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Electric Charge and Coulomb's Law

Charge quantisation; conservation; Coulomb's law F = kQq/r²; electric field

Tutor AI Team
Tutor AI Team

# Electric Charge and Coulomb's Law — IB Physics

Electric charge is one of the fundamental properties of matter. Coulomb's law describes the force between charges, and understanding electric fields is essential for IB Physics.

1. Electric Charge

  • Two types: positive and negative
  • Elementary charge: e=1.6×10−19e = 1.6 \times 10^{-19} C
  • Quantisation: charge comes in multiples of ee: Q=neQ = ne
  • Conservation: total charge in a closed system is constant

2. Coulomb's Law

F=kQ1Q2r2\boxed{F = k\frac{Q_1 Q_2}{r^2}}

Where k=8.99×109k = 8.99 \times 10^9 N m² C⁻² = 1/(4πε0)1/(4\pi\varepsilon_0)

  • Like charges repel; unlike charges attract
  • Inverse square law (like gravity)

3. Electric Field

E=Fq\boxed{E = \frac{F}{q}}

Units: N/C or V/m. Direction: the force on a positive test charge.

For a point charge: E=kQ/r2E = kQ/r^2

For uniform field (parallel plates): E=V/dE = V/d

Field Lines

  • Start on positive, end on negative
  • Never cross
  • Closer lines = stronger field
  • Perpendicular to surface of conductor

4. Electric Potential

V=kQrV = \frac{kQ}{r}

Electric PE: Ep=kQ1Q2/rE_p = kQ_1Q_2/r

Potential difference: ΔV=W/q\Delta V = W/q (energy per unit charge).

Worked Example: Example 1

Problem

Two charges: +3 μC and −2 μC are 10 cm apart. Find the force.

F=8.99×109×3×10−6×2×10−6/(0.1)2=5.39F = 8.99 \times 10^9 \times 3 \times 10^{-6} \times 2 \times 10^{-6}/(0.1)^2 = 5.39 N (attractive)

Solution

Worked Example: Example 2

Problem

Field strength between parallel plates with PD = 200 V, separation = 5 mm.

E=V/d=200/0.005=40,000E = V/d = 200/0.005 = 40{,}000 V/m

Solution

6. Practice Questions

    1. State Coulomb's law and compare it to Newton's law of gravitation. (3 marks)
    1. Two protons are 5×10−155 \times 10^{-15} m apart. Calculate the electrostatic force. (2 marks)
    1. Sketch field lines for (a) a positive point charge, (b) two equal and opposite charges. (2 marks)

    Answers

    1. F=kQ1Q2/r2F = kQ_1Q_2/r^2. Like gravity: both inverse square, both act between two entities. Unlike gravity: electric can be attractive or repulsive; gravity is always attractive.
    1. F=8.99×109×(1.6×10−19)2/(5×10−15)2=9.21F = 8.99 \times 10^9 \times (1.6 \times 10^{-19})^2/(5 \times 10^{-15})^2 = 9.21 N.

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Summary

  • Charge quantised (Q=neQ = ne) and conserved
  • Coulomb: F=kQ1Q2/r2F = kQ_1Q_2/r^2
  • E=F/q=kQ/r2E = F/q = kQ/r^2 (point); E=V/dE = V/d (uniform)
  • Field lines: + to −, never cross
Tags:
#ib#physics#electricity#charge#coulombs-law#electric-field

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