S5 Physics | Electric Field and Potential prep guide (for tutors) | Field superposition, null point, potential and plates

2026-11-01

Tutor lesson-prep guides · Secondary Physics

Students treat potential as another way of writing field strength, so they report the null point as zero potential. This lesson separates the two: field strength is a vector and can cancel to zero, while potential is a scalar that adds, so at the null point it is still 9.0e+04 V.

Objectives — Students can: (1) compute the field of a point charge with E = kQ/r2 and superpose two fields as vectors; (2) locate the null point between two unlike charges and explain that zero field is not zero potential; (3) add potentials with V = kQ/r and use W = qV for moving a charge; (4) use E = V/d in a uniform field, compute qEd and convert to electronvolts.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Coulomb’s law	Two point charges 0.10 m apart: if the distance halves, how does the force change?	Four times: F is proportional to 1/r2
Vector addition	+x 6 N and -x 2 N act on one point: what is the resultant?	+x 4 N: opposite directions subtract
Work and energy	In W = Fd, what sign does the work take when F opposes the displacement?	Negative: the field removes energy

02 Lesson flow (50 minutes)

Time	Stage	Tutor does	Student does
0-6 min	Opening question	At the midpoint between two charges, what are the field and the potential?	Two guessed answers written down
6-12 min	Prerequisite check	Coulomb law, vector subtraction, sign of work	All three correct before moving on
12-22 min	Act 1	Field of each charge, then superposition	2.0e+06 N/C
22-30 min	Act 2	Null point: set the two fields equal and solve for r	20 cm from the +4.0 microC charge
30-40 min	Act 3	Scalar potentials, W = qV, plates with E = V/d	V = 1.8e+05 V, E = 8000 V/m
40-45 min	Worked examples	One start-up routine for each of the four question types	Students copy the routine
45-50 min	Exit check	Three questions marked on the spot	Answered independently

03 Teaching script (read it out)

Scene 1: Field strength is a vector: draw arrows first

Q1 = +4.0 microC and Q2 = -1.0 microC are 30 cm apart, so the midpoint is 15 cm from each. E1 = 1.60e+06 N/C pointing from the positive charge towards the negative one, and E2 = 4.00e+05 N/C pointing the same way, because field lines end on a negative charge. The two vectors are parallel, so E = 2.00e+06 N/C.
Make students draw both arrows at the midpoint before touching the calculator. If the two charges had the same sign the arrows would oppose and the fields would subtract, so run that case once as a contrast.

Task: change Q2 to +1.0 microC, redraw the arrows and compute the midpoint field (answer 0, equal and opposite).

Scene 2: The null point: solve, do not guess the middle

The null point lies on the line joining the charges, closer to the smaller charge. With r measured from Q1, kQ1/r2 = kQ2/(0.30 – r)2 gives r/(0.30 – r) = sqrt(Q1/Q2) = 2, so r = 0.20 m, that is 20 cm from Q1 and 10 cm from Q2.
Then ask for the potential there: V = 9.00e+04 V, not zero. Leave this for students to compute; the arithmetic is what convinces them. Use the hill analogy: a flat stretch has zero slope but its height is not zero.

Task: two positive charges +4.0 microC and +1.0 microC, 30 cm apart. Where is the null point? (20 cm from the +4.0 microC charge, between them.)

Scene 3: Potential and work: add scalars, then multiply by the charge

Potentials add directly because they are scalars. The midpoint potential is 1.80e+05 V. Moving +2.0 microC from the midpoint to the null point changes the potential by -9.00e+04 V, so W = qV = -0.18 J; the negative sign means the field does positive work and the potential energy falls.
Parallel plates: 120 V across 15 mm gives E = 8000 V/m everywhere between the plates. A proton crossing 5 mm sees 40 V and gains 40 eV. Make students do both routes (qV and qEd) and confirm they agree; the electronvolt is only joules divided by 1.6e-19.

Task: raise the plate voltage to 240 V with the same gap. How much energy does the proton gain over 5 mm? (80 eV.)

04 Question types and worked steps

Type	Example	Steps	Answer
Superposition at a point	E = kQ1/r2 +/- kQ2/r2	Draw arrows, add if parallel, subtract if opposite	2.0e+06 N/C
Null point position	kQ1/r2 = kQ2/(d-r)2	Take the square root, solve the linear equation	0.20 m
Potential and W = qV	V = sum kQ/r; W = q x (V final – V initial)	Keep the signs attached to each charge	-0.18 J
Uniform field and eV	E = V/d; W = qEd	Field is constant, energy converts straight to eV	40 eV

Sanity check (teach this): The null point sits between the charges and nearer the smaller one: 20 cm against 10 cm, a 2:1 ratio matching the square roots of 4:1. The midpoint field 2.0e+06 N/C must exceed either single field (1.6e+06 and 4.0e+05) because they add; a smaller answer means the student subtracted. The null-point potential is positive, so an answer of zero signals confusion between field and potential. Energy cross-check: 40 V of potential difference must give a proton 40 eV.

05 Independent practice (3 questions)

Question	Answer	What to watch
Charges +9.0 microC and +4.0 microC are 0.50 m apart. How far is the null point from +9.0 microC?	0.30 m (3:2 ratio)	Same-sign charges put the null point between them
A uniform field has 300 V across 20 mm. What is E, and which way does an electron feel force?	15000 V/m; towards the positive plate	Electrons feel force opposite to the field
Midpoint potential is 1.8e5 V. Move -2.0 microC to the null point: what work does the field do?	+0.18 J	For a negative charge the sign flips

06 Common mistakes and what to say

Mistake	Student writes	Say this
Writing zero potential at the null point	Treating field and potential as one quantity	“Field is slope, potential is height: flat ground has zero slope but is not at sea level.”
Subtracting the fields at the midpoint	Reflex of subtracting whenever two charges appear	“Draw the arrows: both point at the negative charge, so they add.”
Field lines drawn from negative to positive	Misremembered convention	“Lines leave positive and enter negative: arrows always run + to -.”
Dropping the sign in W = qV	Using potential as a magnitude	“Potential carries a sign, so a negative charge falling in potential has positive work done on it.”

07 Homework (paper, section and question numbers)

Secondary Physics chapter index (motion, forces, heat, waves, electricity, magnetism) — Do the work-energy chapter first, then questions 1, 3 and 5 of the electricity set
Secondary prep guides overview (all subjects) — Revise in subject order

08 Exit check (last 5 minutes)

Question	Answer	Decision
Is the potential zero at the null point?	No: 9.0e4 V	Wrong means Act 2 again
In E = V/d, which length is d?	The perpendicular separation of the plates	Wrong means Act 3 again
A negative charge moves to lower potential: does its potential energy rise or fall?	Fall	Two or more wrong means back to Act 1

09 Tutor FAQ

Which year is this for?

S5, the electricity unit of HKDSE physics; also useful for S6 revision before past papers.

How many formulas must be memorised?

Four: E = kQ/r2, V = kQ/r, W = qV and E = V/d; everything else follows.

What if the arithmetic keeps going wrong?

Require units to be written before numbers, convert microC to 1e-6 first, and only then use the calculator.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.