Electrostatics

Review: Charge and the Elementary Charge

Electric charge is a fundamental property of matter. Protons carry a positive charge and electrons carry a negative charge. The smallest unit of charge that exists freely is the elementary charge:

e = 1.6 × 10⁻¹⁹ C

Conservation of charge: the total charge in an isolated system remains constant. Charge can be transferred but never created or destroyed. Conductors allow charge to flow freely; insulators resist the flow of charge.

Coulomb's Law

The electrostatic force between two point charges Q₁ and Q₂ separated by a distance r is given by Coulomb's Law:

F = kQ₁Q₂ / r²
k = 9 × 10⁹ N·m²·C⁻²

Symbol	Quantity	Unit
F	Electrostatic force	Newton (N)
k	Coulomb's constant	N·m²·C⁻²
Q₁, Q₂	Charges	Coulomb (C)
r	Distance between charges	Metre (m)

Sign convention: If F is positive, the force is repulsive (like charges). If F is negative, the force is attractive (opposite charges). When solving problems, determine the direction separately using the rule: opposite charges attract, like charges repel.

Superposition of Forces

When more than two charges are present, the net force on any one charge is the vector sum of the individual forces exerted by every other charge. This is the principle of superposition.

Worked example — 3 charges in a line:
Q₁ = +4 µC at x = 0; Q₂ = −3 µC at x = 0.3 m; Q₃ = +2 µC at x = 0.5 m.
Find the net force on Q₂.

F₁₂ = k|Q₁||Q₂|/r₁₂² = (9×10⁹)(4×10⁻⁶)(3×10⁻⁶)/(0.3)² = 1.2 N (attractive, so directed toward Q₁, i.e., to the left, negative x)
F₃₂ = k|Q₃||Q₂|/r₃₂² = (9×10⁹)(2×10⁻⁶)(3×10⁻⁶)/(0.2)² = 1.35 N (attractive, so directed toward Q₃, i.e., to the right, positive x)
F_net = F₃₂ − F₁₂ = 1.35 − 1.2 = 0.15 N to the right

Electric Field (E)

The electric field at a point is the electrostatic force per unit positive test charge placed at that point:

E = F / q (unit: N·C⁻¹)
E = kQ / r² (field due to a point charge Q)

The electric field is a vector quantity — it has both magnitude and direction.
Direction: away from a positive charge, toward a negative charge.
Fields from multiple charges add by vector superposition.

Uniform Field Between Parallel Plates

When a potential difference V is applied across two parallel plates separated by distance d, the field between the plates is uniform:

E = V / d

This means the field strength is the same at every point between the plates (ignoring edge effects).

Electric Field Lines

Start on positive charges; end on negative charges
Never cross each other
Density of lines is proportional to field strength
Always perpendicular to the surface of a conductor
The tangent to a field line at any point gives the direction of E at that point

Potential Difference and Work

Work is done moving a charge through a potential difference:

W = qV V = W / q
W = qEd (uniform field, distance d along field)

IEB Extension — Electric Potential & Advanced Concepts

Electric potential V at a point due to a point charge Q:

V = kQ / r

Equipotential surfaces are surfaces on which every point is at the same electric potential. They are always perpendicular to electric field lines. No work is done moving a charge along an equipotential surface.

Gauss's Law (qualitative): The total electric flux through any closed surface equals the enclosed charge divided by ε₀. Flux Φ_E = E·A for a uniform field perpendicular to area A. This powerful law allows E to be found for symmetric charge distributions without integrating Coulomb's law directly.

Capacitance

A capacitor stores charge. Capacitance C is the charge stored per volt of potential difference:

C = Q / V (unit: Farad, F)
Energy stored: U = ½CV²
Parallel plate: C = ε₀A / d

Increasing plate area A → increases C
Increasing separation d → decreases C
Inserting a dielectric → increases C (dielectric reduces effective field)

Key fact: ε₀ = 8.85 × 10⁻¹² F·m⁻¹ (permittivity of free space)

Answer all questions. Show full working for calculations. Include units in every answer.

Question 1

Three charges are placed in a line: Q₁ = +6 µC at x = 0, Q₂ = −4 µC at x = 0.4 m, Q₃ = +5 µC at x = 0.7 m. Calculate the magnitude and direction of the net electrostatic force on Q₂.

Question 2

Draw and label the electric field line pattern for (a) an isolated positive charge, (b) two equal and opposite charges (electric dipole), and (c) two parallel plates with opposite charges. Describe two properties of electric field lines.

Question 3

Two parallel plates are separated by 0.02 m and connected to a 120 V supply. (a) Calculate the electric field strength between the plates. (b) A proton (charge = 1.6 × 10⁻¹⁹ C, mass = 1.67 × 10⁻²⁷ kg) is released from rest near the positive plate. Calculate the force on the proton and its acceleration.

Question 4

How much work is done moving an electron (charge = −1.6 × 10⁻¹⁹ C) through a potential difference of 500 V? State whether energy is gained or lost by the electron, and explain why.

Question 5

A parallel plate capacitor has plates of area 0.04 m² separated by 0.005 m. (a) Calculate the capacitance using C = ε₀A/d (ε₀ = 8.85 × 10⁻¹² F·m⁻¹). (b) If connected to a 12 V battery, calculate the charge stored and the energy stored in the capacitor.

Question 6

A small test charge q = 2 × 10⁻⁶ C is placed at increasing distances from a fixed point charge Q, and the electrostatic force it experiences is measured:

Distance, r (m)	Force, F (N)
0.1	3.6
0.2	0.9
0.3	0.4
0.4	0.225

(a) Calculate the value of F × r² for each row of the table. What do you notice about these values, and what does this confirm about the relationship between F and r? (2 marks)
(b) Using k = 9 × 10⁹ N·m²·C⁻², your answer to (a), and Coulomb's Law F = kQq/r², calculate the magnitude of the fixed charge Q. (3 marks)
(c) Predict the force this test charge would experience at r = 0.5 m. (1 mark)

Review: Charge and the Elementary Charge

Coulomb's Law

Superposition of Forces

Electric Field (E)

Uniform Field Between Parallel Plates

Electric Field Lines

Potential Difference and Work

Capacitance

Coulomb Calculator — Superposition of 3 Charges