Physics › Electric fields › Coulomb's law and electric field strength

Coulomb's law and electric field strength

Charge has its own inverse-square law and its own field strength, both close analogues of the gravitational versions. It also has something gravity does not: the almost perfectly uniform field between two parallel plates, in which a moving charge follows the same parabola as a projectile.

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.

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}}ON THE AQA DATA SHEET

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.

Coulomb's law: two charges push or pull on each other with exactly equal force, however unequal the charges+big Q₁+small Q₂F = Q₁Q₂/4πε₀r², on each of thema charged sphere acts from its centre
FIG. 1The Coulomb force acts equally on both charges, however unequal they are, and a charged sphere counts from its centre.

Unlike gravity, this force can point either way. Like charges repel and unlike charges attract. Either way it acts equally on both partners, and it 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 that the size makes sense. It looks tiny in newtons, but acting on a proton's mass of order 10−27 kg it produces an acceleration of order 1019 m s−2. On the atomic scale that is an enormous force.

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}ON THE AQA DATA SHEET

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, an excellent description away from the edges of the plates where the field bows outward, and its strength is

E=VdE = \frac{V}{d}ON THE AQA DATA SHEET
Between parallel plates the field is uniform: parallel, equally spaced lines, with strength E = V/d+V0 VdE = V/d, the same everywhere between the platesfield direction: from + towards 0, the push on a positive charge
FIG. 2Between parallel plates: parallel, equally spaced field lines, and E = V/d everywhere in the gap.

The derivation is asked for, and it balances two expressions for one quantity of work. Carry a charge Q from one plate to the other and the field does work force times distance, W = Fd; the same work written through potential difference is W = QΔV, with ΔV here the magnitude of the pd between the plates so that both sides count the same unsigned work. Set Fd = QΔV, divide both sides by Qd, and F/Q = ΔV/d follows, with E on the left. The unit V m−1 that comes with it 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. A positive charge is always pushed along the field lines.

A charge crossing the field

A charge entering the field at right angles follows a parabola: constant force in one direction, exactly like a projectile+e⁻parabola:like a projectileit bends towards the positive plate, projectile-style
FIG. 3A charge entering a uniform field at right angles follows a parabola: the projectile problem with the field playing gravity.

Fire a charged particle into the uniform field at right angles to the field lines and it keeps its entry speed along one axis while a constant force accelerates it along the other. The path is a parabola, identical to projectile motion with the electric force cast as the weight. This picture 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 map the force at each point, and only a charge released from rest onto a straight field line happens to move 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}}ON THE AQA DATA SHEET
The radial field of a positive point charge: field lines point away, weakening as the inverse square+for a positive charge, the field points awayE = Q/4πε₀r²: the inverse square, again
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 the special local case; the radial field is the general picture, and it will pair with the gravitational one in the closing 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. Work it through.

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 swing a light hanging sphere visibly, which is exactly the tabletop electrostatics you have seen.

ASSESSMENT FOCUS

  • Quote Coulomb's law with its conditions attached. Point charges in a vacuum, with air counting as a vacuum and a sphere acting from its centre. Those conditions carry marks of their own.
  • r is the separation of the centres, and doubling it quarters the force. Set the squared ratio up before you reach for numbers.
  • The E = V/d derivation earns its marks in two lines. Write Fd = QΔV, then divide through by Qd. Practise writing it out; recognising it is a different skill from reproducing it.
  • The trajectory answer has fixed vocabulary. Constant velocity parallel to the plates, constant acceleration perpendicular to them, and so a parabola. Name both parts.
  • Direction errors are expensive. An electron deflects towards the positive plate, opposite to the field-line arrows, so write down the sign of the charge before you commit to any direction.
  • Every equation in this lesson is printed on the AQA data sheet, so no recall is being tested. What earns marks is choosing the right 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

Uniform field first, then force per charge read backwards.

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.

Tiny in newtons and enormous per kilogram. On an electron's mass this force produces an acceleration of around 1015 m s−2.

Coulomb's law is the inverse square with charge in the seats.

Between plates E = V/d, and right-angle entry draws 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.