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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 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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

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=1.60×10−19 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. Between two large parallel plates the lines are straight and evenly spaced, giving a uniform field, which is the link to the parallel-plate dot point.

How SACE assesses this

SACE Stage 2 questions here are nearly always direct substitutions into Coulomb's law, F=kq1q2r2F = \dfrac{kq_1 q_2}{r^2}, to find the force between two point charges, often using atomic-scale data (an electron and a nucleus). The two recurring traps are squaring the very small separation correctly and handling the powers of ten in scientific notation. Several "show that" parts give you the target force, so quote the formula, substitute with units, evaluate the numerator and the squared denominator as separate steps, and confirm the printed value. When asked about direction, use the signs to decide attraction or repulsion but substitute magnitudes for the size of the force. A field part may follow, where E=FqE = \dfrac{F}{q} or E=kQr2E = \dfrac{kQ}{r^2} gives the field, and F=qEF = qE then recovers the force on any other charge placed there.

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.

SACE 20232 marksTwo charged particles q1q_1 and q2q_2 have their centres 0.500 m0.500\ \text{m} apart. The magnitude of q1q_1 is 7.00×10−6 C7.00\times10^{-6}\ \text{C} and that of q2q_2 is 3.00×10−6 C3.00\times10^{-6}\ \text{C}. Calculate the magnitude of the electric force on q1q_1 due to q2q_2.
Show worked answer →

Apply Coulomb's law, F=kq1q2r2F = \dfrac{k q_1 q_2}{r^2}, with k=8.99×109 N m2C−2k = 8.99\times10^9\ \text{N m}^2\text{C}^{-2}.

F=(8.99×109)(7.00×10−6)(3.00×10−6)(0.500)2=0.18880.250=0.755 N.F = \dfrac{(8.99\times10^9)(7.00\times10^{-6})(3.00\times10^{-6})}{(0.500)^2} = \dfrac{0.1888}{0.250} = 0.755\ \text{N}.

1 mark for correct substitution, 1 mark for the answer of about 0.755 N0.755\ \text{N}. By Newton's third law the force on q2q_2 due to q1q_1 has the same magnitude.

SACE 20251 marksElectron X is located 2.64×10−11 m2.64\times10^{-11}\ \text{m} from a nucleus of charge magnitude 9.60×10−19 C9.60\times10^{-19}\ \text{C}. Show that the magnitude of the electric force between the electron and the nucleus is 1.98×10−6 N1.98\times10^{-6}\ \text{N}.
Show worked answer →

Apply Coulomb's law with k=8.99×109 N m2C−2k = 8.99\times10^9\ \text{N m}^2\text{C}^{-2} and the electron charge 1.60×10−19 C1.60\times10^{-19}\ \text{C}.

Numerator: (8.99×109)(9.60×10−19)(1.60×10−19)=1.381×10−27(8.99\times10^9)(9.60\times10^{-19})(1.60\times10^{-19}) = 1.381\times10^{-27}.
Denominator: (2.64×10−11)2=6.97×10−22(2.64\times10^{-11})^2 = 6.97\times10^{-22}.

F=1.381×10−276.97×10−22=1.98×10−6 N.F = \dfrac{1.381\times10^{-27}}{6.97\times10^{-22}} = 1.98\times10^{-6}\ \text{N}.

1 mark for correct substitution and the answer of 1.98×10−6 N1.98\times10^{-6}\ \text{N}. The force is attractive because the charges are opposite.

Openly AI-built from the official SACE Board syllabus, past papers and marking guides. Not individually reviewed; always check SACE Boardand your school's materials for current rules.

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