Charged objects exert forces on each other even when they are not touching. This action-at-a-distance is explained by the concept of the electric field. In this chapter you will quantify electrostatic forces using Coulomb's law, define the electric field, and learn to draw and interpret electric field line patterns.
3.1 Coulomb's Law
In Grade 10 you learned that like charges repel and unlike charges attract. But HOW STRONG is this force? Charles-Augustin de Coulomb (1736–1806) used a torsion balance to measure electrostatic forces precisely. His results showed a beautiful mathematical pattern: the electrostatic force is similar in form to gravity — it depends on the product of the charges and decreases with the SQUARE of the distance between them. Notice the similarity to Newton's Universal Gravitation: F = Gm₁m₂/d². For charges, we replace mass with charge and G with k.
Definition
Coulomb's law
Coulomb's law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. F = kQ₁Q₂/r²
Formula
Coulomb's law
F = electrostatic force (N), k = Coulomb's constant = 9,0 × 10⁹ N·m²·C⁻², Q₁ and Q₂ = magnitudes of the charges (C), r = distance between the charges (m)
SI unit: N
Note
Coulomb's constant: k = 9,0 × 10⁹ N·m²·C⁻². This is given on the CAPS data sheet. Note: k = 1/(4πε₀) where ε₀ = 8,85 × 10⁻¹² C²·N⁻¹·m⁻² is the permittivity of free space — but for Grade 11 you only need to use k = 9,0 × 10⁹ N·m²·C⁻².
Coulomb's Law vs Newton's Gravitational Law
| Property | Coulomb's Law (electrostatic) | Newton's Law (gravitational) |
|---|---|---|
| Formula | F = kQ₁Q₂/r² | F = Gm₁m₂/d² |
| Constant | k = 9,0 × 10⁹ N·m²·C⁻² | G = 6,67 × 10⁻¹¹ N·m²·kg⁻² |
| Source quantity | Charge Q (in coulombs, C) | Mass m (in kilograms, kg) |
| Force direction | Attractive (unlike) OR repulsive (like) | Always attractive |
| Can be shielded? | Yes (Faraday cage blocks electric fields) | No (gravity cannot be shielded) |
| Inverse square? | Yes — halve r → force × 4 | Yes — halve d → force × 4 |
Watch Out
COMMON MISTAKE — Coulomb's law gives the MAGNITUDE of the force. You must then determine the DIRECTION separately: unlike charges → attractive force (toward each other); like charges → repulsive force (away from each other). Never forget to state the direction in your answer.
Worked Example
Two point charges Q₁ = +3 μC and Q₂ = −5 μC are 0,30 m apart. Calculate the electrostatic force between them. State whether it is attractive or repulsive.
Given
- Q₁ = +3 μC = 3 × 10⁻⁶ C
- Q₂ = −5 μC = 5 × 10⁻⁶ C (use magnitude)
- r = 0,30 m
- k = 9,0 × 10⁹ N·m²·C⁻²
Find
F (magnitude and nature)
Solution
- 1F = kQ₁Q₂/r²
- 2F = (9,0 × 10⁹)(3 × 10⁻⁶)(5 × 10⁻⁶) / (0,30)²
- 3Numerator: 9,0 × 10⁹ × 15 × 10⁻¹² = 135 × 10⁻³ = 0,135
- 4Denominator: (0,30)² = 0,09
- 5F = 0,135 / 0,09 = 1,5 N
- 6Q₁ is positive and Q₂ is negative (unlike charges) → ATTRACTIVE force.
Worked Example
The electrostatic force between two charges is 4,0 N. The charges are moved so that the distance between them is tripled. What is the new force?
Given
- F₁ = 4,0 N
- r₂ = 3r₁ (distance tripled)
Find
F₂
Solution
- 1Coulomb's law: F ∝ 1/r². If r triples, r² increases by 3² = 9.
- 2F₂ = F₁/9 = 4,0/9 = 0,44 N
Practice Question
Two positive point charges, Q₁ = +8 μC and Q₂ = +2 μC, are placed 0,20 m apart. (a) Calculate the electrostatic force between them. (b) Is the force attractive or repulsive? (c) If Q₁ is halved and the distance is halved, by what factor does the force change? Calculate the new force.
(10 marks)
3.2 Electric Field Concept and Calculation
How does a charged object exert a force on another charged object without touching it? The answer is through the ELECTRIC FIELD. A charged object modifies the space around it, creating an electric field. Any other charge placed in this field feels a force. Think of it like the Sun's gravity: the Sun creates a gravitational field throughout the solar system. A planet placed anywhere in that field feels a force pulling it toward the Sun. Similarly, a charge Q creates an electric field in all directions around it — any test charge placed in that field feels an electrostatic force.
Definition
Electric field
A region of space in which an electric charge will experience a force. The direction of the field at a point in space is the direction in which a positive test charge would move if placed at that point.
Definition
Electric field strength
The magnitude of the electric field, E, at a point can be quantified as the force per unit charge: E = F/q, where F is the Coulomb force exerted by a charge on a test charge q. The units of the electric field are newtons per coulomb: N·C⁻¹.
Formula
Electric field from force
E = electric field strength (N·C⁻¹), F = force on test charge (N), q = magnitude of test charge (C)
SI unit: N·C⁻¹
We can also express the electric field directly in terms of the source charge Q and the distance r from it. Substitute Coulomb's law F = kQq/r² into E = F/q: E = (kQq/r²)/q = kQ/r². The test charge q cancels! The electric field at distance r from a source charge Q depends ONLY on Q and r — not on the size of the test charge we use to measure it.
Formula
Electric field of a point charge
E = electric field strength (N·C⁻¹), k = 9,0 × 10⁹ N·m²·C⁻², Q = source charge (C), r = distance from source charge (m)
SI unit: N·C⁻¹
Watch Out
COMMON MISTAKE — The electric field E = F/q uses q as the TEST charge (the small charge placed in the field to measure it). The electric field E = kQ/r² uses Q as the SOURCE charge (the charge CREATING the field). Do not confuse Q (source) and q (test charge). The test charge must be so small that it does not disturb the field created by Q.
Worked Example
A test charge of +2 × 10⁻⁶ C experiences a force of 0,018 N to the right when placed at point P. (a) Calculate the electric field at point P. (b) What is the direction of the electric field at P?
Given
- q = +2 × 10⁻⁶ C (test charge)
- F = 0,018 N to the right
Find
(a) E at P (b) Direction of E
Solution
- 1(a) E = F/q = 0,018 / (2 × 10⁻⁶) = 9 000 N·C⁻¹
- 2(b) The test charge is positive. A positive test charge moves in the direction of the field. The force on it is to the right → the electric field at P is directed to the RIGHT.
Worked Example
A point charge Q = +4 μC is located in space. Calculate (a) the electric field strength at a point 0,30 m from Q, and (b) the force on a −6 μC charge placed at that point.
Given
- Q = +4 μC = 4 × 10⁻⁶ C
- r = 0,30 m
- k = 9,0 × 10⁹ N·m²·C⁻²
- q = −6 μC = 6 × 10⁻⁶ C (for part b)
Find
(a) E at 0,30 m from Q (b) Force on q
Solution
- 1(a) E = kQ/r² = (9,0 × 10⁹)(4 × 10⁻⁶) / (0,30)²
- 2E = 36 × 10³ / 0,09 = 36 000 / 0,09 = 4,0 × 10⁵ N·C⁻¹
- 3Direction: away from Q (Q is positive, so field points outward from Q).
- 4(b) F = Eq = (4,0 × 10⁵)(6 × 10⁻⁶) = 2,4 N
- 5q is negative → force is opposite to E direction → force is toward Q (attractive, as expected for unlike charges).
Exam Tip
EXAM TIP — The electric field is a VECTOR. Always state its direction. For a positive source charge Q, E points AWAY from Q (outward). For a negative source charge Q, E points TOWARD Q (inward). These are the directions a positive test charge would move.
Practice Question
A +5 nC (5 × 10⁻⁹ C) point charge is at the origin. (a) Calculate the electric field strength at a point 0,10 m from the charge. (b) A charge of −3 nC is placed at that point. Calculate the force it experiences and state its direction. (c) In which direction does the electric field at that point point?
(8 marks)
3.3 Electric Field Lines — Patterns and Properties
Michael Faraday introduced the idea of ELECTRIC FIELD LINES (also called lines of force) as a way to visualise the electric field. These lines are imaginary — they cannot be seen — but they give us an immediate visual picture of the direction and relative strength of the field at every point in space. The rules for drawing field lines are precise and must be followed exactly in exam questions.
Rules (properties) of electric field lines
- Electric field lines start on POSITIVE charges (or at infinity) and end on NEGATIVE charges (or at infinity).
- Field lines NEVER cross each other. If two lines crossed, the field would have two directions at that point — which is impossible.
- The DENSITY of field lines (how closely packed they are) indicates the STRENGTH of the field: closely packed lines = strong field; widely spaced lines = weak field.
- Field lines are always perpendicular (at 90°) to the surface of a conductor.
- A positive test charge placed on a field line would move in the direction the line points (from + toward −).
- Field lines are smooth, continuous curves (not zigzag or looped).
Electric Field Patterns for Common Charge Configurations
| Property | Charge configuration | Field line pattern description |
|---|---|---|
| Single positive charge +Q | Lines radiate outward in all directions (symmetric) | Field strongest near the charge, weakens with distance (∝ 1/r²) |
| Single negative charge −Q | Lines converge inward from all directions (symmetric) | Same strength pattern but opposite direction to +Q |
| Unlike charges (+Q and −Q) | Lines leave +Q and end on −Q, curving in between | Strong field between charges, lines densely packed in gap |
| Like charges (+Q and +Q) | Lines leave both charges, repel outward, none cross | Zero field at midpoint between equal charges |
Watch Out
COMMON MISTAKE — Do NOT draw field lines crossing each other — this is always wrong. Do NOT draw field lines that start OR end in empty space between charges — they must start on positive charges and end on negative charges (or go to infinity for isolated charges). Do NOT draw field lines that are not smooth curves.
Worked Example
Describe and sketch the electric field pattern between a positive charge +2Q and a negative charge −Q, where the positive charge has twice the magnitude. State two differences between this pattern and the pattern for two equal opposite charges.
Given
- Q₁ = +2Q (positive, larger)
- Q₂ = −Q (negative, smaller)
Find
Description and sketch of field line pattern + two differences from equal-charge dipole
Solution
- 1Twice as many lines leave the positive charge (+2Q) as enter the negative charge (−Q) — because it has twice the charge.
- 2The excess lines from +2Q that do not end on −Q go off to infinity (they continue outward indefinitely).
- 3Between the charges, lines still go from +2Q to −Q (curving from left to right).
- 4Difference 1: For equal charges, ALL lines from + end on −. Here, only HALF the lines end on −Q; the rest continue to infinity.
- 5Difference 2: The neutral point (zero field) is NOT at the midpoint — it is closer to the smaller charge (−Q), displaced from centre.
Practice Question
(a) State THREE properties of electric field lines. (b) Draw the electric field pattern for two equal and opposite charges (+Q and −Q) placed 3 cm apart. (c) At which point between them is the field strongest, and how can you tell from the field line diagram?
(9 marks)
Exam Tip
EXAM TIP — When drawing field lines between two charges, draw at least 4–6 lines per charge, make sure they curve smoothly from positive to negative, and make lines near each charge radiate symmetrically. The more lines you draw on the diagram, the more credit you get — as long as they are correct.
Worked Example
A point charge Q = −8 μC is fixed at the origin. (a) Calculate E at a distance of 0,20 m from Q. (b) A proton (charge = +1,6 × 10⁻¹⁹ C) is placed at 0,20 m from Q. Calculate the force on the proton and state its direction. (c) Describe the direction of the electric field at a point 0,20 m to the RIGHT of Q.
Given
- Q = −8 μC = 8 × 10⁻⁶ C (magnitude)
- r = 0,20 m
- q = +1,6 × 10⁻¹⁹ C (proton)
- k = 9,0 × 10⁹ N·m²·C⁻²
Find
(a) E (b) Force on proton (c) Direction of E
Solution
- 1(a) E = kQ/r² = (9,0 × 10⁹)(8 × 10⁻⁶) / (0,20)² = 72 × 10³ / 0,04 = 1,8 × 10⁶ N·C⁻¹
- 2(b) F = Eq = 1,8 × 10⁶ × 1,6 × 10⁻¹⁹ = 2,88 × 10⁻¹³ N
- 3Q is negative, proton is positive (unlike) → attractive → force on proton is TOWARD Q (to the left if proton is to the right of Q).
- 4(c) Q is negative → field lines point TOWARD Q. A point to the right of Q: the field points to the LEFT (toward Q).
3.4 Combining Electric Fields (Superposition)
When more than one charge creates an electric field at a point, the TOTAL electric field is found by adding the individual fields as VECTORS. This is called the superposition principle for electric fields. Similarly, the net force on a charge is found by adding all the individual Coulomb forces as vectors. In 1D problems (all charges on a line), this simply means adding or subtracting forces depending on their directions.
Note
SUPERPOSITION PRINCIPLE FOR ELECTRIC FIELDS: The total electric field at any point is the vector sum of the electric fields produced by each charge independently. Each charge produces its own field as if the other charges were not there.
Worked Example
Two positive charges are placed on a horizontal line: Q₁ = +6 μC at position x = 0, and Q₂ = +4 μC at position x = 0,50 m. Calculate the electric field at point P, which is at x = 0,20 m (between the two charges).
Given
- Q₁ = +6 μC = 6 × 10⁻⁶ C at x = 0
- Q₂ = +4 μC = 4 × 10⁻⁶ C at x = 0,50 m
- Point P at x = 0,20 m
- k = 9,0 × 10⁹ N·m²·C⁻²
Find
Total E at P (magnitude and direction)
Solution
- 1Distance from Q₁ to P: r₁ = 0,20 m
- 2Distance from Q₂ to P: r₂ = 0,50 − 0,20 = 0,30 m
- 3E due to Q₁: E₁ = kQ₁/r₁² = (9 × 10⁹)(6 × 10⁻⁶)/(0,20)² = 54 × 10³/0,04 = 1 350 000 N·C⁻¹ = 1,35 × 10⁶ N·C⁻¹
- 4Direction of E₁: Q₁ is positive → field points AWAY from Q₁ → toward the RIGHT (positive x).
- 5E due to Q₂: E₂ = kQ₂/r₂² = (9 × 10⁹)(4 × 10⁻⁶)/(0,30)² = 36 × 10³/0,09 = 400 000 N·C⁻¹ = 4,0 × 10⁵ N·C⁻¹
- 6Direction of E₂: Q₂ is positive → field at P points AWAY from Q₂ → toward the LEFT (negative x).
- 7Taking rightward as positive: E_total = E₁ − E₂ = 1 350 000 − 400 000 = +950 000 N·C⁻¹
Worked Example
Three charges lie on the x-axis: Q₁ = +3 μC at x = 0, Q₂ = −3 μC at x = 0,30 m. Calculate the net force on Q₂ due to Q₁. (Treat as a two-charge system.)
Given
- Q₁ = +3 μC = 3 × 10⁻⁶ C at x = 0
- Q₂ = −3 μC = 3 × 10⁻⁶ C (magnitude) at x = 0,30 m
- k = 9,0 × 10⁹ N·m²·C⁻²
Find
Force on Q₂ due to Q₁
Solution
- 1F = kQ₁Q₂/r² = (9,0 × 10⁹)(3 × 10⁻⁶)(3 × 10⁻⁶) / (0,30)²
- 2F = (9,0 × 10⁹ × 9 × 10⁻¹²) / 0,09
- 3F = (81 × 10⁻³) / 0,09 = 0,9 N
- 4Q₁ is positive, Q₂ is negative (unlike charges) → attractive force.
- 5Force on Q₂ is directed toward Q₁ → to the LEFT (toward x = 0).
Watch Out
COMMON MISTAKE — When calculating the net electric field or net force from multiple charges, you MUST treat each individual contribution as a vector. On a line, this means keeping track of signs (positive = right, negative = left). Simply adding magnitudes without considering direction will give a wrong answer.
Practice Question
Two charges are on the x-axis: Q₁ = +5 μC at x = 0 and Q₂ = +5 μC at x = 0,40 m. (a) Calculate the electric field at the midpoint (x = 0,20 m) due to each charge. (b) Calculate the net electric field at the midpoint. (c) Explain physically why your answer to (b) makes sense.
(8 marks)