Electrostatics and Coulomb's Law
Electrostatics and Coulomb's Law: Comprehensive WAEC Guide
1. Definitions
Electrostatics is the branch of physics that deals with the study of electric charges at rest (static electricity), their forces, fields, and potentials.
Coulomb's Law states that the electrostatic force of attraction or repulsion 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.
2. Mathematical Formula
Mathematically, Coulomb's Law is expressed as:
Where:
- F = Electrostatic force between the charges (measured in Newtons, N)
- q₁ , q₂ = Magnitudes of the two point charges (measured in Coulombs, C)
- r = Distance separating the centers of the two charges (measured in meters, m)
- k = Coulomb's constant. In a vacuum or air:
(where ε₀ is the permittivity of free space = 8.85 × 10⁻¹² C²/N·m²)
3. WAEC Theory Questions & Solutions
Question 1: State two methods by which an insulated neutral metallic sphere can be charged. Explain briefly how it can be charged positively using a negatively charged rubber rod.
Two methods of charging:
- Charging by Induction
- Charging by Conduction (Contact)
Explanation (Charging positively by Induction):
- Bring the negatively charged rubber rod close to (but not touching) the neutral insulated metallic sphere. The negative charges on the rod repel free electrons to the far side of the sphere, leaving the near side positively charged.
- Earth the far side of the sphere momentarily by touching it with a finger while the rod is still held in place. Free electrons flow away from the sphere into the ground.
- Remove the ground connection (your finger).
- Finally, remove the negatively charged rod. The induced positive charges will redistribute themselves uniformly over the surface of the sphere.
Question 2: Two point charges, Q₁ = +4.0 × 10⁻⁶ C and Q₂ = -2.0 × 10⁻⁶ C, are placed in air at a distance of 30 cm apart. Calculate the magnitude and state the nature of the electrostatic force between them. [Take k = 9.0 × 10⁹ N·m²/C²]
Given data:
q₁ = 4.0 × 10⁻⁶ C, q₂ = 2.0 × 10⁻⁶ C, r = 30 cm = 0.3 m, k = 9.0 × 10⁹
Using Coulomb's Formula:
F = (k × q₁ × q₂) / r²
F = (9.0 × 10⁹ × 4.0 × 10⁻⁶ × 2.0 × 10⁻⁶) / (0.3)²
F = 0.072 / 0.09
F = 0.8 N
Nature of force: Since the charges are opposite (+ and -), the force is attractive.
Question 3: Define electric field intensity at a point in an electric field. Write down its formula and SI unit.
Definition: Electric field intensity at a point is the electrostatic force experienced per unit positive charge placed at that point within the electric field.
Formula: E = F / q or E = kQ / r²
SI Unit: Newton per Coulomb (N/C) or Volt per meter (V/m).
Question 4: State three properties of electric lines of force.
- They start from positive charges and end on negative charges.
- They never cross or intersect each other.
- They are closer together where the electric field is strong, and wider apart where it is weak.
4. WAEC Objective Questions & Solutions
1. Which of the following parameters remains unchanged when a conductor is charged by induction?
- A. The type of charge on the charging rod
- B. The total amount of charge on the charging rod
- C. The distribution of charge on the conductor
- D. The potential of the conductor
Correct Answer: B
Explanation: During induction, the charging rod never touches the object being charged, so it experiences no loss or gain of electrons.
2. If the distance between two point charges is halved, the electrostatic force between them becomes:
- A. Halved
- B. Doubled
- C. Quartered
- D. Quadrupled (Four times)
Correct Answer: D
Explanation: Since F ∝ 1/r², halving the distance (1/2) means the inverse square yields (2)² = 4 times the original force value.
3. The SI unit of electric permittivity (ε₀) is:
- A. N·m²/C²
- B. C²/N·m²
- C. N·C/m²
- D. C/N·m
Correct Answer: B
Explanation: Derived by making ε₀ the subject of the formula: ε₀ = q₁q₂ / (4πFr²), giving C² / (N·m²).
4. A glass rod rubbed with silk gains a positive charge because:
- A. Protons are transferred from silk to glass
- B. Electrons are transferred from glass to silk
- C. Protons are transferred from glass to silk
- D. Electrons are transferred from silk to glass
Correct Answer: B
Explanation: Static charging relies on electron migration. Glass drops electrons easily, letting silk capture them.
5. The expression for the electric potential V at a distance r from a point charge Q is:
- A. kQ / r
- B. kQ / r²
- C. kQ² / r
- D. kQ² / r²
Correct Answer: A
Explanation: Electric potential scales inversely with distance (1/r), whereas electric field strength scales with 1/r².
6. Which of the following instruments is used to detect the presence of static electric charge?
- A. Ammeter
- B. Gold-leaf electroscope
- C. Galvanometer
- D. Voltmeter
Correct Answer: B
Explanation: The gold-leaf electroscope divergence identifies charge presence and magnitude safely.
7. Two identical charges of magnitude 2.0 × 10⁻⁶ C experience a repulsive force of 10 N in a vacuum. Calculate the separation distance between them.
- A. 0.06 m
- B. 0.19 m
- C. 0.36 m
- D. 0.60 m
Correct Answer: A
Explanation: r² = kQ²/F = (9×10⁹ × 4×10⁻¹²) / 10 = 0.0036. Square root of 0.0036 evaluates to exactly 0.06 m.
8. Electric field lines always point in which direction?
- A. From negative to positive
- B. From positive to negative
- C. In circular paths around a single charge
- D. Randomly based on temperature
Correct Answer: B
Explanation: Field line vectors represent trajectories chosen by hypothetical positive test charges, leaving positives and rushing to negatives.
9. A lightning conductor protects tall buildings by:
- A. Repelling clouds away from the building
- B. Absorbing the heat energy from lightning safely
- C. Conducting electric discharge safely to the earth
- D. Isolating the building structurally from rain
Correct Answer: C
Explanation: It serves as an ultra-low resistance copper detour routing surging atmospheric ions safely underground.
10. What happens to the capacitance of a parallel plate capacitor when a dielectric material is introduced?
- A. It decreases
- B. It increases
- C. It remains constant
- D. It becomes zero
Correct Answer: B
Explanation: Substituting vacuum space with solid dielectric insulation raises permittivity factors (ε), amplifying charge storage potential.
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