21  Electrostatics

21.1 Syllabus inquiry question

  • How do electric charges create forces and fields?
Feynman Insight

From The Feynman Lectures on Physics, Vol II, Chapter 1:

We explain electric forces by introducing the field. The field is not a trick. It is the physical agent that stores energy and transmits changes from one charge to another.

21.2 Learning Objectives

  • Describe electric charge, quantisation, and conservation.
  • Compare charging by friction, conduction, and induction.
  • Apply Coulomb’s law to calculate forces between charges.
  • Define electric field strength and interpret field lines.
  • Relate electric potential, potential difference, and electrical energy.

21.3 Content

21.3.1 Electric Charge and Conservation

Electric charge is a fundamental property of matter that comes in two types: positive and negative.

Fundamental Principles
  • Like charges repel; unlike charges attract
  • Charge is conserved—it cannot be created or destroyed, only transferred
  • Charge is quantised in units of the elementary charge: \[e = 1.60 \times 10^{-19}\ \text{C}\]

All observable charges are integer multiples of \(e\). Protons carry \(+e\) and electrons carry \(-e\).

21.3.2 Charging Methods

Friction: When two surfaces are rubbed together, electrons transfer from one material to the other. The material that gains electrons becomes negatively charged.

Conduction: A charged object touches a neutral object, sharing charge directly through contact.

Induction: A nearby charge redistributes charges within an object without contact. Earthing (grounding) the object can leave it with a net charge opposite to the inducing charge.

21.3.3 Interactive: Charge Distribution

Visualising how charges redistribute by induction:

21.3.4 Coulomb’s Law

The electrostatic force between two point charges is:

\[F = k\frac{|q_1 q_2|}{r^2}\]

where: - \(F\) = force (N) - \(k = 8.99 \times 10^{9}\ \text{N m}^2/\text{C}^2\) (Coulomb’s constant) - \(q_1, q_2\) = charges (C) - \(r\) = separation distance (m)

Direction of Force

The formula gives the magnitude of the force. Direction is along the line joining the charges: attractive for opposite signs, repulsive for like signs.

21.3.5 Interactive: Two-Charge System

Explore the electric field from two point charges:

Key observation: Field lines originate from positive charges and terminate on negative charges. The density of lines indicates field strength.

21.3.6 Electric Field and Field Lines

Electric field strength is the force per unit positive test charge:

\[E = \frac{F}{q}\]

For a point charge \(Q\):

\[E = k\frac{Q}{r^2}\]

Field line conventions: - Lines start on positive charges and end on negative charges - Line density indicates field strength - Lines never cross - Field direction is tangent to the line at any point

21.3.7 Interactive: Uniform Electric Field

A uniform field exists between parallel charged plates:

Between parallel plates, the field is uniform with strength: \[E = \frac{V}{d}\]

21.3.8 Electric Potential and Voltage

Electric potential difference (voltage) is work done per unit charge:

\[V = \frac{W}{q}\]

Electrical potential energy is related to voltage by:

\[U = qV\]

Key Relationships
  • Voltage is energy per charge: \(V = W/q\)
  • In a uniform field: \(E = V/d\)
  • Equipotential lines are surfaces of constant \(V\), always perpendicular to field lines

21.4 Worked Examples

21.4.1 Example 1: Coulomb force

Two charges \(q_1 = +2.0\ \mu\text{C}\) and \(q_2 = -3.0\ \mu\text{C}\) are separated by \(0.20\ \text{m}\). Find the force.

Solution:

  1. Use Coulomb’s law: \(F = k|q_1 q_2|/r^2\)

  2. Convert units: \(q_1 = 2.0 \times 10^{-6}\) C, \(q_2 = 3.0 \times 10^{-6}\) C

  3. Calculate: \(F = \frac{8.99 \times 10^9 \times 2.0 \times 10^{-6} \times 3.0 \times 10^{-6}}{(0.20)^2}\)

  4. \(F = \frac{8.99 \times 10^9 \times 6.0 \times 10^{-12}}{0.040} = 1.35\ \text{N}\)

  5. Force is attractive (opposite charges)

21.4.2 Example 2: Electric field strength

A point charge of \(+5.0\ \text{nC}\) is \(0.30\ \text{m}\) away. Find \(E\).

Solution:

  1. Use \(E = kQ/r^2\)

  2. Convert: \(Q = 5.0 \times 10^{-9}\) C

  3. \(E = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-9}}{(0.30)^2} = \frac{45.0}{0.090}\)

  4. \(E \approx 5.0 \times 10^{2}\ \text{N/C}\)

21.4.3 Example 3: Electrical energy

A \(2.0\ \mu\text{C}\) charge moves through a potential difference of \(120\ \text{V}\). Find the energy change.

Solution:

  1. Use \(U = qV\)

  2. \(U = 2.0 \times 10^{-6} \times 120\)

  3. \(U = 2.4 \times 10^{-4}\ \text{J}\) (or 0.24 mJ)

21.4.4 Example 4: Uniform field between plates

Parallel plates are separated by \(5.0\ \text{mm}\) with \(200\ \text{V}\) across them. Find the field strength.

Solution:

  1. Use \(E = V/d\)

  2. Convert: \(d = 5.0 \times 10^{-3}\) m

  3. \(E = \frac{200}{5.0 \times 10^{-3}} = 4.0 \times 10^{4}\ \text{V/m}\) (or N/C)

21.5 Common Misconceptions

Common Misconceptions

  • Misconception: Charge can be created or destroyed. Correction: Charge is conserved; it can only be transferred between objects.

  • Misconception: Field lines are paths that charges follow. Correction: Field lines show the direction of force on a positive test charge, not trajectories.

  • Misconception: Voltage and energy are the same thing. Correction: Voltage is energy per unit charge (\(V = W/q\)).

  • Misconception: Electric forces require direct contact. Correction: Electric forces act through fields across empty space.

  • Misconception: A neutral object has no charges. Correction: A neutral object has equal amounts of positive and negative charge.

21.6 Practice Questions

21.6.1 Easy (2 marks)

Two charges of \(+1.0\ \mu\text{C}\) are \(0.50\ \text{m}\) apart. Find the force.

  • Apply Coulomb’s law correctly (1)
  • Correct magnitude with units: \(F = 0.036\) N (1)

Answer: 0.036 N (repulsive)

21.6.2 Medium (4 marks)

A \(0.30\ \mu\text{C}\) charge experiences a force of \(0.15\ \text{N}\) in a uniform field. Find the field strength and state its unit.

  • Use \(E = F/q\) correctly (2)
  • Correct value: \(E = 0.15 / (3.0 \times 10^{-7}) = 5.0 \times 10^{5}\) (1)
  • Correct unit: N/C or V/m (1)

Answer: \(5.0 \times 10^{5}\) N/C (or V/m)

21.6.3 Hard (5 marks)

Parallel plates are \(4.0\ \text{mm}\) apart with \(200\ \text{V}\) across them. An electron (\(q = -1.6 \times 10^{-19}\ \text{C}\)) is placed between the plates. Find the field strength and the force on the electron.

  • Use \(E = V/d\) correctly (1)
  • Convert mm to m: \(d = 0.004\) m (1)
  • \(E = 200/0.004 = 5.0 \times 10^{4}\) V/m (1)
  • Force from \(F = qE = 1.6 \times 10^{-19} \times 5.0 \times 10^{4}\) (1)
  • \(F = 8.0 \times 10^{-15}\) N, toward positive plate (1)

Answer: \(E = 5.0 \times 10^{4}\) V/m; \(F = 8.0 \times 10^{-15}\) N toward the positive plate

21.7 Multiple Choice Questions

Test your understanding with these interactive questions:

21.8 Summary

Key Takeaways
  • Charge is conserved and quantised in units of \(e = 1.60 \times 10^{-19}\) C
  • Coulomb’s law: \(F = k|q_1 q_2|/r^2\)
  • Electric field strength: \(E = F/q = kQ/r^2\) for a point charge
  • Between parallel plates: \(E = V/d\) (uniform field)
  • Voltage is work per charge: \(V = W/q\)
  • Energy change: \(U = qV\)

21.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: