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Coulomb's law and electric field strength

Coulomb's law gives the force between two point charges as Q₁Q₂/4πε₀r², repulsive for two like charges and attractive for two unlike ones. Electric field strength is the force per unit positive charge, E = F/Q, uniform at E = V/d between plates and radial around a point charge. A charge entering that uniform field at right angles follows a parabola.

Coulomb's law and electric field strength

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Year 13AQA 3.7.3.1, 3.7.3.2CIE 18.1.1, 18.1.2, 18.1.3, 18.2.1, 18.2.2, 18.3.1, 18.3.2, 18.4.1OCR A 6.2.1(a), 6.2.1(b), 6.2.1(c), 6.2.1(d), 6.2.2(a), 6.2.2(b), 6.2.3(a), 6.2.3(c)EDEXCEL 109, 110, 111, 113

Builds on The field concept and Projectile motion.

IN THIS TOPIC

  • Use Coulomb's law for point charges, with charged spheres acting from their centres.
  • Use E = F/Q, and the radial field of a point charge.
  • Derive the uniform field E = V/d from Fd = QΔV, and use it.
  • Predict the parabolic path of a charge entering a uniform field at right angles.

COMMON MISCONCEPTION

Field lines are the paths charges follow.

Field lines show the direction of the force on a positive charge, not trajectories. A charge entering a uniform field at right angles follows a parabola, crossing the field lines.

Coulomb's law

The force between two point charges in a vacuum has the same shape as Newton's law of gravitation.

F=14πε0Q1Q2r2F = \frac{1}{4\piε_{0}}\,\frac{Q_{1}Q_{2}}{r^{2}}IN THE AQA DATA BOOKLET

where ε0, the permittivity of free space, is 8.85 × 10−12 F m−1, printed in the data booklet. Two working simplifications come with the law. Air can be treated as a vacuum when you calculate forces between charges, and a charged sphere acts as if its whole charge sat at its centre, the exact twin of the rule for spherical masses.

Two positive charges of different sizes with equal-length force arrows pushing them apart: the Coulomb force acts equally on both, and a charged sphere counts as a point charge at its centre.
FIG. 1The Coulomb force acts equally on both charges, whatever their magnitudes, and a charged sphere acts as if its whole charge were at its centre.

Unlike gravity, this force acts in either direction: like charges repel and unlike charges attract. In both cases the forces on the two charges are equal in magnitude and opposite in direction, and each follows the inverse square, so doubling the separation quarters the force.

WORKED EXAMPLE

A force between two protons

Two protons sit 1.0 × 10−10 m apart, about one atomic spacing. Find the electrostatic force between them.

Both are point charges, so Coulomb's law applies directly, and the booklet supplies e = 1.60 × 10−19 C and ε0 = 8.85 × 10−12 F m−1; one reciprocal turns the latter into 1/(4πε0) = 8.99 × 109 m F−1.

F = 8.99 × 109 × (1.60 × 10−19)2/(1.0 × 10−10)2 = 8.99 × 109 × 2.56 × 10−38/(1.0 × 10−20) = 2.3 × 10−8 N, repulsive, since both charges are positive.

Check the magnitude. The force is small in newtons, but acting on a proton of mass of order 10−27 kg it produces an acceleration of order 1019 m s−2, which is very large on the atomic scale.

Field strength, and the uniform field

The electric field strength at a point is the force per unit charge on a small positive test charge placed there.

E=FQE = \frac{F}{Q}IN THE AQA DATA BOOKLET

measured in N C−1. Between two parallel plates a distance d apart with a potential difference V across them, the field is modelled as uniform, which is accurate except near the edges of the plates, where the field bows outward. Its strength is

E=VdE = \frac{V}{d}IN THE AQA DATA BOOKLET
Two horizontal plates, the top at plus V and the bottom at zero, with five parallel equally spaced field arrows running straight down between them and the separation marked d.
FIG. 2Between parallel plates: parallel, equally spaced field lines, and E = V/d everywhere in the gap.

The derivation is required, and it equates two expressions for the same work. Moving a charge Q from one plate to the other, the work done by the field is force × distance, W = Fd; the same work written in terms of potential difference is W = QΔV, where ΔV is the magnitude of the pd between the plates so that both expressions are unsigned. Setting Fd = QΔV and dividing both sides by Qd gives F/Q = ΔV/d, and F/Q is E. The unit V m−1 is N C−1 rearranged, not a different unit.

GUIDED PRACTICE

Holding a drop still

A small drop carries a charge of +3.2 × 10−19 C, two electrons' worth, and weighs 6.4 × 10−15 N. A vertical uniform field holds it stationary. The electric force must balance the weight: start from E = F/Q, find the field strength, and decide which way the field must point.

Show the working

E = F/Q = 6.4 × 10−15/(3.2 × 10−19) = 2.0 × 104 N C−1, or the same number in V m−1.

The force on the drop must point upward, and the charge is positive, so the field itself points upward too. The force on a positive charge always acts in the direction of the field.

A charge crossing the field

An electron enters horizontally between charged plates and its path bends steadily upward towards the positive plate in a parabola, the dashed line showing where it would have gone: projectile motion with the field playing gravity.
FIG. 3A charge entering a uniform field at right angles follows a parabola: constant velocity parallel to the plates, constant acceleration perpendicular to them.

A charged particle entering the uniform field at right angles to the field lines keeps its entry velocity along one axis while a constant force accelerates it along the other, so the path is a parabola, with the same structure as projectile motion and the electric force in place of the weight. This also settles the misconception at the top of the page: the particle enters moving across the field lines and never travels along them. Field lines show the direction of the force at each point, and only a charge released from rest on a straight field line moves along one.

The radial field

Around a point charge, or outside a charged sphere, the field is radial, and its magnitude is

E=14πε0Qr2E = \frac{1}{4\piε_{0}}\,\frac{Q}{r^{2}}IN THE AQA DATA BOOKLET
Eight cyan arrows radiate outward from a positive charge, the mirror image of the gravitational picture, with the radial field strength falling as the inverse square of distance.
FIG. 4The radial field of a positive charge points away from it and weakens as the inverse square.

with the lines pointing away from a positive charge and towards a negative one, because the direction convention follows the force on a positive test charge. The uniform field is a special case; the radial field is the general one, and it is compared with the gravitational radial field in the last lesson of this unit.

INDEPENDENT PRACTICE

Two charged spheres

Two small spheres each carry +30 nC and their centres are 5.0 cm apart. Find the force between them, and state the fact that lets you use Coulomb's law for spheres at all.

Show the working

A charged sphere behaves as if its whole charge sat at its centre, so the two count as point charges 0.050 m apart.

F = 8.99 × 109 × (3.0 × 10−8)2/(0.050)2 = 8.99 × 109 × 9.0 × 10−16/(2.5 × 10−3) = 3.2 × 10−3 N, repulsive.

A few millinewtons is enough to deflect a light suspended sphere visibly, which is the scale of bench-top electrostatics experiments.

ASSESSMENT FOCUS

  • Quote Coulomb's law with its conditions: point charges in a vacuum, with air treated as a vacuum and a sphere acting from its centre. The conditions carry credit in their own right.
  • r is the separation of the centres, and doubling it quarters the force. Set up the squared ratio before substituting numbers.
  • The E = V/d derivation takes two lines: write Fd = QΔV, then divide through by Qd. Practise writing it out, since reproducing it is a different skill from recognising it.
  • The trajectory answer uses fixed terms: constant velocity parallel to the plates, constant acceleration perpendicular to them, and therefore a parabola. Name both components.
  • State the sign of the charge before giving any direction. An electron deflects towards the positive plate, opposite to the field-line arrows.
  • AQA prints all of these equations on the data sheet, so recall is not tested. What is tested is choosing the correct geometry, since E = V/d and the radial E = Q/4πε₀r2 describe different arrangements and are not interchangeable.

CHECK YOURSELF

Two plates 4.0 cm apart carry a potential difference of 500 V. Find the field strength between them and the force on an electron in the gap.

Show a hint

Use E = V/d first, then F = EQ.

Show the answer

E=V/dE = V/d = 500 / 0.040 = 1.25 × 104 V m−1.

F=EQF = EQ = 1.25 × 104 × 1.60 × 10−19 = 2.0 × 10−15 N, directed towards the positive plate because the electron's charge is negative.

The force is small in newtons but large per unit mass: on an electron it produces an acceleration of around 1015 m s−2.

Coulomb's law has the same inverse-square dependence as Newton's law of gravitation, with charge replacing mass.

Between parallel plates E = V/d, and a charge entering at right angles follows a parabola.

WORKBOOK

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  • Use Coulomb's law for point charges, with charged spheres acting from their centres.
  • Use E = F/Q, and the radial field of a point charge.
  • Derive the uniform field E = V/d from Fd = QΔV, and use it.
  • Predict the parabolic path of a charge entering a uniform field at right angles.

Open the full revision checklist to track your progress across the whole unit.