Parallel plates 2.0\ cm apart have a potential difference of 300\ V. Find the field strength. [2 marks]

OCR OCR 2019-style practice: Two point charges of +3.0\ nC and +5.0\ nC are placed 4.0\ cm apart in a vacuum. Calculate the force between them. Take (1)/(4π_0) = 8.99 × 10^9\ N m^2\ C^-2.

Coulomb's law: F = (1)/(4π_0)(Q_1 Q_2)/(r^2). F = (8.99 × 10^9)(3.0 × 10^-9)(5.0 × 10^-9)(0.040)^2. Numerator: (8.99 × 10^9)(1.5 × 10^-17) = 1.35 × 10^-7. Denominator: (0.040)^2 = 1.6 × 10^-3. F = 1.35 × 10^-71.6 × 10^-3 = 8.4 × 10^-5\ N, repulsive (both positive). Markers reward Coulomb's law, converting to metres, and the force about 8.4 × 10^-5\ N with a statement that it is repulsive.

OCR OCR 2021-style practice: Two parallel plates 5.0\ mm apart have a potential difference of 200\ V across them. Calculate the electric field strength between the plates and the force on an electron placed there. Take e = 1.6 × 10^-19\ C.

Uniform field strength: E = (V)/(d) = 2005.0 × 10^-3 = 4.0 × 10^4\ V m^-1. Force on the electron: F = EQ = (4.0 × 10^4)(1.6 × 10^-19) = 6.4 × 10^-15\ N. Markers reward E = (V)/(d) giving 4.0 × 10^4\ V m^-1, and F = EQ giving about 6.4 × 10^-15\ N.

EnglandPhysicsSyllabus dot point

How do electric charges exert forces at a distance, and how does an electric field compare with gravity?

Electric fields: Coulomb's law, electric field strength for radial and uniform fields, electric potential, and the comparison between electric and gravitational fields.

A focused answer to the OCR H556 electric fields content, covering Coulomb's law for the force between point charges, electric field strength for radial and uniform fields, the motion of charges in a uniform field, electric potential, and the parallels and contrasts between electric and gravitational fields.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Coulomb's law gives the force between two point charges, $F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$ , attractive for opposite charges and repulsive for like charges. Electric field strength is the force per unit positive charge, $E = \frac{F}{Q}$ , equal to $\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$ for a radial field and $E = \frac{V}{d}$ for a uniform field between parallel plates. Electric potential is the work done per unit charge bringing a small positive charge from infinity, $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$ . Electric and gravitational fields share an inverse-square form, but gravity is always attractive while the electric force can attract or repel, and gravity is far weaker between fundamental particles.

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What this dot point is asking
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What this dot point is asking

OCR wants you to state and use Coulomb's law, define and calculate electric field strength for radial and uniform fields, analyse the motion of charges in a uniform field, define electric potential, and compare electric fields with gravitational fields.

The answer

Coulomb's law

Electric field strength

A charge in a uniform field experiences a constant force $F = EQ$ , so it accelerates uniformly (like a projectile in gravity). This is how electron beams are deflected in old cathode-ray tubes and in mass spectrometers.

Electric potential

Comparing electric and gravitational fields

Speed of an electron accelerated across plates

An electron is accelerated from rest through a potential difference of $500\ \text{V}$ . Find its final speed. Take $e = 1.6 \times 10^{-19}\ \text{C}$ and $m = 9.11 \times 10^{-31}\ \text{kg}$ .

step 1: Energy gained

Work done on the charge equals $W = QV = (1.6 \times 10^{-19})(500) = 8.0 \times 10^{-17}\ \text{J}$ .

step 2: Equate to kinetic energy

$\frac{1}{2}mv^2 = 8.0 \times 10^{-17}$ , so $v^2 = \frac{2 \times 8.0 \times 10^{-17}}{9.11 \times 10^{-31}} = 1.76 \times 10^{14}$ .

step 3: Solve

$v = \sqrt{1.76 \times 10^{14}} = 1.3 \times 10^{7}\ \text{m s}^{-1}$ .

Examples in context

Electric fields steer charged particles in particle accelerators, cathode-ray tubes, inkjet printers and mass spectrometers, where the deflection depends on charge and mass. Electrostatic precipitators charge smoke particles so a field can collect them, cleaning industrial exhaust. The parallel-plate field underlies the capacitor, and the comparison with gravity is a favourite synoptic theme, highlighting why the electric force dominates atomic structure while gravity governs planets and stars.

Try this

Q1. State Coulomb's law. [2 marks]

Cue. The force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation.

Q2. Parallel plates $2.0\ \text{cm}$ apart have a potential difference of $300\ \text{V}$ . Find the field strength. [2 marks]

Cue. $E = \frac{V}{d} = \frac{300}{0.020} = 1.5 \times 10^{4}\ \text{V m}^{-1}$ .

Q3. State one similarity and one difference between electric and gravitational fields. [2 marks]

Cue. Both obey an inverse-square law for force; gravity is always attractive whereas the electric force can attract or repel.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksTwo point charges of

+3.0\ \text{nC}

and

+5.0\ \text{nC}

are placed

4.0\ \text{cm}

apart in a vacuum. Calculate the force between them. Take

\frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2}

Show worked answer →

Coulomb's law: $F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$ .

$F = (8.99 \times 10^{9})\frac{(3.0 \times 10^{-9})(5.0 \times 10^{-9})}{(0.040)^2}$ .

Numerator: $(8.99 \times 10^{9})(1.5 \times 10^{-17}) = 1.35 \times 10^{-7}$ . Denominator: $(0.040)^2 = 1.6 \times 10^{-3}$ .

$F = \frac{1.35 \times 10^{-7}}{1.6 \times 10^{-3}} = 8.4 \times 10^{-5}\ \text{N}$ , repulsive (both positive).

Markers reward Coulomb's law, converting to metres, and the force about $8.4 \times 10^{-5}\ \text{N}$ with a statement that it is repulsive.

OCR 20213 marksTwo parallel plates

5.0\ \text{mm}

apart have a potential difference of

200\ \text{V}

across them. Calculate the electric field strength between the plates and the force on an electron placed there. Take

e = 1.6 \times 10^{-19}\ \text{C}

Show worked answer →

Uniform field strength: $E = \frac{V}{d} = \frac{200}{5.0 \times 10^{-3}} = 4.0 \times 10^{4}\ \text{V m}^{-1}$ .

Force on the electron: $F = EQ = (4.0 \times 10^{4})(1.6 \times 10^{-19}) = 6.4 \times 10^{-15}\ \text{N}$ .

Markers reward $E = \frac{V}{d}$ giving $4.0 \times 10^{4}\ \text{V m}^{-1}$ , and $F = EQ$ giving about $6.4 \times 10^{-15}\ \text{N}$ .

Related dot points

Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)