Electrostatics
Objectives:
- Identify positive and negative charges in matter.
- Examine the uses of an electroscope.
- Apply Coulomb’s inverse square law of electrostatics to solve problems.
- Deduce expressions for electric field intensity and potential difference.
- Identify electric field flux patterns of isolated and interacting charges.
- Analyse the distribution of charges on a conductor and how it is used in lightning conductors.
Key Concepts and Formulas
Existence of Charges: Matter contains positive and negative charges. Like charges repel; unlike charges attract.
Charging Methods:
- Friction: Rubbing two different materials transfers electrons.
- Contact: Direct contact allows charge sharing.
- Induction: A charged object can induce a charge separation in a nearby conductor.
Electroscope: A device that detects electric charge through the divergence of its metal leaves.
Coulomb’s Law: F = k |q₁q₂|/r², where k ≈ 9×10⁹ N·m²/C², q₁ and q₂ are charges, and r is the separation distance.
Electric Field (E): E = F/q, and Electric Potential (V): V = kq/r. The potential difference is ΔV = Vₐ - V_b.
Electric Discharge and Lightning: When charge accumulates excessively on a conductor, it discharges as lightning. Lightning conductors help distribute charges to safely dissipate energy.
Worked Examples in Electrostatics
The following 6 worked examples (each in a green box) illustrate key concepts and numerical problems in electrostatics.
Example 1:
Concept: Identification of charges.
When a glass rod is rubbed with silk, the glass rod becomes positively charged and the silk becomes negatively charged.
Example 2:
Charging by Contact: A metal sphere initially neutral touches a positively charged rod and acquires a net negative charge (by electron transfer). (Conceptual explanation)
Example 3:
Coulomb’s Law Problem: Calculate the force between two point charges of 2 µC and 3 µC separated by 0.5 m.
Solution: F = (9×10⁹)×|2×10⁻⁶ × 3×10⁻⁶|/(0.5)² = (9×10⁹×6×10⁻¹²)/(0.25) = (54×10⁻³)/0.25 = 0.216 N.
Example 4:
Electric Field: Find the electric field at a distance of 0.3 m from a point charge of 5×10⁻⁶ C.
Solution: E = kq/r² = (9×10⁹×5×10⁻⁶)/(0.3²) = (45×10³)/(0.09) = 500,000 N/C.
Example 5:
Electric Potential Difference: Calculate the potential difference between two points 0.4 m and 0.6 m from a point charge of 4×10⁻⁶ C.
Solution: V = kq/r. V₁ = (9×10⁹×4×10⁻⁶)/0.4 = 90,000 V; V₂ = (9×10⁹×4×10⁻⁶)/0.6 = 60,000 V; ΔV = 90,000 - 60,000 = 30,000 V.
Example 6:
Lightning and Discharge: Explain how a lightning conductor works.
Solution: A lightning conductor is a metal rod mounted on top of a building. It provides a low-resistance path for the accumulated charges to disperse into the earth, thus protecting the building from lightning strikes.
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