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Lesson 6.4 • Competency 6.4Complete Lesson
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Capacitors, Dielectrics and Electrostatic Energy

Capacitance C = Q/V, parallel-plate capacitor formula C = ε₀εᵣA/d, series and parallel combinations, stored energy U = 1/2 CV² = Q²/2C = 1/2 QV, energy density u = 1/2 εE², and dielectric polarization.

Marking a lesson complete represents finishing your study of the topic; it does not denote tested mastery.

Learning Objectives

  • Derive parallel-plate capacitance C = ε₀εᵣA/d and analyze how altering area A, separation d, and dielectric constant εᵣ affects C, Q, V, and E under connected vs disconnected battery states.
  • Calculate equivalent capacitance for series (1/C_s = Σ 1/C_i) and parallel (C_p = Σ C_i) networks and solve multi-capacitor circuit problems.
  • Derive stored electrostatic energy (U = 1/2 CV² = Q²/2C = 1/2 QV) and electrostatic energy density (u = 1/2 εE²), explaining energy dissipation during capacitor charge sharing.

Core Theory & Concepts

A capacitor is a device designed to store electrostatic potential energy in the electric field between two opposing conductors. By inserting insulating dielectrics, capacitance can be increased substantially while preventing electrical breakdown.

Parallel Plate Capacitance

Capacitance of two parallel plates of area A separated by distance d with dielectric εᵣ: C = ε₀ εᵣ A / d.

C = \frac{\varepsilon_0 \varepsilon_r A}{d}

Capacitor Combinations

Series: 1/C_s = 1/C₁ + 1/C₂ + ... (all store same charge Q). Parallel: C_p = C₁ + C₂ + ... (all experience same voltage V).

\frac{1}{C_s} = \sum \frac{1}{C_i}, \quad C_p = \sum C_i

Stored Electrostatic Energy

Total work done establishing the electric field: U = 1/2 CV² = Q² / (2C) = 1/2 QV. The energy resides in the field with density u = 1/2 ε E².

U = \frac{1}{2} C V^2 = \frac{Q^2}{2C} = \frac{1}{2} Q V, \quad u = \frac{1}{2} \varepsilon E^2

Common Misconceptions

Misconception

When a dielectric slab is inserted into a charged capacitor, its stored energy always decreases.

Physical Clarification:It depends on whether the battery remains connected! If disconnected (Q = const), U = Q²/(2C) decreases as C increases (the dielectric is pulled in by electrostatic forces, doing mechanical work). But if the battery is CONNECTED (V = const), U = 1/2 CV² INCREASES as C increases because the battery draws additional charge from its terminals.

Syllabus Topic Outline

Competency 6.4
  • 1Concept and definition of capacitance: C = Q / V (Farads)
  • 2Parallel plate capacitor derivation: C = ε₀ εᵣ A / d
  • 3Dielectric polarization, bound surface charges, and dielectric strength
  • 4Effects of dielectric insertion: battery connected (V const) vs disconnected (Q const)
  • 5Series capacitor combinations: charge is identical, voltages divide
  • 6Parallel capacitor combinations: voltage is identical, charges add
  • 7Derivation of stored electrostatic energy: U = 1/2 CV² = Q² / (2C)
  • 8Electrostatic field energy density: u = 1/2 ε₀ εᵣ E²
  • 9Energy loss during charge sharing between connected capacitors