InkPhysics

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

Coulomb's law and electric field strength

Charge gets its own inverse-square law, its own field strength, and one new trick gravity never showed you: a perfectly uniform field between two plates, where a moving charge replays projectile motion with the field cast as gravity.

Year 13AQA 3.7.3.1, 3.7.3.2

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, the uniform field E = V/d with its derivation, and the radial field of a point charge.
  • Predict the parabolic path of a charge entering a uniform field at right angles.

WHAT YOU PROBABLY THINK

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 = 14πε0 Q1Q2r2ON YOUR 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 calculating 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, the force here can point either way: like charges repel, unlike charges attract. Either way it is mutual and equal on both, and it carries the inverse square, so doubling the separation quarters the 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 = FQON YOUR 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 uniform, and its strength is

E = VdON YOUR 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 + toward 0, the way a positive charge is pushed
FIG. 2Between parallel plates: parallel, equally spaced field lines, and E = V/d everywhere in the gap.

The spec asks for the derivation, and it is one balancing of two expressions for the same work. Carrying a charge Q from one plate to the other, the field does work force times distance, W = Fd; the same work written through potential difference is W = QΔV. Setting Fd = QΔV and dividing both sides by Qd gives F/Q = ΔV/d, and the left side is E. The unit V m−1 that falls out is exactly N C−1 in different clothes.

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+parabola:like a projectileit bends toward 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 recipe for a parabola, identical to projectile motion with the electric force cast as the weight. This picture also settles the lie above. 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 in a uniform field 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 = 14πε0 Qr2ON YOUR 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 toward 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.

THE EXAM BIT

  • Quote Coulomb's law with its conditions: point charges, vacuum, and note that air counts as a vacuum and a sphere acts from its centre. The conditions carry marks.
  • r is the separation of the centres, and doubling it quarters the force: set up the squared ratio before reaching for numbers.
  • The E = V/d derivation earns its marks in two lines: Fd = QΔV, then divide by Qd. Practise writing it rather than merely recognising it.
  • The trajectory answer has fixed vocabulary: constant velocity parallel to the plates, constant acceleration perpendicular to them, hence a parabola. Name both parts.
  • Direction errors cost dearly: an electron deflects toward the positive plate, opposite to the field-line arrows. State the sign of the charge before stating the direction.

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/d = 500 / 0.040 = 1.25 × 104 V m−1.

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

Tiny in newtons, enormous per kilogram: on an electron's mass this force produces an acceleration around 1015 m s−2.

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

Uniform field: E = V/d, and right-angle entry means a parabola.

No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.