Capacitance and Electric Energy Storage

Capacitance Overview

  • Capacitance: Ability of a capacitor to store electric charge.
  • Capacitors: Composed of two conductors close together, not touching.
    • Connected to a battery, the charge on plates is proportional to voltage.

Key Concepts

  • Capacitance (C): Defined as the charge (Q) per unit voltage (V).
  • Unit of Capacitance: Farad (F) - 1 F = 1 C/V.

Determination of Capacitance

  • Parallel-Plate Capacitor:
    • Electric field (E) between plates: E=Qε<em>0AE = \frac{Q}{\varepsilon<em>0 A} where ( \varepsilon0 ) is the permittivity of empty space.
    • Potential difference (V) derived from:
      V<em>ba=Qdε</em>0AV<em>{ba} = \frac{Q d}{\varepsilon</em>0 A}
  • Formula for Capacitance: C=QVC = \frac{Q}{V}
    • Substitute V:
      C=ε0AdC = \frac{\varepsilon_0 A}{d}
Example Calculation - Parallel-Plate Capacitor
  1. Geometry: Plates are 20 cm x 3.0 cm apart with a 1.0 mm air gap.
  2. Capacitance Calculation: Use the formula for capacitance with the area and distance values.
  3. Charge Calculation: Using a 12-V battery, the charge on each plate can be calculated:
    Q=C×VQ = C \times V
  4. Electric Field Calculation: Evaluating the electric field between plates.
  5. Capacitance for 1 F: Estimate the area required to achieve a capacitance of 1 F.

Capacitors in Series and Parallel

  • Parallel: Each capacitor experiences the same voltage.
    • Effective Capacitance:
      C<em>eq=C</em>1+C<em>2+C</em>3C<em>{eq} = C</em>1 + C<em>2 + C</em>3
  • Series: Each capacitor carries the same charge:
    • Effective Capacitance:
      1C<em>eq=1C</em>1+1C<em>2+1C</em>3\frac{1}{C<em>{eq}} = \frac{1}{C</em>1} + \frac{1}{C<em>2} + \frac{1}{C</em>3}
Example Calculation - Series and Parallel Capacitors
  • Determine capacitance for given configurations to find equivalent capacitance and charge/voltage across each.

Electric Energy Storage

  • Charged capacitors store electric energy as work is done to charge them.
  • Energy Stored (U): U=12CV2U = \frac{1}{2} C V^2
    • For a flash unit with a 150-μF capacitor at 200 V, calculate energy and power output when discharged over 1 ms.
Energy Changes with Plate Separation
  • Discuss how energy changes if plate separation is increased while maintaining charge.
  • Compare initial and final energies when plates pull apart with static charge.
Dielectrics
  • Definition: An insulator characterized by dielectric constant K.
    • Increases capacitance when inserted into a capacitor.
    • Dielectric Strength: Maximum field a dielectric can withstand without breakdown.
Effect of Dielectric on Capacitors
  • Two scenarios for inserting a dielectric:
    1. Connected to Battery: Constant voltage; capacitance and charge increase.
    2. Disconnected: Constant charge; capacitance increases, and voltage drops.
Example Calculation with Dielectrics
  • Given a dielectric with constant K = 3.4, determine new capacitance, charge, field strength, and energy after dielectric is removed with charge preserved.

Summary of Important Equations

  • Capacitance: C=QVC = \frac{Q}{V}
  • Capacitors in Series: 1C<em>eq=1C</em>1+1C2+\frac{1}{C<em>{eq}} = \frac{1}{C</em>1} + \frac{1}{C_2} + …
  • Capacitors in Parallel: C<em>eq=C</em>1+C2+C<em>{eq} = C</em>1 + C_2 + …
  • Energy Stored: U=12CV2U = \frac{1}{2} C V^2
  • Electric Field with Dielectric: If introduced, the field reduces due to induced polarization.