What is electric field strength?

Around a point charge the field is radial: E = (1)/(4π_0)(Q)/(r^2), pointing away from a positive charge. Between two parallel charged plates the field is uniform: E = (V)/(d), where V is the potential difference and d the plate separation, with field lines running straight from the positive to the negative plate.

What is electric potential?

Potential is a scalar, so potentials from several charges simply add. Unlike field strength (inverse square), potential falls off as (1)/(r). The work done moving a charge between two points depends only on the potential difference, not the path taken, because the electrostatic field is conservative.

EnglandPhysicsSyllabus dot point

How do charges exert forces and store energy in a field?

Coulomb's law, electric field strength for radial and uniform fields, electric potential, and the motion of charged particles in a uniform field.

A focused answer to the Edexcel 9PH0 electric fields content, covering Coulomb's law, electric field strength in radial and uniform fields, electric potential, and the motion of charged particles in a uniform field.

Generated by Claude Opus 4.811 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 point charges, $F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$ , an inverse-square law like gravitation but able to attract or repel. Electric field strength is force per unit positive charge, $E = \frac{F}{Q}$ ; it is radial ( $E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$ ) around a point charge and uniform ( $E = \frac{V}{d}$ ) between parallel plates. Electric potential is the work done per unit charge from infinity, $V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}$ , and the field is the negative potential gradient. A charge in a uniform field accelerates as $a = \frac{EQ}{m}$ , following a parabolic path just like a projectile.

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What this dot point is asking
The answer
Examples in context
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What this dot point is asking

Edexcel wants you to apply Coulomb's law to point charges, define and calculate electric field strength for radial and uniform fields, work with electric potential and the link between field and potential gradient, and analyse the motion of a charged particle moving through a uniform field, drawing the parallel with projectile motion.

The answer

Coulomb's law

This is an inverse-square law: doubling the separation quarters the force. It is structurally identical to Newton's law of gravitation, except that charge comes in two signs (so the force can repel) and the constant is vastly larger, making the electrostatic force between fundamental particles enormously stronger than the gravitational one.

Electric field strength

Around a point charge the field is radial: $E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$ , pointing away from a positive charge. Between two parallel charged plates the field is uniform: $E = \frac{V}{d}$ , where $V$ is the potential difference and $d$ the plate separation, with field lines running straight from the positive to the negative plate.

Electric potential

Potential is a scalar, so potentials from several charges simply add. Unlike field strength (inverse square), potential falls off as $\frac{1}{r}$ . The work done moving a charge between two points depends only on the potential difference, not the path taken, because the electrostatic field is conservative.

Motion of a charged particle in a uniform field

A particle of charge $Q$ in a uniform field feels a constant force $F = EQ$ and hence a constant acceleration $a = \frac{EQ}{m}$ in the direction of the field (for a positive charge) or against it (for a negative charge). If it enters at right angles to the field with speed $u$ , it keeps a constant velocity along the entry direction and accelerates uniformly across the field, tracing a parabola exactly as a projectile does under gravity.

Deflection in a uniform field

An electron enters midway between two parallel plates $4.0$ cm long, $2.0$ cm apart, with a potential difference of $200$ V and a horizontal speed of $3.0 \times 10^{7}$ m per second. Find its vertical deflection on leaving the plates.

step 1: Field and acceleration

$E = \frac{V}{d} = \frac{200}{0.020} = 1.0 \times 10^{4}$ V per metre. $a = \frac{EQ}{m} = \frac{1.0 \times 10^{4} \times 1.6 \times 10^{-19}}{9.11 \times 10^{-31}} = 1.76 \times 10^{15}$ m per second squared.

step 2: Time in the field

$t = \frac{\text{length}}{u} = \frac{0.040}{3.0 \times 10^{7}} = 1.33 \times 10^{-9}$ s.

step 3: Vertical deflection

$s = \frac{1}{2}at^2 = \frac{1}{2} \times 1.76 \times 10^{15} \times (1.33 \times 10^{-9})^2 = 1.6 \times 10^{-3}$ m, about $1.6$ mm.

Examples in context

Inkjet printers deflect charged ink droplets through a uniform field to steer them onto the page, the deflection set by the applied voltage. The cathode-ray tube in older oscilloscopes used the same parabolic deflection of an electron beam. Lightning conductors exploit the very strong radial field at a sharp point to ionise air and bleed charge safely to earth. Electrostatic precipitators charge smoke particles and pull them onto collector plates in a strong field, cleaning industrial flue gases.

Try this

Q1. Define electric field strength. [1 mark]

Cue. The force per unit positive charge at a point, $E = \frac{F}{Q}$ .

Q2. Two parallel plates $5.0$ mm apart have a potential difference of $100$ V. Find the field strength between them. [2 marks]

Cue. $E = \frac{V}{d} = \frac{100}{5.0 \times 10^{-3}} = 2.0 \times 10^{4}$ V per metre.

Q3. State how electric field strength and electric potential each depend on distance from a point charge. [2 marks]

Cue. Field strength varies as $\frac{1}{r^2}$ (inverse square); potential varies as $\frac{1}{r}$ .

Exam-style practice questions

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

Edexcel 20184 marksTwo point charges of

+2.0

nC and

+3.0

nC are separated by

5.0

cm in a vacuum. Calculate the magnitude of the electrostatic force between them.

Show worked answer →

Coulomb's law: $F = \frac{1}{4\pi\varepsilon_0}\frac{Q_1 Q_2}{r^2}$ , with $\frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}$ N m squared per C squared.

$F = 8.99 \times 10^{9} \times \frac{2.0 \times 10^{-9} \times 3.0 \times 10^{-9}}{(0.050)^2} = 8.99 \times 10^{9} \times \frac{6.0 \times 10^{-18}}{2.5 \times 10^{-3}}$ .

$F = 8.99 \times 10^{9} \times 2.4 \times 10^{-15} = 2.2 \times 10^{-5}$ N (repulsive).

Markers reward correct conversion of nC and cm to SI, use of the Coulomb constant, and the value about $2.2 \times 10^{-5}$ N.

Edexcel 20214 marksShow that an electron entering a uniform field of strength

2.0 \times 10^{4}

V per metre experiences an acceleration of order

10^{15}

m per second squared, and state the direction of its acceleration relative to the field.

Show worked answer →

Force on the electron: $F = EQ = 2.0 \times 10^{4} \times 1.6 \times 10^{-19} = 3.2 \times 10^{-15}$ N.

Acceleration: $a = \frac{F}{m} = \frac{3.2 \times 10^{-15}}{9.11 \times 10^{-31}} = 3.5 \times 10^{15}$ m per second squared, which is of order $10^{15}$ .

Direction: the electron is negative, so the force (and acceleration) is opposite to the field direction.

Markers reward $F = EQ$ , division by the electron mass, and stating the acceleration is opposite to the field.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Physics (9PH0) specification — Pearson Edexcel (2015)