Coulomb’s Law (Grade 11 NSC Matric Physical Sciences): Revision Notes

Worked Example: Basic Force Between Two Charges

Problem: Two point-like charges carrying charges of +3 × 10⁻⁹ C and -5 × 10⁻⁹ C are 2 m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

Step 1: Determine what is required We need to find the electrostatic force between two point charges given their charges and separation distance.

Step 2: Choose the appropriate formula We can use Coulomb's law: $F = \frac{kQ_1Q_2}{r^2}$

Step 3: Identify the given information

• Q₁ = +3 × 10⁻⁹ C
• Q₂ = -5 × 10⁻⁹ C
• r = 2 m
• k = 9.0 × 10⁹ N⋅m²⋅C⁻²

Step 4: Calculate the magnitude

• $F = \frac{kQ_1Q_2}{r^2} = \frac{(9.0 \times 10^9)(3 \times 10^{-9})(5 \times 10^{-9})}{(2)^2}$
• $F = 3.37 \times 10^{-8} \text{ N}$

Step 5: Determine the nature of the force

• Since the charges have opposite signs (+3 × 10⁻⁹ C and -5 × 10⁻⁹ C), the force will be attractive.

Answer: The magnitude of the force is 3.37 × 10⁻⁸ N, and the force is attractive.

Worked Example: Electrostatic vs Gravitational Forces

Problem: Compare the electrostatic force and gravitational force between two electrons that are 10⁻¹⁰ m apart (typical distance inside an atom).

Given information:

• Charge of electron: Q₁ = Q₂ = 1.6 × 10⁻¹⁹ C
• Mass of electron: m₁ = m₂ = 9.1 × 10⁻³¹ kg
• Distance: r = 1 × 10⁻¹⁰ m

Electrostatic force calculation:

• $F_e = \frac{kQ_1Q_2}{r^2} = \frac{(9.0 \times 10^9)(1.60 \times 10^{-19})(1.60 \times 10^{-19})}{(10^{-10})^2}$
• $F_e = 2.30 \times 10^{-8} \text{ N}$

Gravitational force calculation:

• $F_g = \frac{Gm_1m_2}{d^2} = \frac{(6.67 \times 10^{-11})(9.11 \times 10^{-31})(9.11 \times 10^{-31})}{(10^{-10})^2}$
• $F_g = 5.54 \times 10^{-51} \text{ N}$

Result: The electrostatic force (2.30 × 10⁻⁸ N) is much larger than the gravitational force (5.54 × 10⁻⁵¹ N). Since electrons carry like charges, the electrostatic force is repulsive, while gravitational force is always attractive.

Worked Example: Three Charges in a Line

Problem: Three point charges are arranged in a straight line. Their charges are Q₁ = +2 × 10⁻⁹ C, Q₂ = +1 × 10⁻⁹ C, and Q₃ = -3 × 10⁻⁹ C. The distance between Q₁ and Q₂ is 2 × 10⁻² m, and the distance between Q₂ and Q₃ is 4 × 10⁻² m. What is the net electrostatic force on Q₂?

Step 1: Calculate individual forces

Force on Q₂ due to Q₁: $F_1 = \frac{k(Q_1Q_2)}{r_{12}^2} = \frac{(9.0 \times 10^9)(2 \times 10^{-9})(1 \times 10^{-9})}{(2 \times 10^{-2})^2} = 4.5 \times 10^{-5} \text{ N}$

Force on Q₂ due to Q₃: $F_3 = \frac{k(Q_2Q_3)}{r_{23}^2} = \frac{(9.0 \times 10^9)(1 \times 10^{-9})(3 \times 10^{-9})}{(4 \times 10^{-2})^2} = 1.69 \times 10^{-5} \text{ N}$

Step 2: Determine force directions

• Force between Q₁ and Q₂ is repulsive (both positive), so F₁ pushes Q₂ to the right
• Force between Q₂ and Q₃ is attractive (opposite signs), so F₃ pulls Q₂ to the right

Step 3: Add the forces vectorially

• Since both forces act in the same direction (to the right):
• $F_R = F_1 + F_3 = 4.5 \times 10^{-5} \text{ N} + 1.69 \times 10^{-5} \text{ N} = 6.19 \times 10^{-5} \text{ N}$

Answer: The resultant force on Q₂ is 6.19 × 10⁻⁵ N to the right.

Worked Example: Triangular Charge Configuration

Problem: Three point charges form a right-angled triangle. Q₁ = +4 nC, Q₂ = +6 nC, and Q₃ = -3 nC. The distance between Q₁ and Q₂ is 5 × 10⁻² m, and the distance between Q₁ and Q₃ is 3 × 10⁻² m. What is the net electrostatic force on Q₁?

Step 1: Calculate force magnitudes

Force on Q₁ due to Q₂: $F_2 = \frac{k(Q_1Q_2)}{r_{12}^2} = \frac{(9.0 \times 10^9)(4 \times 10^{-9})(6 \times 10^{-9})}{(5 \times 10^{-2})^2} = 8.630 \times 10^{-5} \text{ N}$

Force on Q₁ due to Q₃: $F_3 = \frac{k(Q_1Q_3)}{r_{13}^2} = \frac{(9.0 \times 10^9)(4 \times 10^{-9})(3 \times 10^{-9})}{(3 \times 10^{-2})^2} = 1.199 \times 10^{-4} \text{ N}$

Step 2: Determine force directions

• F₂ is repulsive (both positive charges), pushing Q₁ away from Q₂
• F₃ is attractive (opposite charges), pulling Q₁ toward Q₃

Step 3: Calculate resultant using Pythagoras' theorem

• Since the forces act at right angles:
• $F_R^2 = F_2^2 + F_3^2$
• $F_R = \sqrt{[(8.630 \times 10^{-5})^2 + (1.199 \times 10^{-4})^2]} = 1.48 \times 10^{-4} \text{ N}$

Step 4: Find the direction using trigonometry

• $\tan(\theta_R) = \frac{F_3}{F_2} = \frac{1.199 \times 10^{-4}}{8.630 \times 10^{-5}}$
• $\theta_R = \tan^{-1}\left(\frac{1.199 \times 10^{-4}}{8.630 \times 10^{-5}}\right) = 54.25°$

Answer: The final resultant force acting on Q₁ is 1.48 × 10⁻⁴ N at an angle of 54.25°.