H2 Physics Electric Fields & Capacitance Notes | 9478

Study guide14 Jul 2025, 00:00 ZUpdated 21 Aug 2026

Q: What does A-Level Physics: 14) Electric Fields Guide cover?
A: From Coulomb's law to capacitor energy graphs, this post unpacks Section V Topic 14 of the 2026 H2 Physics syllabus for IP students and parents.

TL;DR
Electric field questions are the hinge between mechanics and circuits. Nail the three k-values (force, field, potential), the V/d shortcut for plates and the $\tfrac{1}{2}$ -factor for capacitor energy, and you will harvest marks across Papers 1-3 and the practical.

Concrete example: how to choose the formula

If the question says "between parallel plates", start with $E=V/d$ and then use $F=qE$ . If it says "point charge", start with the radial field formula. That one decision prevents most electric field formula swaps.

Electric-fields decision map

Question clue	First check	Main relation	Trap to avoid
Two point charges exert force on each other	Both charges and their separation are known	Coulomb's law for $F$	Forgetting that force is a vector and direction depends on charge signs.
One source charge creates a field at a point	Test charge is not needed unless force is asked	$E=kQ/r^2$

Misconception check: Electric potential is a scalar. Electric field is a vector. At a symmetry point, fields may cancel while potentials still add algebraically.

Need the rest of the electromagnetism refresh (currents, circuits, EMF)? Jump to our free H2 Physics notes to stay in sequence with Topics 15-18 and grab the shared practice decks. For the full topic map and paper weightings, see our H2 Physics Syllabus 2026-27 overview.

1 Coulomb's law: the force glue

The syllabus demands you recall and use

$F = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{r^2}.$

1.1 Quick-fire cues

SI unit for charge is $\pu{C}$ .
Sign mistakes: repulsive if charges share sign, attractive otherwise.

1.2 Mini-drill

Two $\pu{+2.0 nC}$ charges sit $\pu{5.0 cm}$ apart in air. Calculate $F$ . Answer: $\pu{1.4e-5 N}$

2 Radial electric fields & potentials

Electric potential $V$ at a point is defined as the work done per unit charge by an external force in bringing a small positive test charge from infinity to that point. Because the reference is set at infinity (where potential energy is zero), $V$ is negative near a negative source charge and positive near a positive one. Integrating Coulomb's law from infinity to distance $r$ gives the formula for $V$ around a point charge $Q$ .

Because $E = \dfrac{F}{q}$ , the field around a point charge is

$E = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2}.$

Integrating this field gives the scalar potential

$V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}.$

2.1 Zero reference

Infinity is the zero-potential reference unless the question states otherwise.

2.2 Potential energy pair

Two point charges form a system with

$U_E = \frac{1}{4 \pi \varepsilon_0} \frac{Q_1 Q_2}{r}.$

Superposition checkpoint: field versus potential

When more than one source charge is present, decide whether the question asks for a vector or a scalar before adding contributions.

source charges
  -> calculate contribution from each charge
  -> add fields with direction
  -> add potentials with sign only

Quantity asked for	What each charge contributes	How to combine	Common trap
Electric field strength at a point	A vector field pointing away from positive charge or towards negative charge	Resolve directions, then add vector components	Cancelling magnitudes without checking directions.
Electric potential at a point	A scalar value $V = kQ/r$ with the sign of $Q$	Add algebraically, including positive and negative signs

Worked check: at the midpoint between equal positive charges, the two electric fields are equal in magnitude and opposite in direction, so the net field is zero. The potentials are both positive scalars, so they add and the net potential is not zero.

Misconception check: field cancellation is directional cancellation. Potential cancellation needs opposite-signed scalar contributions.

3 Potential gradients & equipotentials

Field lines show direction of $\vec{E}$ .
Equipotential surfaces are always perpendicular to field lines.
Mathematically, $E = - \dfrac{\mathrm{d}V}{\mathrm{d}r}$

Exam cue: check sign conventions when taking gradients.

Field-potential sign checkpoint

When a question gives a potential graph or equipotential map, separate the field direction from the motion of the charge.

Clue in the question	First sign check	What it means	Common trap
Potential decreases as distance increases	Gradient is negative, so field points in the positive distance direction	A positive test charge would accelerate that way if released	Saying the field points "down the graph" instead of along the physical distance axis.
Potential increases as distance increases	Gradient is positive, so field points in the negative distance direction	A positive test charge accelerates opposite to the increasing coordinate	Forgetting the minus sign in $E = -\mathrm{d}V/\mathrm{d}r$ .
Equipotential lines are close together

Worked check: if a potential graph falls from $\pu{12 V}$ to $\pu{4 V}$ over $\pu{0.20 m}$ , the gradient is negative. Since $E$ is the negative gradient, the electric field points in the positive distance direction and has magnitude $\pu{40 V.m-1}$

4 Uniform electric fields

Between parallel plates:

$E = \frac{V}{d}.$

4.1 Force & motion

A charge $q$ experiences $F = qE$ .
With constant $a = \frac{F}{m}$

4.2 Mini-drill

An electron enters midway between a $\pu{600 V}$ plate pair $\pu{8.0 mm}$ apart, with horizontal speed $2.0\times 10^7\,\pu{m.s-1}$

Answer: $E=V/d=7.50\times 10^4\,\pu{N.C-1}$ , $a=eE/m_e\approx 1.32\times 10^{16}\,\pu{m.s-2}$

5 Capacitance fundamentals

Definition: $C = \frac{Q}{V}$ . Unit: farad $( \pu{F} = \pu{C.V-1} )$

5.1 Energy store via the Q-V graph

Because $V = Q/C$ , potential difference is proportional to charge stored. The Q-V graph is a straight line through the origin with gradient $1/C$ . The energy stored equals the work done charging the capacitor, which is $\int_0^Q V , \mathrm{d}Q$

Axis / point	Label	Value
Horizontal axis	Charge stored	$Q$
Vertical axis	Potential difference across capacitor	$V$
Line through origin	Gradient	$1/C$ (since $V = Q/C$

Substituting $Q = CV$ gives $U = \tfrac{1}{2} CV^2$

5.2 Exam-class graph sketch

Draw the Q-V axes, plot the straight line from the origin to the operating point $(Q,,V)$ , shade the triangular area, and label it $U = \tfrac{1}{2} QV$ . State which substitution you will use before applying numbers.

5.3 Discharge sequence checkpoint

Use this as a bridge to the Topic 16 RC treatment linked later in the note. In discharge questions, separate the starting values from the changing values. The capacitor does not lose charge at a constant rate because the current gets smaller as the p.d. falls.

Step	What to write first	Relation to use	Trap to avoid
Initial current	Treat the fully charged capacitor like the supply voltage across the resistor.	$I_0 = V_0/R$	Using the final current, which is zero after a long time.

Worked check: after one time constant, $V = V_0e^{-1} \approx 0.37V_0$ . The capacitor is not empty after $RC$

Misconception check: the resistor dissipates the stored energy as thermal energy, but the rate of dissipation falls during discharge because both current and p.d. fall.

6 Three timing hacks for WA & prelims

Copy units first - minimizes careless mistakes.
Vector-check every force sum.
Leave capacitor energy algebra until numbers are parked.

7 IP-specific pitfalls

Pitfall	Fix
Treating $\varepsilon_0$ as 0	Memorise $8.85 \times 10^{-12} \space \pu{F.m-1}$

Need structured practice on Electric Fields? Our H2 Physics tuition programme covers this topic with weekly problem sets and Paper 4 practical drills.

Comprehensive revision pack

9478 Section V, Topic 14 Syllabus outcomes

Candidates should be able to:

(a) recall and use Coulomb's law in the form $F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q_1 Q_2}{r^2}$

Boundary note: Combined capacitance in series and parallel arrangements is placed in SEAB 9478 Topic 16 (Circuits) outcome (k), not Topic 14. RC charging and discharging behaviour is Topic 16 outcome (l). See the Circuits chapter for those derivations and worked examples.

Concept map (in words)

Start with Coulomb's inverse-square law. Divide by charge to get field strength; integrate to find potential. Between plates you can treat the field as uniform $E = \frac{V}{d}$ . Once you know $E$ , the force on a charge is $qE$ and motion follows mechanics. Capacitors store charge/energy; their energy formulas mirror triangular areas on Q-V graphs.

Key relations

Quantity	Expression / note
Coulomb's law	$F = \dfrac{1}{4 \pi \varepsilon_0} \dfrac{Q_1 Q_2}{r^2}$

Series / parallel capacitor combinations and RC transients live under Topic 16 (Circuits) outcomes (k) and (l) in SEAB 9478 - see that chapter for the derivations.

Derivations & reasoning to master

Field-potential link: integrate Coulomb's law to obtain potential; differentiate back to recover field.
Parallel plate field: derive $E = \dfrac{V}{d}$ by combining force and work definitions.
Energy stored in capacitor: integrate V with respect to Q to show factor 1/2.
Charged particle deflection: combine $F = qE$ with SUVAT relations to predict trajectories in oscilloscopes or mass-spectrometer velocity selectors.

Worked example 1 - potential due to multiple charges

Two charges $\pu{+4.0 nC}$ and $\pu{-2.0 nC}$ are $\pu{12 cm}$ apart. Find the potential and field at a point midway between them.

Approach: potential adds algebraically, field adds vectorially; choose direction from positive to negative charge, compute magnitude using inverse-square.

At the midpoint, $r=\pu{0.060 m}$ . Using $k=\dfrac{1}{4\pi\varepsilon_0}$

$V = k\left(\dfrac{4.0\times 10^{-9}}{0.060} + \dfrac{-2.0\times 10^{-9}}{0.060}\right) = k\left(\dfrac{2.0\times 10^{-9}}{0.060}\right) \approx \pu{3.0e2 V}.$

The field points from $+$ to $-$ (both contributions in the same direction):

$E = k\left(\dfrac{4.0\times 10^{-9}}{0.060^2} + \dfrac{2.0\times 10^{-9}}{0.060^2}\right) \approx \pu{1.5e4 N.C-1}.$

Worked example 2 - capacitor discharge

A $C = \pu{220 \times 10^{-6} F}$ capacitor charged to $V = \pu{12 V}$ discharges through $R = \pu{4.7 \times 10^3 \Omega}$

Solution: $U = \tfrac{1}{2} C V^2$ , $I_0 = \frac{V}{R}$

$U = \tfrac{1}{2}(220\times 10^{-6})(12^2) = \pu{1.58e-2 J},\qquad I_0 = \dfrac{12}{4.7\times 10^3} = \pu{2.55 mA}.$

Practical & data tasks

Map equipotentials using conductive paper and voltmeter; sketch electric field lines from data.
Use apparatus to show Millikan-style oil drop balancing (qualitative) linking gravitational and electric forces.
Build RC circuit and record V(t) to verify exponential decay; extract time constant.

Common misconceptions & exam traps

Treating potential as vector quantity; it is scalar.
Forgetting that equipotentials are perpendicular to field lines.
Mixing units (mm vs m) when using V/d.
Neglecting the $\tfrac{1}{2}$ factor in capacitor energy or forgetting charge changes when voltage changes.

Quick self-check quiz

What is the electric field midway between two equal positive charges? - Zero by symmetry (fields cancel).
A proton accelerated through $\pu{500 V}$ gains how much kinetic energy? - $500 \ \text{eV} \approx 8.0 \times 10^{-17} \text{ J}$ .
Doubling plate separation while keeping voltage constant does what to field strength? - Halves it.
A capacitor with $C = \pu{100 \mu F}$

Revision workflow

Re-derive Coulomb-potential-field relationships weekly to cement mathematical links.
Solve combined problems involving charged particle motion in uniform fields and capacitor energy.
Prepare flashcards linking capacitor energy formulae to area-under-graph interpretations on $V$ - $Q$ diagrams.
After revising capacitance, move to Topic 16 (Circuits) for combined capacitance and RC discharge practice - they build directly on Topic 14 outcomes.

8 Further reading

Last updated 14 Jul 2025. Next review when SEAB issues the 2027 draft syllabus.

Practice Quiz

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Reviewed by

Chee Wei Jie·Academic Advisor (Physics)