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 21-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
Reviewed by
Chee Wei Jie·Academic Advisor (Physics)
. 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/r2
Mixing the source charge Q with the test charge q.
Potential or potential energy is requested
Potential is scalar, field is vector
V=kQ/r or UE=kQ1Q2/r
Charged particle moves between plates
Field is approximately uniform
E=V/d, then F=qE, then SUVAT
Using horizontal speed as the vertical speed.
Capacitor energy is requested
Decide whether Q, V, or C is fixed
Area under the V-Q graph
Dropping the 21
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=4πε01r2Q1Q2.
1.1 Quick-fire cues
SI unit for charge is C.
Sign mistakes: repulsive if charges share sign, attractive otherwise.
1.2 Mini-drill
Two +2.0nC charges sit 5.0cm apart in air. Calculate F. Answer: 1.4⋅10−5N (repulsive).
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=qF, the field around a point charge is
E=4πε01r2Q.
Integrating this field gives the scalar potential
V=4πε01rQ.
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
UE=4πε01rQ1Q2.
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
Setting potential to zero just because the fields cancel.
Force on a test charge
First find the net field, then use F=qE
Direction follows the sign of the test charge
Reversing the source field when the test charge is negative.
Potential energy of two charges
A scalar system energy UE=kQ1Q2/r
Use the signs of both charges in the product
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 E.
Equipotential surfaces are always perpendicular to field lines.
Mathematically, E=−drdV.
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=−dV/dr.
Equipotential lines are close together
Magnitude of potential gradient is large
Field is stronger where equipotentials are closer
Treating equal potential spacing as equal physical spacing.
Particle has negative charge
Field direction stays fixed by the source charges
Force and acceleration are opposite to E
Reversing the electric field itself instead of reversing the force on the particle.
Worked check: if a potential graph falls from 12V to 4V over 0.20m, the gradient is negative. Since E is the negative gradient, the electric field points in the positive distance direction and has magnitude 40V⋅m−1. An electron released there accelerates in the opposite direction because its charge is negative.
4 Uniform electric fields
Between parallel plates:
E=dV.
4.1 Force & motion
A charge q experiences F=qE. With constant a=mF, kinematics mirrors vertical projectile motion.
4.2 Mini-drill
An electron enters midway between a 600V plate pair 8.0mm apart, with horizontal speed 2.0×107m⋅s−1. Find the magnitude of its vertical displacement after 1.0cm of travel through the field.
Answer:E=V/d=7.50×104N⋅C−1, a=eE/me≈1.32×1016m⋅s−2, t=0.010/(2.0×107)=5.0×10−10s, so y=21at2≈1.65mm.
5 Capacitance fundamentals
Definition:C=VQ. Unit: farad (F=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 ∫0QV,dQ. Geometrically this is the area of the triangle under that straight line, giving the factor of 21. Substituting Q=CV or V=Q/C converts between the three equivalent forms.
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=21QV. 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.
I0=V0/R
Using the final current, which is zero after a long time.
Time constant
Multiply the circuit values before using an exponential.
τ=RC
Treating RC as a final time when discharge is complete.
Voltage or charge after time t
Decide whether the question asks for V or Q.
V=V0e−t/RC
Energy left
Recalculate from the remaining voltage or charge.
U=21CV2
Halving voltage and then saying energy halves; energy depends on V2
Worked check: after one time constant, V=V0e−1≈0.37V0. The capacitor is not empty after RC; it has about 37% of its initial p.d., charge, and current magnitude left.
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 ε0 as 0
Memorise 8.85×10−12F⋅m−1.
Confusing F=qE and E=dV
Write a “fields toolkit” flashcard.
Forgetting 21 in energy
Draw the Q-V triangle before calculating.
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=4πε01r2Q1Q2 for the electric force between two point charges in free space or air.
(b) recall and use E=4πε01r2Q
(c) define electric potential at a point as the work done per unit charge by an external force in bringing a small positive test charge from infinity to that point.
(d) use the equation V=4πε01rQ
(e) show an understanding that the electric potential energy of a system of two point charges is UE=4πε01rQ1Q2
(f) recall that electric field strength at a point is equal to the negative potential gradient at that point and use this to solve problems.
(g) calculate the field strength of the uniform electric field between charged parallel plates in terms of the potential difference and plate separation.
(h) calculate the force on a charge in a uniform electric field.
(i) describe the effect of a uniform electric field on the motion of a charged particle.
(j) define capacitance as the ratio of the charge stored to the potential difference and use C=VQ to solve problems.
(k) recall that the electric potential energy stored in a capacitor is given by the area under the graph of potential difference against charge stored, and use this and the equations U=21QV, U=21CQ2
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=dV. 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=4πε01r2Q1Q2
Electric field (point charge)
E=qF=4πε01r2Q
Electric potential
V=4πε01rQ
Potential gradient relation
E=−drdV
Uniform field
E=dV
Work/energy of charge
W=qV
Capacitance definition
C=VQ
Capacitor energy
U=21CV2=21QV=21CQ2
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=dV 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 +4.0nC and −2.0nC are 12cm 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.
A C=220×10−6F capacitor charged to V=12V discharges through R=4.7×103Ω. Calculate the initial energy stored and the initial discharge current. Discuss how energy is dissipated.
Solution: U=21CV2, I0=RV. Mention exponential decay of current and conversion to thermal energy.
