Capacitance, energy storage, and Gauss's law applications
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This unit explores the behavior of conductors in electrostatic equilibrium, the storage of energy in electric fields via capacitors, and the modification of fields and capacitance by dielectric materials. In electrostatic equilibrium, a conductor has E = 0 everywhere inside, any net charge resides entirely on the surface, and the surface is an equipotential. Just outside a conductor, the electric field is perpendicular to the surface with magnitude E = σ/ε₀, where σ is the local surface charge density. Cavities inside conductors shield their interiors from external fields; a charge placed inside a cavity induces an equal and opposite charge on the cavity wall and an equal charge on the outer surface. A capacitor is a device that stores charge and energy by maintaining two oppositely charged conductors separated by an insulator. Capacitance is defined as C = Q/V, where Q is the magnitude of charge on one plate and V is the potential difference between plates. For a parallel-plate capacitor, C = ε₀A/d. You can derive this using Gauss's Law to find E between the plates and then integrating to find V. Capacitors store energy in the electric field; the energy is U = ½CV² = ½QV = Q²/(2C). The energy density (energy per unit volume) in any electric field is u = ½ε₀E². When a dielectric material is inserted between the plates, it polarizes, creating bound surface charges that partially cancel the field from the free charges. This reduces the effective field by a factor of the dielectric constant κ: E = E₀/κ. Consequently, the potential difference drops and the capacitance increases: C = κC₀. If the capacitor remains connected to a battery, V stays constant and Q increases. If isolated, Q stays constant and V decreases. On the AP Exam, you must derive capacitance for standard geometries, analyze energy changes when dielectrics are inserted, and understand how charge redistributes when capacitors are connected.
C = Q / VC = ε0 * A / d (parallel plate)U = ½ * C * V² = ½ * Q * V = Q² / (2*C)u = ½ * ε0 * E²C = κ * C0E = E0 / κV = V0 / κ (with dielectric, Q constant)C_series: 1/C_eq = Σ1/C_iC_parallel: C_eq = ΣC_iAnswer each question one at a time. Click an option to select your answer.
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2025 AP Physics C: Electricity and Magnetism Full Review (EVERYTHING YOU NEED TO KNOW!!) by Prepworks Education
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Quick reference for Conductors, Capacitors, Dielectrics. Print this out and review before the exam!
C = Q / VC = ε0 * A / d (parallel plate)U = ½ * C * V² = ½ * Q * V = Q² / (2*C)u = ½ * ε0 * E²C = κ * C0E = E0 / κC_series: 1/C_eq = Σ1/C_iC_parallel: C_eq = ΣC_iDownload official review materials for this unit.
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