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SAPhysicsSyllabus dot point

How does the electric force between charges depend on their size and separation, and what is meant by an electric field?

Apply Coulomb's law to the force between point charges and describe the electric field around a charge.

Coulomb's inverse-square law for the force between point charges, the meaning of electric field strength, and how field lines represent the field, with worked examples.

Generated by Claude Opus 4.78 min answer

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  1. What this dot point is asking
  2. Coulomb's law
  3. The electric field
  4. Field lines

What this dot point is asking

You need to apply Coulomb's law to find the force between point charges and describe the electric field surrounding a charge.

Coulomb's law

The structure is identical to Newton's law of gravitation - an inverse-square law proportional to the product of the "source quantities" (here charge instead of mass). Key differences:

  • Charge comes in two signs, so the electric force can attract or repel: like charges repel, unlike charges attract. Gravity is always attractive.
  • The electric force is vastly stronger than gravity for everyday particles.

The elementary charge is e=1.60×1019 Ce = 1.60\times10^{-19} \text{ C} (the charge of a proton; an electron carries e-e).

The electric field

Rather than describing the force between every pair of charges, we say each charge creates an electric field in the space around it. A second charge then feels a force from that field.

Once you know the field EE at a point, the force on any charge qq placed there is simply F=qEF = qE.

Field lines

We draw electric fields with field lines:

  • Lines point away from positive charges and toward negative charges.
  • The field is stronger where lines are closer together.
  • Lines never cross (the field has one direction at each point).

A single positive charge has radial lines pointing outward; a pair of opposite charges (a dipole) has lines curving from positive to negative.

Exam-style practice questions

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

2023 SACE Stage 22 marksTwo charged particles q1 and q2 are positioned so that the centre of q1 is 0.500 m from the centre of q2. The magnitude of q1 is 7.00 x 10^-6 C and the magnitude of q2 is 3.00 x 10^-6 C. Calculate the magnitude of the electric force acting on q1 due to q2.
Show worked answer →

Apply Coulomb's law: F = k q1 q2 / r squared, with the Coulomb constant k = 8.99 x 10^9 N m2 C-2.

F = (8.99 x 10^9)(7.00 x 10^-6)(3.00 x 10^-6) / (0.500) squared.

Numerator = (8.99 x 10^9)(7.00 x 10^-6)(3.00 x 10^-6) = 0.1888.
Denominator = (0.500) squared = 0.250.

F = 0.1888 / 0.250 = 0.755 N.

1 mark for correct substitution into Coulomb's law, 1 mark for the answer of about 0.755 N. By Newton's third law the force on q2 due to q1 has the same magnitude.

2025 SACE Stage 21 marksElectron X is located 2.64 x 10^-11 m from the positively charged nucleus of an atom. The nucleus has a charge of magnitude 9.60 x 10^-19 C. Show that the magnitude of the electric force between the electron and the nucleus is 1.98 x 10^-6 N.
Show worked answer →

Apply Coulomb's law: F = k q1 q2 / r squared, with k = 8.99 x 10^9 N m2 C-2 and the electron charge 1.60 x 10^-19 C.

F = (8.99 x 10^9)(9.60 x 10^-19)(1.60 x 10^-19) / (2.64 x 10^-11) squared.

Numerator = (8.99 x 10^9)(9.60 x 10^-19)(1.60 x 10^-19) = 1.381 x 10^-27.
Denominator = (2.64 x 10^-11) squared = 6.97 x 10^-22.

F = 1.381 x 10^-27 / 6.97 x 10^-22 = 1.98 x 10^-6 N.

1 mark for correct substitution and the answer of 1.98 x 10^-6 N. The force is attractive because the charges are opposite.