Fields Problem Solving

This lesson brings together the key ideas from the entire topic: electric fields, Coulomb’s law, capacitors, RC circuits, magnetic forces, and electromagnetic induction. The problems here require you to combine concepts from multiple areas, just as you will need to in the exam. Each worked example demonstrates a multi-step approach.

Strategy for Multi-Step Problems

Identify the physics — Which concepts are involved? Electric field, capacitor, magnetic force, induction?
List the relevant equations — Write down the formulas you will need
Check units — Convert everything to SI before substituting
Work in steps — Calculate intermediate quantities before the final answer
Check your answer — Does the magnitude make sense? Are the units correct?

Complete Formulae Reference

Topic	Key Formulae
Uniform electric field	E = V/d, F = EQ, F = VQ/d
Coulomb’s law	F = kQ₁Q₂/r², E = kQ/r²
Electric potential	V = kQ/r, Eₚ = kQ₁Q₂/r, W = QΔV
Capacitance	C = Q/V, C = ε₀A/d, C = ε₀εᵣA/d
Capacitor energy	E = ½QV = ½CV² = ½Q²/C
Series capacitors	1/Cₜ = 1/C₁ + 1/C₂
Parallel capacitors	Cₜ = C₁ + C₂
RC discharge	Q = Q₀e^(−t/RC), τ = RC, t½ = 0.693RC
RC charging	Q = Q₀(1 − e^(−t/RC))
Magnetic force (wire)	F = BIL sinθ
Magnetic force (charge)	F = BQv sinθ
Circular motion in B	r = mv/(BQ), T = 2πm/(BQ)
Magnetic flux	Φ = BA cosθ
Flux linkage	NΦ = NBA cosθ
Faraday’s law	ε = NΔΦ/Δt
Generator EMF	ε = NBAω sin(ωt)
Straight conductor	ε = BLv
Transformer	Vₛ/Vₚ = Nₛ/Nₚ
Transmission loss	P_loss = I²R = P²R/V²

Comparing Electric and Gravitational Fields

The parallels between electric and gravitational fields run deep. Both follow inverse-square laws, both have a field strength and a potential, and both have associated potential energy. The key differences are:

Concept	Electric	Gravitational
Source property	Charge (Q)	Mass (m)
Field strength	E = F/Q = kQ/r²	g = F/m = GM/r²
Potential	V = kQ/r	V = −GM/r
Potential energy	Eₚ = kQ₁Q₂/r	Eₚ = −Gm₁m₂/r
Force law	F = kQ₁Q₂/r²	F = Gm₁m₂/r²
Attractive/repulsive	Both (depends on sign)	Attractive only
Constant	k = 8.99 × 10⁹	G = 6.67 × 10⁻¹¹

Note the negative signs in gravitational potential and energy — these arise because gravitational fields are always attractive, so the potential is always negative (taking infinity as zero).

Worked Example 1: Electric Field and Projectile Motion

An electron is fired horizontally at 2.0 × 10⁷ m s⁻¹ between two horizontal parallel plates separated by 2.0 cm, with a potential difference of 500 V. The plates are 5.0 cm long. Does the electron hit a plate before emerging from the other side?

Step 1: Find the electric field: E = V/d = 500 / 0.020 = 25 000 V m⁻¹

Step 2: Find the vertical force on the electron: F = eE = 1.6 × 10⁻¹⁹ × 25 000 = 4.0 × 10⁻¹⁵ N

Step 3: Find the vertical acceleration: a = F/m = 4.0 × 10⁻¹⁵ / 9.11 × 10⁻³¹ = 4.39 × 10¹⁵ m s⁻²

Step 4: Time to cross the plates horizontally: t = L/v = 0.050 / 2.0 × 10⁷ = 2.5 × 10⁻⁹ s

Step 5: Vertical displacement in this time: s = ½at² = ½ × 4.39 × 10¹⁵ × (2.5 × 10⁻⁹)² = ½ × 4.39 × 10¹⁵ × 6.25 × 10⁻¹⁸ = 0.0137 m = 1.37 cm

Since the maximum vertical displacement (half the plate separation) is 1.0 cm and the electron deflects 1.37 cm, the electron hits the plate before emerging.

Worked Example 2: Capacitor Energy and Charge Sharing

A 2200 μF capacitor is charged to 15 V and then disconnected. It is connected across a 1000 μF uncharged capacitor. Find the final voltage, the charge on each capacitor, and the energy lost.

Initial charge: Q₀ = CV = 2200 × 10⁻⁶ × 15 = 0.033 C

After connection (capacitors in parallel): C_total = 2200 + 1000 = 3200 μF

Charge is conserved: Q_total = 0.033 C

Final voltage: V = Q/C = 0.033 / 3200 × 10⁻⁶ = 10.3 V

Charge on each:

Q₁ = C₁V = 2200 × 10⁻⁶ × 10.3 = 0.0227 C
Q₂ = C₂V = 1000 × 10⁻⁶ × 10.3 = 0.0103 C

Energy before: E₁ = ½CV² = ½ × 2200 × 10⁻⁶ × 15² = 0.2475 J

Energy after: E₂ = ½ × 3200 × 10⁻⁶ × 10.3² = 0.170 J

Energy lost: 0.2475 − 0.170 = 0.078 J ≈ 0.08 J (dissipated as heat in the connecting wires)

Key insight: Energy is ALWAYS lost when charge is shared between capacitors at different voltages. The fraction of energy lost depends on the ratio of capacitances. For two equal capacitors, exactly half the energy is lost.

Worked Example 3: RC Circuit Timing with Logarithms

A smoke detector uses a capacitor circuit. A 47 μF capacitor charges through a 220 kΩ resistor. How long does it take for the voltage across the capacitor to reach 90% of the supply voltage?

Time constant: τ = RC = 220 000 × 47 × 10⁻⁶ = 10.34 s

Charging equation: V = V₀(1 − e^(−t/RC))

When V = 0.90V₀: 0.90 = 1 − e^(−t/RC)

e^(−t/RC) = 0.10

−t/RC = ln(0.10) = −2.303

t = 2.303 × 10.34 = 23.8 s ≈ 24 s

This is approximately 2.3 time constants. Useful reference: reaching a given percentage of V₀ during charging:

Target %	Number of time constants	ln calculation
50%	0.693τ	ln(0.50) = −0.693
63.2%	1.000τ	ln(0.368) = −1.000
75%	1.386τ	ln(0.25) = −1.386
90%	2.303τ	ln(0.10) = −2.303
95%	2.996τ	ln(0.05) = −2.996
99%	4.605τ	ln(0.01) = −4.605

Worked Example 4: Magnetic Force, Circular Motion, and Energy

A doubly-ionised helium atom (He²⁺, charge = 2e = 3.2 × 10⁻¹⁹ C, mass = 6.64 × 10⁻²⁷ kg) is first accelerated from rest through a potential difference of 5000 V, then enters a magnetic field of 0.50 T perpendicular to its velocity. Find the speed, the radius of its circular path, and the time for one complete revolution.

Step 1: Find the speed

W = QΔV = 3.2 × 10⁻¹⁹ × 5000 = 1.6 × 10⁻¹⁵ J

½mv² = 1.6 × 10⁻¹⁵

v = √(2 × 1.6 × 10⁻¹⁵ / 6.64 × 10⁻²⁷) = √(4.82 × 10¹¹) = 2.20 × 10⁵ m s⁻¹

Step 2: Find the radius

r = mv / (BQ) = (6.64 × 10⁻²⁷ × 2.20 × 10⁵) / (0.50 × 3.2 × 10⁻¹⁹) = 1.46 × 10⁻²¹ / 1.6 × 10⁻¹⁹ = 9.1 × 10⁻³ m ≈ 9.1 mm

Step 3: Find the period

T = 2πm / (BQ) = 2π × 6.64 × 10⁻²⁷ / (0.50 × 3.2 × 10⁻¹⁹) = 4.17 × 10⁻²⁶ / 1.6 × 10⁻¹⁹ = 2.6 × 10⁻⁷ s ≈ 0.26 μs

Note: the period is independent of speed. If the accelerating voltage were doubled, the speed would increase by √2, the radius would increase by √2, but the period would remain 0.26 μs.

Worked Example 5: Electromagnetic Induction and Energy Conservation

A rectangular coil of 80 turns, length 0.10 m and width 0.05 m, is pulled at constant velocity 0.20 m s⁻¹ out of a magnetic field of 0.60 T. The coil is connected to a 4.0 Ω resistor.

EMF while the coil exits: While being pulled out, one side of length L = 0.05 m (the width) cuts field lines:

ε = NBLv = 80 × 0.60 × 0.05 × 0.20 = 0.48 V

Current: I = ε / R = 0.48 / 4.0 = 0.12 A

Force needed to maintain constant velocity: The current in the wire inside the field experiences a force opposing the motion (Lenz’s law):

F = NBIL = 80 × 0.60 × 0.12 × 0.05 = 0.288 N ≈ 0.29 N

Power dissipated in resistor: P = I²R = 0.12² × 4.0 = 0.058 W

Energy conservation check: Power input = Fv = 0.288 × 0.20 = 0.058 W ✔

The mechanical work done pulling the coil equals the electrical energy dissipated. This is a beautiful demonstration of energy conservation connecting mechanics and electromagnetism.

Worked Example 6: Combined Coulomb’s Law and Circular Motion

Two protons are fixed at positions 0.10 m apart. An electron orbits the midpoint in a circular path perpendicular to the line joining the protons. If the orbit radius is 0.04 m, find the speed of the electron.

Step 1: Distance from each proton to the electron: d = √(0.05² + 0.04²) = √(0.0025 + 0.0016) = √(0.0041) = 0.064 m

Step 2: Coulomb force from each proton on the electron: F = ke²/d² = 8.99 × 10⁹ × (1.6 × 10⁻¹⁹)² / (0.064)² = 2.30 × 10⁻²⁸ / 4.10 × 10⁻³ = 5.62 × 10⁻²⁶ N

Step 3: The component of each force directed radially inward (towards the centre of the orbit): sinθ = 0.04/0.064 = 0.625 F_radial (each proton) = 5.62 × 10⁻²⁶ × 0.625 = 3.51 × 10⁻²⁶ N

Total centripetal force = 2 × 3.51 × 10⁻²⁶ = 7.02 × 10⁻²⁶ N

Step 4: Set equal to centripetal force: mv²/r = 7.02 × 10⁻²⁶ v² = 7.02 × 10⁻²⁶ × 0.04 / 9.11 × 10⁻³¹ = 3.08 × 10⁴ v = 175 m s⁻¹

This is extremely slow for an electron (normally ~10⁶ m s⁻¹) because the orbit radius is macroscopic, not atomic.

Worked Example 7: Transformer and Transmission Losses Combined

A 50 MW power station uses a step-up transformer (95% efficient) to increase the voltage from 25 kV to 400 kV for transmission. The cables have a total resistance of 8.0 Ω. Calculate: (a) the power entering the transmission cables (b) the current in the cables (c) the power lost in the cables (d) the overall efficiency from power station to end of cables

(a) Power after transformer: 50 × 10⁶ × 0.95 = 47.5 MW

(b) I = P/V = 47.5 × 10⁶ / 400 000 = 118.75 A ≈ 119 A

(d) Power delivered = 47.5 − 0.113 = 47.39 MW Overall efficiency = 47.39 / 50.0 × 100% = 94.8%

The transformer loss (5%) dominates. The cable loss is only 0.24% of the transmitted power.

Worked Example 8: Capacitor Discharge with Energy Analysis

A defibrillator charges a 32 μF capacitor to 5000 V. It discharges through the patient (modelled as a 50 Ω resistor) when triggered.

(a) Energy stored: E = ½CV² = ½ × 32 × 10⁻⁶ × 5000² = 400 J

(b) Time constant: τ = RC = 50 × 32 × 10⁻⁶ = 1.6 × 10⁻³ s = 1.6 ms

(d) Current after 3 ms: I = 100 × e^(−0.003/0.0016) = 100 × e^(−1.875) = 100 × 0.153 = 15.3 A

(e) Energy remaining after 3 ms: V = 5000 × e^(−1.875) = 5000 × 0.153 = 767 V E = ½ × 32 × 10⁻⁶ × 767² = 9.4 J

Energy delivered to patient: 400 − 9.4 = 390.6 J ≈ 391 J

Most of the energy (97.6%) is delivered within 3 ms — approximately 2 time constants. This rapid energy delivery is essential for the defibrillator to work.

Exam Strategy: Common Multi-Step Combinations

The exam frequently combines concepts in these ways:

Accelerate → Deflect: Charge accelerated through potential difference (W = QΔV → KE) then enters electric or magnetic field (calculate deflection or radius)
Charge → Share: Capacitor charged, disconnected, then connected to another capacitor (conservation of charge, energy lost)
Induction → Circuit: Conductor moves in field (EMF = BLv or NΔΦ/Δt), drives current through resistor (I = ε/R), force opposes motion (energy conservation)
Transformer → Transmission: Transformer changes voltage, then calculate cable losses (P = I²R)

For each, the key is to identify the linking quantity — usually energy, charge, or current — that connects the two parts of the problem.

Common Exam Mistakes

Not converting units: The single most common error. Always convert to SI (metres, seconds, coulombs, farads, ohms) before calculating.
Confusing energy formulas: E = ½CV² for a capacitor, but Eₚ = kQ₁Q₂/r for potential energy of two charges. These are different energies.
Forgetting conservation of charge: When capacitors are connected after disconnection from the supply, charge is conserved but energy is not.
Sign errors with Lenz’s law: Always check the direction of the induced current opposes the change.
Using the wrong formula for the wrong situation: E = V/d is for uniform fields only. E = kQ/r² is for point charges only.

Key Points

Multi-step problems require identifying which equations apply and working systematically
Electric and gravitational fields have parallel structures but differ in sign conventions and the possibility of repulsion
Capacitor problems often involve conservation of charge (when disconnected) or constant voltage (when connected to a battery)
RC timing problems use logarithms to solve for time
Magnetic force problems may combine F = BQv with circular motion (BQv = mv²/r)
Energy conservation links mechanical work, electrical energy, and thermal energy in induction problems
Always check units and verify your answer makes physical sense

A-Level Deep Dive: Fields Problem Solving

Spec mapping

Edexcel 9PH0 specification, Topic 7 — Electric and Magnetic Fields, synoptic capstone strand integrates the field, capacitor and electromagnetic-induction sub-topics that have run through the course (refer to the official specification document for exact wording). Synoptic problem solving is not a separate content area; it is the assessed expectation that, given a multi-step physical scenario, a candidate can select equations from across electric fields (Coulomb, parallel-plate, potential), capacitors (energy storage, RC discharge), magnetic fields (force on a moving charge, Lenz/Faraday induction), and mechanics or particle-physics carry-over (circular motion, work–energy, kinetic energy of accelerated ions). It surfaces in Paper 3 — General and Practical Principles in Physics, where extended-response items pull together two or more strands of Topic 7, and recurs in Paper 1 under shorter structured questions. The Edexcel data and formulae booklet supplies $F = BQv$ , $E = V/d$ , $U = \tfrac{1}{2}CV^2$ , $\tau = RC$ and $\varepsilon = -\dfrac{d(N\Phi)}{dt}$ ; the synoptic skill tested is the selection of which equation(s) apply, not their recall.

Worked example with full mark scheme

Question (12 marks): A velocity selector consists of crossed uniform electric and magnetic fields in the same region. Singly ionised particles enter horizontally with a range of speeds. The plates are separated by $d = 8.0\,\text{mm}$ with a potential difference $V = 1.6\,\text{kV}$ across them, producing a vertical electric field. A horizontal magnetic field of magnitude $B = 0.40\,\text{T}$ is applied perpendicular to both the velocity and the electric field.

(a) Show that only ions of one specific speed pass through undeflected, and calculate that speed. (4)

(b) Ions that emerge enter a second region containing only the magnetic field $B = 0.40\,\text{T}$ , where they follow a circular arc of radius $r = 0.052\,\text{m}$ . Determine the mass-to-charge ratio $m/Q$ of the selected ions. (4)

(c) Before injection, the ions were accelerated from rest through a potential difference $V_a$ supplied by a $220\,\mu\text{F}$ capacitor initially charged to $5.0\,\text{kV}$ . Estimate $V_a$ , stating one assumption. (4)

Solution with mark scheme:

(a) Step 1 — equate forces for undeflected motion. The electric force $F_E = QE$ acts vertically; the magnetic force $F_B = BQv$ acts vertically in the opposite direction for the correct sign of charge. For straight-line motion the net force is zero:

$QE = BQv \implies v = \dfrac{E}{B}$

M1 — equating electric and magnetic forces in opposite senses. Note the charge cancels, so the selector is mass- and charge-independent: it filters speed only.

Step 2 — compute $E$ . $E = V/d = 1600 / 0.0080 = 2.0 \times 10^5\,\text{V m}^{-1}$ .

M1 — correct evaluation of $E$ from plate geometry.

Step 3 — compute $v$ . $v = E/B = 2.0 \times 10^5 / 0.40 = 5.0 \times 10^5\,\text{m s}^{-1}$ .

A1 — numerical answer.

A1 — explanation that only this speed gives $F_E = F_B$ ; faster ions are deflected by the dominant magnetic force, slower ions by the dominant electric force. Without this comment the final A is withheld.

(b) Step 1 — set up circular motion. The magnetic force provides centripetal force: $BQv = \dfrac{mv^2}{r}$ .