the constant force $F = eE$F=eE acting through distance $d$d. By the work-energy theorem, the work done on the electron equals its gain in kinetic energy:

$W = F \cdot d = eE \cdot d = e \cdot \dfrac{V}{d} \cdot d = eV$W=F⋅d=eE⋅d=e⋅dV​⋅d=eV

Hence

$\tfrac{1}{2}m_e v^2 = eV \implies v = \sqrt{\dfrac{2eV}{m_e}}$21​me​v2=eV⟹v=me​2eV​​

Substituting $e = 1.60 \times 10^{-19}$e=1.60×10−19 C, $V = 320$V=320 V, $m_e = 9.11 \times 10^{-31}$me​=9.11×10−31 kg:

$v = \sqrt{\dfrac{2(1.60 \times 10^{-19})(320)}{9.11 \times 10^{-31}}} \approx 1.1 \times 10^7 \text{ m s}^{-1}$v=9.11×10−312(1.60×10−19)(320)​​≈1.1×107 m s−1

Examiner commentary: Full marks (6/6). The derivation that $W = eV$W=eV — independent of $d$d — is the key piece of physics insight, and the candidate writes it explicitly. The result is given to two significant figures matching the data; the comment about ignoring fringe effects shows awareness of the model's limits. At $1.1 \times 10^7$1.1×107 m s⁻¹, the speed is about $3.7\%$3.7% of $c$c — still safely non-relativistic, though an A* candidate would mention this as a sanity check on a longer question.

9-mark question

Question: Two parallel plates separated by $2.0$2.0 cm sit horizontally, with the upper plate held at $+1500$+1500 V relative to the lower plate. A small charged oil droplet of mass $3.2 \times 10^{-15}$3.2×10−15 kg is observed to be stationary in the field. (a) State the direction of the electric force on the droplet. (2) (b) Calculate the magnitude and sign of the charge on the droplet. (5) (c) Express this charge as a multiple of the elementary charge $e = 1.60 \times 10^{-19}$e=1.60×10−19 C and comment on the result. (2)

Grade A response (~360 words):*

(a) For the droplet to be stationary in the gravitational field, the net force on it must be zero. Gravity acts downward, so the electric force must act upward to balance it.

(b) The electric field between the plates points from the positive (upper) plate to the negative (lower) plate — i.e. downward. For the electric force $F_E = QE$FE​=QE to point upward (opposite to $E$E), the charge $Q$Q must be negative.

The magnitudes balance:

$|Q| E = mg \implies |Q| = \dfrac{mg}{E} = \dfrac{mg \cdot d}{V}$∣Q∣E=mg⟹∣Q∣=Emg​=Vmg⋅d​

Substituting $m = 3.2 \times 10^{-15}$m=3.2×10−15 kg, $g = 9.81$g=9.81 m s⁻², $d = 0.020$d=0.020 m, $V = 1500$V=1500 V:

$|Q| = \dfrac{(3.2 \times 10^{-15})(9.81)(0.020)}{1500} = \dfrac{6.28 \times 10^{-16}}{1500} \approx 4.18 \times 10^{-19} \text{ C}$∣Q∣=1500(3.2×10−15)(9.81)(0.020)​=15006.28×10−16​≈4.18×10−19 C

So the charge is $Q \approx -4.2 \times 10^{-19}$Q≈−4.2×10−19 C.

(c) Dividing by $e$e:

$\dfrac{|Q|}{e} = \dfrac{4.18 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 2.6$e∣Q∣​=1.60×10−194.18×10−19​≈2.6

This is not a clean integer, but it is close to $3$3. Within the precision of the experiment (uncertainties of a few percent in mass and field strength), the result is consistent with a droplet carrying $3e$3e of negative charge — i.e. three excess electrons. The slightly low value is most plausibly explained by experimental rounding. Historically, this was the structure of Millikan's oil-drop experiment, and the integer-multiple result was the central evidence that charge is quantised.

Examiner commentary: Full marks (9/9). Part (a) (2/2): explicit balance of forces and direction. Part (b) (5/5): the candidate combines $E = V/d$E=V/d, $F_E = QE$FE​=QE and the equilibrium condition cleanly, gives a magnitude correct to two significant figures, and signs the charge correctly with an explicit physical justification (force opposite to field implies negative charge). Part (c) (2/2): not just a numerical division but a physical interpretation connecting the result to the historical experiment. Tying the calculation back to charge quantisation is exactly the AO3 evaluation that pushes a 7/9 response to 9/9.

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A-Level-depth misconceptions

The errors that distinguish A from A* on electric-field questions:

1. "The field points in the direction the charge moves." Wrong for negative charges. The field $E$E is defined as the direction of the force on a positive test charge. A negative charge (e.g. an electron) experiences force opposite to $E$E and therefore accelerates opposite to $E$E. Forgetting the sign reverses the deflection direction in every parallel-plates question.

2. Confusing radial and uniform field formulae. $E = kQ/r^2$E=kQ/r2 applies only to a radial field around a point charge (or, by symmetry, outside a uniformly charged sphere). $E = V/d$E=V/d applies only to a uniform field between parallel plates. Mixing them — using $V/d$V/d for a point charge or $kQ/r^2$kQ/r2 for parallel plates — produces nonsense.

3. Treating $V$V, $E$E and $F$F as interchangeable. Potential difference $V$V (units: J C⁻¹ = V) is energy per unit charge; field strength $E$E (units: V m⁻¹ = N C⁻¹) is force per unit charge; force $F$F (units: N) is force. They are linked by $E = V/d$E=V/d (uniform field) and $F = EQ$F=EQ, but they are not the same quantity. Tracking units protects against this collapse.

4. Inverse-square versus inverse-linear thinking. Field strength $E \propto 1/r^2$E∝1/r2; potential $V \propto 1/r$V∝1/r; force between two charges $F \propto 1/r^2$F∝1/r2. A* candidates can quote and distinguish all three. A common slip is to halve the field when the distance doubles — the field actually quarters.

5. Treating the test charge as the source charge in $E = kQ/r^2$E=kQ/r2. The $Q$Q in this formula is the source charge creating the field, not the test charge sampling it. To find the force on the test charge, calculate $E$E from the source first, then apply $F = EQ_\text{test}$F=EQtest​.

6. Forgetting that field is a vector and superposes by vector addition. When two source charges are present, the net field at a point is the vector sum of the two individual fields — not the algebraic sum of magnitudes. Questions involving two charges on a horizontal axis are easier (the directions are colinear), but oblique geometries require resolving into components.

7. Misinterpreting $E = V/d$E=V/d as "field equals voltage divided by anything". The $d$d in this formula is the plate separation, measured along the field direction. For a point partway between the plates, the field is still $V/d$V/d (the field is uniform); the potential at that point varies linearly with distance from one plate, but the field strength does not.

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Common errors and mark-loss patterns on field calculations

Three patterns repeatedly cost candidates marks on Paper 2 field questions. They are all about reading the question and tracking units.

• Unit-conversion slip. Plate separations are routinely quoted in millimetres or centimetres; voltages in kilovolts; charges in nanocoulombs or picocoulombs. Substituting the quoted numerical value (e.g. $d = 8$d=8 for an 8 mm separation) without conversion to SI produces an error of $10^3$103 or worse. Cure: write a "convert to SI" line at the top of every solution before substituting.
• Sign of the force from the sign of the charge. Coulomb's law $F = kQ_1Q_2/r^2$F=kQ1​Q2​/r2 with both charges substituted as signed quantities gives a signed force whose physical meaning is unclear. Examiners want magnitude and a separate direction statement: "attractive" (opposite signs) or "repulsive" (same signs). Substituting magnitudes only and stating the direction explicitly avoids confusion.
• Confusing field strength with force on an electron. "Calculate the force on the electron" in a uniform field requires two steps: $E = V/d$E=V/d, then $F = eE$F=eE. Candidates who give the field as their final answer lose at least one mark; those who write $F = V/d$F=V/d directly lose two.
• Quoting too many significant figures. Data quoted to two or three significant figures cannot support an answer to five. Edexcel tolerates one extra figure in working, but the final answer should match the input precision. $1.05674 \times 10^7$1.05674×107 when the data are 2 s.f. throughout reads as careless.

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Going further — university and academic signposting

Electric fields point directly toward several undergraduate trajectories:

• Maxwell's equations (Year 2 physics): the four equations that unify electricity and magnetism. Gauss's law for the electric field, $\oint \mathbf{E} \cdot d\mathbf{A} = Q_\text{enc}/\varepsilon_0$∮E⋅dA=Qenc​/ε0​, contains Coulomb's law as a special case and lets you derive the field of any charge distribution (sphere, infinite plane, infinite line) in two or three lines. The radial-field result $E = kQ/r^2$E=kQ/r2 is what Gauss's law gives you for a point charge.
• Vector calculus and field theory (Year 1/2 maths and physics): an electric field is a vector field, and the relationship $E = -\nabla V$E=−∇V between field and potential is one of the first non-trivial uses of the gradient operator. The fact that $\nabla \times \mathbf{E} = 0$∇×E=0 for an electrostatic field is what makes $V$V well-defined as a scalar potential.
• Quantum electrodynamics (Year 4 / postgraduate): the modern theory of electromagnetic interactions describes the electric force as the exchange of virtual photons between charged particles. Coulomb's $1/r^2$1/r2 law emerges in the low-energy limit of QED; the agreement between QED predictions and measurements of, e.g., the electron's magnetic moment is the most accurate confirmation of any physical theory in history.
• Plasma physics and fusion (Year 3+): controlling charged particles in fields is the central engineering challenge of magnetic-confinement fusion (tokamaks) and inertial-confinement approaches. The same parallel-plates physics scales up to the Z-pinch and the field-shaping coils of ITER.

Oxbridge interview prompt: "Coulomb's law and Newton's law of gravitation share the same $1/r^2$1/r2 form. Why? And what would be different about the universe if they had a $1/r^3$1/r3 form instead?"

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Required practical reference

Topic 7 is supported by deflection-tube practicals that let students see the parabolic trajectory of charged particles in a uniform electric field. In a typical school deflection tube, electrons accelerated through a few kilovolts pass between two horizontal parallel plates with a deflecting potential difference of a few hundred volts; the beam strikes a phosphor-coated screen and traces a visibly parabolic curve. Measuring the deflection at known horizontal positions and applying $y = \tfrac{1}{2}(QE/m)(x/v_x)^2$y=21​(QE/m)(x/vx​)2 gives a route to the charge-to-mass ratio $e/m_e$e/me​ for the electron. The same apparatus, with a magnetic field added, supports Thomson-style determinations and forms the conceptual basis of all cathode-ray devices.

A second canonical reference is Millikan's oil-drop experiment (1909), in which charged oil droplets are suspended between horizontal parallel plates; balancing gravity against the electric force $mg = QE = QV/d$mg=QE=QV/d allows the charge to be deduced. Repeated measurements show the charge always to be an integer multiple of a fundamental quantum $e \approx 1.6 \times 10^{-19}$e≈1.6×10−19 C — the experimental basis for charge quantisation. Modern A-Level practicals do not reproduce Millikan directly, but the analytical method (force balance, $E = V/d$E=V/d, $F = QE$F=QE) appears repeatedly in Paper 3 questions and is the structural model for Topic 7 calculation work.

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Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Physics (9PH0) specification, Paper 2 — Advanced Physics II, Topic 7: Electric and Magnetic Fields. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

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Visual summary

graph TD
    A["Charge configuration"] --> B{"Geometry?"}
    B -->|"Two point charges"| C["Coulomb's law<br/>F = kQ₁Q₂ / r²"]
    B -->|"Single point charge<br/>(field at a distance)"| D["Radial field<br/>E = kQ / r²"]
    B -->|"Parallel plates"| E["Uniform field<br/>E = V / d"]
    C --> F{"Same sign?"}
    F -->|"Yes"| G["Repulsive"]
    F -->|"No"| H["Attractive"]
    D --> I["Direction:<br/>radially outward<br/>(positive source)"]
    E --> J["Direction:<br/>positive plate<br/>to negative plate"]
    G --> K["Force on test charge:<br/>F = E·Q_test"]
    H --> K
    I --> K
    J --> K
    K --> L{"Charge sign<br/>of test particle?"}
    L -->|"Positive"| M["Force in direction of E"]
    L -->|"Negative<br/>(e.g. electron)"| N["Force opposite to E"]
    M --> O["Apply F = ma<br/>and kinematics"]
    N --> O

    style C fill:#3498db,color:#fff
    style D fill:#3498db,color:#fff
    style E fill:#3498db,color:#fff
    style O fill:#27ae60,color:#fff