This unit is ≈16% of the A-Level Physics, across 6 lessons. Full syllabus
Lesson 2 of 6 · Fields and their consequences
Electric fields, potential and charged particles
9 min read · about 2 h with practice3 quick checks≈3% of the testStretch: Stretch: harder material that separates the top grades
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Electric fields are the model for every other field at A-Level: the same ideas of field strength, potential, field lines and equipotentials return in gravitation, capacitors and particle physics. Examiners test them in four ways: quick ratio MCQs on Coulomb’s law, structured calculations on radial fields and potential, electrons deflected between parallel plates (projectile motion in disguise), and an extended comparison of electric and gravitational fields.
By the end you’ll be able to
Use F = Qq/4πε₀r² and E = Q/4πε₀r², and combine fields from several charges as vectors
Use E = V/d for parallel plates and analyse deflection of charged particles (projectile-like motion)
Define electric potential, use V = Q/4πε₀r and relate E to the potential gradient; sketch equipotentials
Calculate work done moving charges between points of different potential
Compare gravitational and electric fields (similarities, differences, relative strengths)
What the exam asks
Use E=QF, Coulomb’s law F and , and .
vii.Check your understanding
3 questions on electric fields, potential and charged particles. Every option is explained once you answer.
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PromptCard 1 of 3
Define electric field strength at a point.
=
4πε0r2Q1Q2
E=4πε0r2Q
add fields from several charges as vectors
Use E=dV for parallel plates and analyse a charged particle that is accelerated by, or deflected in, a uniform field.
Define electric potential, use V=4πε0rQ (a scalar), and link field to potential gradient. Read V–r graphs (the gradient gives E) and E–r graphs (the area gives ΔV).
Calculate work done, ΔW=QΔV, and electric potential energy EP=4πε0rQq.
Sketch field lines and equipotentials, and compare electric and gravitational fields.
Core ideas
Field strength and field lines
Electric field strength at a point is the force per unit positive charge on a small test charge placed there: E=F/Q, in N C⁻¹ (identical to V m⁻¹). It is a vector pointing the way a positive charge would be pushed.
Field-line rules for sketches:
Lines start on positive charges and end on negative charges, and they never cross.
Closer lines mean a stronger field.
Lines meet a conducting surface at 90°.
Between parallel plates the field is uniform: straight, parallel, equally spaced lines, curving outwards only at the edges.
Coulomb’s law and radial fields
F=4πε0r2Q1Q
ε0=8.85×10−12F m is the permittivity of free space. Air can be treated as a vacuum.
Uniform fields: E=V/d
Moving charge Q from one plate to the other takes work W=QV. The force is constant, so W=Fd. Then Fd, which gives . This is also why N C⁻¹ = V m⁻¹. Convert from mm to m before you divide.
Electric potential and potential energy
Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point. It is zero at infinity.
V=4πε0r
Potential is a scalar. Near a positive charge it is positive; near a negative charge it is negative. For several charges, add the potentials with their signs. No vectors are needed.
Field is the negative of the potential gradient: E=−ΔrΔV. On a V– graph, the gradient’s magnitude is . On an – graph, the area between and is .
Charged particles in uniform fields
Acceleration along the field (from rest):21mv2=QV.
Entering at right angles to the field: this is a projectile problem.
Direction
Motion
Equation
Parallel to plates
constant velocity vx
t=L/vx
The path between the plates is a parabola. Once the particle leaves the field it travels in a straight line at angle θ, where tanθ=vy/vx. For electrons and ions, gravity is negligible: is about times in a typical deflection tube. For a charged oil drop, gravity matters, and a drop held stationary has .
Electric versus gravitational fields
Feature
Gravitational (point mass)
Electric (point charge)
Force law
F=r2Gm1, inverse square
Both have radial field lines, a potential that is zero at infinity, and field equal to minus the potential gradient.
Worked examples
Exam technique
Name the field type first. Plates mean uniform: E=V/d and constant force. A point or sphere means radial: inverse square, and potential goes as 1/r.
Ratio MCQs: write F∝Q or and scale. Doubling both charges and doubling leaves unchanged.
Common mistakes
Quick recap
E=F/Q is a vector in N C⁻¹ (= V m⁻¹). Field lines run from + to −, and closer lines mean a stronger field.
For a point charge, F= and . Add fields from several charges as vectors.
Q
0
r
Qq
V/d
recall
V=Q/4πε0r
EP=Qq/4πε0r
9
N m2
C−2
2
E
=
4πε0r2Q4πε01≈
8.99×
109N m2C−2
−
1
Outside a uniformly charged sphere, the field is the same as if all the charge were at its centre, so r is always measured from the centre.
Like charges repel and unlike charges attract. A positive F from the formula means repulsion.
With several charges, find each field separately, then add them as vectors: resolve into components and use Pythagoras. Symmetry often cancels one component.
=
QV
E=F/Q=V/d
d
Q
EP
=
qV=
4πε0rQqΔW=
qΔV
r
E
E
r
r1
r2
ΔV
Equipotentials are surfaces of constant V. They are always perpendicular to field lines, and no work is done moving along one. For a point charge they are concentric spheres that get further apart for equal steps of V. Between plates they are equally spaced planes.
A particle of charge e accelerated through V volts gains eV joules. That is the origin of the electronvolt.
Across the field
uniform acceleration a=mQE=mdQV
y=21at2,
eE
1014
mg
QV/d=mg
m2
F=4πε0r2Q1Q2, inverse square
Field strength
force per unit mass, N kg⁻¹
force per unit positive charge, N C⁻¹
Direction of force
always attractive
attractive or repulsive
Potential
V=−GM/r, always negative
V=+Q/4πε0r, sign follows Q
Relative strength (two protons)
1
about 1036 times larger
Dominant scale
planets, stars, galaxies (bulk matter is neutral)
atoms, molecules, nuclei
x28.0=(0.30−x)22.0⇒x2=0.30−x1⇒x=0.20m
Check: each field is 8.99×109×8.0×10−9/0.202=1.8×103N C−1 and 8.99×109×2.0×10−9/0.1 ✓. Taking the square root first avoids solving a quadratic. The null point is always closer to the smaller charge.
Time between the plates: t=0.060/(4.0×107)=1.5×10−9 s.
y=21at2=0.5×2.34×1015×(1.5×10−9)2=2.6×10−3 m (2.6 mm), which is less than 15 mm, so the electron clears the plate.
vy=at=3.5×106m s−1, so θ=tan−1(3.5×106/4.0×10 to its original direction.
Δ
V
=
4πε0Q(0.201−0.501)=
8.99×
109×
30×
10−9×
3.0=
809V
W=qΔV=2.0×10−6×809=1.6×10−3J
Using W=Fd here would be wrong, because the force changes with distance. In radial fields, use potential for energy.
1
Q2
/
r2
V∝Q/r
r
F
Vectors or scalars? Fields and forces are vectors, so draw arrows and resolve. Potentials and energies are scalars, so add them with signs. Using the wrong one is the single biggest source of lost marks.
The V and E shortcut: at a distance r from a point charge, V/E=r. If a question gives both, you can find r and then Q without the constant.
Deflection questions: set out two columns, horizontal and vertical. Only the time links them.
Show that: quote 1/4πε0 or ε0 explicitly and give one more significant figure than the target value.
Sketches: draw equipotentials perpendicular to field lines everywhere. Space them further apart where the field is weaker.
V
1/r
r
centre
2.0×10−9
r
Q1Q2/4πε0r2
E=Q/4πε0r2
Uniform field: E=V/d, with constant force and acceleration.
V=Q/4πε0r is a scalar, zero at infinity, with the same sign as Q. ΔW=qΔV and EP=Qq/4πε0r.
E=−ΔV/Δr. The gradient of a V–r graph gives E, and the area under an E–r graph gives ΔV. Equipotentials are perpendicular to field lines.
Particles: 21mv2=QV from rest. Across plates, the motion is a parabolic projectile with a=QV/md, then a straight line.
Electric forces are about 1036 times gravitational forces between protons and can attract or repel, while gravity only attracts.