A Level · Physics 9702 · Fields & motion

Electric fields: A Level Physics questions with answers

Free Electric fields questions for A Level Physics (9702), part of Fields & motion. Each question has the answer and the marking points an examiner would look for.

Generate a full Electric Fields worksheet →All Fields & motion questions

Electric fields practice questions

Question 1Structured · 9 marksA* challenge

An electron enters the uniform field between two parallel plates, travelling at 3.0 × 107 m s−1 parallel to the plates. The plates are 40 mm apart and 5 cm long, with a p.d. of 1200 V between them.

(a)

Calculate the electric field strength and the force on the electron.

[2]
(b)

Calculate the vertical deflection of the electron as it leaves the plates.

[3]
(c)

Describe the shape of the path in the field and explain why the gravitational force can be ignored.

[2]
(d)

Two point charges of +4.0 nC and +1.0 nC are 0.30 m apart. Calculate the distance from the +4.0 nC charge of the point between them where the field strength is zero.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)30000 V m−1, 4.80 × 10−15 N
E = V ÷ d C1
F = eE, both correct A1
2
(b)7.32 × 10−3 m
t = 1.67 × 10−9 s C1
a = 5.27 × 1015 m s−2 C1
7.32 × 10−3 A1
3
(c)parabolic; weight ~10−29 N is negligible compared with ~10−14 N electric force
parabola B1
weight many orders of magnitude smaller than electric force B1
2
(d)0.20 m
4/x2 = 1/(0.30 − x)2 C1
0.20 m A1
2
Question 2Structured · 5 marksA* challenge

An isolated metal sphere carries a charge of +5.0 × 10−9 C. Point P is 5 cm from the centre of the sphere, outside the sphere.

(a)

Define electric potential at a point.

[2]
(b)

Calculate the electric field strength and the electric potential at P.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)work done per unit positive charge from infinity
work done per unit positive charge B1
bringing a small test charge from infinity to the point B1
2
(b)E = 18000 N C−1; V = 899 V
E = Q/4πε₀r2 C1
18000 N C−1 A1
V = 899 V A1
3
Question 3Structured · 10 marks

An isolated conducting sphere of radius 5.0 cm carries a charge of +6.0 nC. Treat the charge as a point charge at its centre.

(a)

State Coulomb’s law.

[2]
(b)

Calculate the electric field strength at 20 cm from the centre.

[2]
(c)

Calculate the electric potential at 20 cm from the centre.

[2]
(d)

Use your answer to (c) to calculate the work done in moving a charge of +1.0 nC from infinity to this point.

[2]
(e)

State one similarity and one difference between gravitational and electric fields due to point sources.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)force proportional to product of charges, inversely proportional to square of separation
the force between two point charges is proportional to the product of the charges B1
and inversely proportional to the square of their separation B1
2
(b)1350 N C−1
E = Q/(4πε₀r2) C1
1350 N C−1 A1
2
(c)270 V
V = Q/(4πε₀r) C1
270 V A1
2
(d)2.70 × 10−7 J
W = qV C1
2.70 × 10−7 J A1
2
(e)both inverse square; gravity always attractive, electric can repel
both obey an inverse-square law (field ∝ 1/r2) B1
gravitational forces are always attractive, electric forces can be attractive or repulsive B1
2
Question 4Structured · 8 marks

In a Millikan-type experiment a charged oil drop of mass 2.0 × 10−15 kg is held stationary between horizontal plates 1.0 cm apart when the p.d. between them is 307 V.

(a)

State the direction of the electric force on the drop and of the electric field if the drop is negatively charged.

[2]
(b)

Calculate the electric field strength between the plates.

[1]
(c)

Use your answer to (b) to calculate the charge on the drop, and the number of electrons it has gained.

[3]
(d)

Explain how results from many drops showed that charge is quantised.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)force upwards; field downwards
electric force upwards (balancing the weight) B1
field downwards, since the force on a negative charge is opposite to the field B1
2
(b)30700 V m−1
E = V/d = 30700 V m−1 B1
1
(c)6.40 × 10−19 C; 4
qE = mg C1
q = 6.40 × 10−19 C A1
4 electrons A1
3
(d)all charges were integer multiples of one value, 1.6 × 10⁻¹⁹ C
every charge measured was a whole-number multiple of a smallest value B1
1.60 × 10−19 C (the elementary charge) B1
2
Question 5Structured · 9 marks

An electron enters the uniform field between two parallel plates (Fig. 5.1), midway between them and at right angles to the field, with speed 3.0 × 107 m s−1. The plates are 4.0 cm apart, 5.0 cm long, with a potential difference of 500 V between them.

+−+500 V0 Velectron5.0 cm4.0 cm
Fig. 5.1
(a)

Define electric field strength.

[1]
(b)

Calculate the electric field strength between the plates.

[1]
(c)

Calculate the acceleration of the electron between the plates.

[2]
(d)

Use your answer to (c) to calculate the vertical deflection of the electron as it leaves the plates.

[3]
(e)

Describe the shape of the path of the electron between the plates and after it leaves them, and explain.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)force per unit positive charge
force per unit positive charge (on a stationary charge) B1
1
(b)12500 V m−1
E = V/d = 12500 V m−1 B1
1
(c)2.20 × 1015 m s−2
F = eE C1
a = eE/me = 2.20 × 1015 m s−2 A1
2
(d)3.05 × 10−3 m
time between plates = 1.67 × 10−9 s C1
y = ½at2 C1
3.05 × 10−3 m A1
3
(e)parabola inside (constant force perpendicular to velocity); straight line after (no force)
parabolic path between the plates as there is a constant force perpendicular to the initial velocity B1
straight line after leaving, since no force acts B1
2
Question 6Structured · 5 marks

Two parallel plates are 4 cm apart with a potential difference of 500 V between them.

(a)

Calculate the electric field strength between the plates.

[1]
(b)

Calculate the force on an electron between the plates and its acceleration.

[2]
(c)

Describe the path of an electron that enters the field moving parallel to the plates.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)12500 V m−1
12500 V m−1 B1
1
(b)2.00 × 10−15 N; 2.20 × 1015 m s−2
F = eE: 2.00 × 10−15 N A1
a = F/m: 2.20 × 1015 m s−2 A1
2
(c)parabola towards positive plate
parabolic path B1
curving towards the positive plate B1
2

More Fields & motion questions

Want a mentor, not just a paper?

Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.

Meet Riyaz →

Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in