Electrostatics¶
Syllabus mapping
PY-11-03 — Electricity and Magnetism. Covers: charging by friction, conduction and induction; the electric field around a point charge; lightning as static discharge; the inverse square law for field strength and Coulomb's law; the field between parallel plates; electric potential energy and work done moving a charge.
Wording above is paraphrased for teaching use — cross-check against the official syllabus PDF (see Resources) before treating any line as verbatim NESA text.
You need to know: the three charging methods, how to calculate electric field strength and force using Coulomb's law, the difference between a point-charge field and a parallel-plate field, and how to relate work, potential difference and energy for a moving charge.
Assumed knowledge check¶
Before this page makes sense, you should be comfortable with:
- Quantities of Motion — the component method, needed once more than one charge is involved
Core¶
Charging methods¶
| Method | How it works |
|---|---|
| Friction | Rubbing two materials transfers electrons from one to the other (e.g. a balloon rubbed on hair) |
| Conduction | Direct contact between a charged and uncharged conductor transfers charge — both end up with the same sign |
| Induction | A charged object brought near (but not touching) a conductor redistributes its charge, inducing an opposite charge on the near side, without any charge transfer |
Electric field around a point charge¶
🖼️ Diagram needed: field line diagrams for an isolated positive charge, an isolated negative charge, and a dipole. TODO — placeholder until sourced/drawn.
Field lines point away from a positive charge, into a negative charge. The force on any charge \(q\) placed in a field \(\vec{E}\):
Coulomb's law¶
Same inverse square shape you've already seen for gravity (\(F=\dfrac{GMm}{r^2}\), in Motion Relationships' data sheet formulae) and light intensity (in the Advanced box on Light and Sound) — same reasoning too: the influence spreads from a point source over an expanding sphere.
Worked example
Find the force between two charges of \(+2\times10^{-6}\ \text{C}\) and \(-3\times10^{-6}\ \text{C}\), separated by \(0.5\ \text{m}\). Use \(\varepsilon_0 = 8.854\times10^{-12}\ \text{A}^2\text{s}^4\text{kg}^{-1}\text{m}^{-3}\) (from your data sheet).
Opposite charges, so this force is attractive.
Lightning as static discharge¶
Inside a storm cloud, collisions between ice particles separate charge (typically leaving the lower part of the cloud negatively charged). This builds an intense electric field between cloud and ground. When that field exceeds the point at which the surrounding air breaks down and becomes conductive, a massive, sudden discharge — lightning — equalises the charge difference. Genuinely just an electrostatics problem at enormous scale.
The field between parallel plates¶
Unlike a point charge's field (which weakens with distance via the inverse square law), the field between two oppositely charged parallel plates is uniform — same strength and direction everywhere between the plates:
where \(V\) is the potential difference between the plates and \(d\) is their separation. Work done moving a charge \(q\) across this uniform field, a distance \(d\):
Electric potential energy and work¶
The second equation is the electrical version of the work–energy theorem — work done on a charge by an electric field converts directly to kinetic energy, exactly parallel to the mechanical version in Momentum and Energy.
Worked example
An electron (\(q = -1.602\times10^{-19}\ \text{C}\), from your data sheet) is accelerated from rest through a potential difference of \(500\ \text{V}\). Find the kinetic energy gained.
(Using the magnitude of charge here, since we only want the energy gained, not a signed direction.)
Advanced
Multiple point charges. When more than one charge contributes to the field or force at a point, calculate each contribution separately using Coulomb's law, then combine them as vectors using the component method — exactly the same vector addition skill from Quantities of Motion, just applied to electric forces/fields instead of displacement or velocity.
Combining the field and parallel-plate equations — e.g. finding the force on a charge between charged plates given only voltage and separation, without ever computing the field from a point-charge formula.
Extension
Beyond the Physics 11–12 syllabus — won't appear in the HSC, included for interest / depth study inspiration.
📎 Depth study idea
Coulomb's law (\(F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}\)) and Newton's law of gravitation (\(F=\dfrac{GMm}{r^2}\)) have exactly the same mathematical shape — both inverse square laws from a point source, as flagged already in the Extension box on Light and Sound. A genuine depth study: compare the two laws side by side (both forces, both fields, both potential energies) and work out exactly where the analogy holds and where it breaks down — the biggest difference being that gravity is only ever attractive, while electric charge comes in two signs and can attract or repel. Since both mass and charge produce inverse-square fields, this is a rich source of "solve the same style of problem in two different contexts" practice too.
Video/visual resources¶
- 🖥️ PhET Simulation — Charges and Fields — TODO: confirm current PhET link
- 🎥 Khan Academy — TODO: source a Coulomb's law / electric field explainer
- 🎥 Physics High — TODO: check for a NSW-syllabus-aligned electrostatics video
Check yourself¶
-
Find the electric field strength at a distance of \(0.2\ \text{m}\) from a point charge of \(+4\times10^{-6}\ \text{C}\).
Answer
\(E = \dfrac{1}{4\pi(8.854\times10^{-12})}\times\dfrac{4\times10^{-6}}{0.2^2} \approx 8.99\times10^5\ \text{N C}^{-1}\)
-
Two charged parallel plates are \(0.05\ \text{m}\) apart with a potential difference of \(200\ \text{V}\) across them. Find the field strength between them, and the work done moving a \(1\times10^{-9}\ \text{C}\) charge from one plate to the other.
Answer
\(E = \dfrac{V}{d} = \dfrac{200}{0.05} = 4000\ \text{V m}^{-1}\)
\(W = qEd = (1\times10^{-9})(4000)(0.05) = 2\times10^{-7}\ \text{J}\) — or equivalently, more directly, \(W=qV = (1\times10^{-9})(200) = 2\times10^{-7}\ \text{J}\).
-
State which charging method is at work: (a) a plastic comb picks up small pieces of paper after being run through hair, (b) a neutral metal sphere becomes charged after briefly touching a charged rod, (c) a neutral metal sphere develops a charge on its near side when a charged rod is brought close without touching.
Answer
(a) Friction — charge transferred by rubbing.
(b) Conduction — charge transferred by direct contact.
(c) Induction — charge redistributed without contact.