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🎯⭐ INTERACTIVE LESSON

Capacitors and Dielectrics

Learn step-by-step with interactive practice!

Capacitors and Dielectrics - Complete Interactive Lesson

Part 1: Capacitance

⚡ Capacitance

Part 1 of 7 — Capacitance

C=QVC = \frac{Q}{V}

  • Capacitance is the ratio of stored charge to voltage
  • SI unit: Farad (F) = C/V
  • Typical values: μF, nF, pF
  • Capacitance depends on geometry, not on QQ or VV

Worked Example

A capacitor stores 6×10−66 \times 10^{-6} C at 3 V. Find CC.

C=Q/V=6×10−6/3=2×10−6C = Q/V = 6 \times 10^{-6} / 3 = 2 \times 10^{-6} F =2= 2 μF ✅

Concept Check 🎯

Capacitance 🧮

  1. Q=6Q = 6 μC, V=3V = 3 V. CC (μF)?

  2. C=4C = 4 μF, V=3V = 3 V. QQ (μC)?

  3. C=2C = 2 μF, Q=10Q = 10 μC. VV (V)?

Concept Check 🔍

Practice

#KnownFind
1QQ, VVCC
2CC, VVQQ
3CC, QQVV

Challenge Question 📋

Part 2: Parallel-Plate Capacitors

⚡ Parallel-Plate Capacitors

Part 2 of 7 — Parallel-Plate Capacitors

C=epsilon0AdC = \frac{epsilon_0 A}{d}

where AA = plate area, dd = plate separation.

  • Electric field between plates: E=V/d=σ/epsilon0E = V/d = \sigma/epsilon_0
  • Field is uniform between the plates
  • Increasing AA or decreasing dd increases CC

Worked Example

Plates: A=0.01A = 0.01 m², d=0.001d = 0.001 m. Find CC.

C=epsilon0A/d=8.85×10−12×0.01/0.001=88.5C = epsilon_0 A/d = 8.85 \times 10^{-12} \times 0.01 / 0.001 = 88.5 pF ✅

Concept Check 🎯

Parallel-Plate Capacitors 🧮

  1. Plate area is doubled, dd unchanged. CC increases by factor ___

  2. Plate area is doubled and dd is halved. CC increases by factor ___

  3. V=10V = 10 V, d=0.01d = 0.01 m. EE between the plates (V/m)?

Concept Check 🔍

Practice

#ChangeEffect on CC
1Double AACC doubles
2Double ddCC halves
3Both doubleCC stays same

Challenge Question 📋

Part 3: Series & Parallel Combinations

⚡ Capacitors in Series & Parallel

Part 3 of 7 — Series & Parallel Combinations

Parallel: Ceq=C1+C2+⋯C_{eq} = C_1 + C_2 + \cdots (same voltage)

Series: 1Ceq=1C1+1C2+⋯\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots (same charge)

Note: opposite of resistors! Parallel adds, series uses reciprocals.

Worked Example

Two capacitors: 3 μF and 6 μF in series. Find CeqC_{eq}.

1Ceq=13+16=2+16=12\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{1}{2}

Ceq=2C_{eq} = 2 μF ✅

Concept Check 🎯

Series & Parallel Capacitors 🧮

  1. 3 μF and 6 μF in parallel. CeqC_{eq} (μF)?

  2. 3 μF and 6 μF in series. CeqC_{eq} (μF)?

  3. Two 3 μF caps in parallel: CeqC_{eq} (μF)?

Concept Check 🔍

Practice

#ConfigurationFormula
1ParallelCeq=C1+C2C_{eq} = C_1 + C_2
2Series1/Ceq=1/C1+1/C21/C_{eq} = 1/C_1 + 1/C_2
3MixedSimplify step by step

Challenge Question 📋

Part 4: Energy Stored

⚡ Energy Stored in Capacitors

Part 4 of 7 — Energy Stored

U=12CV2=Q22C=12QVU = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV

The energy is stored in the electric field between the plates.

Energy density: u=12epsilon0E2u = \frac{1}{2}epsilon_0 E^2 (J/m3)(J/m^{3})

Worked Example

A 10 μF capacitor is charged to 100 V. Energy stored?

U=12CV2=12(10×10−6)(10000)=0.05U = \frac{1}{2}CV^2 = \frac{1}{2}(10 \times 10^{-6})(10000) = 0.05 J ✅

Concept Check 🎯

Energy Stored 🧮

  1. C=10C = 10 μF, V=100V = 100 V. Energy (mJ)?

  2. VV is doubled while CC stays the same. Energy increases by a factor of ___

  3. C=4C = 4 μF, V=10V = 10 V. U=12(4)(100)=?U = \frac{1}{2}(4)(100) = ? μJ

Concept Check 🔍

Practice

#KnownFormula
1CC, VVU=12CV2U = \frac{1}{2}CV^2
2QQ, CCU=Q2/(2C)U = Q^2/(2C)
3QQ, VVU=12QVU = \frac{1}{2}QV

Challenge Question 📋

Part 5: Dielectrics

⚡ Dielectrics

Part 5 of 7 — Dielectrics

Inserting a dielectric (insulating material) between plates:

C=κC0=κepsilon0AdC = \kappa C_0 = \frac{\kappa epsilon_0 A}{d}

where κ\kappa (kappa) is the dielectric constant (κ>1\kappa > 1).

Effects of a dielectric (battery disconnected):

  • CC increases by factor κ\kappa
  • VV decreases by factor κ\kappa
  • EE decreases by factor κ\kappa
  • QQ stays the same

Worked Example

A 5 μF capacitor has a dielectric with κ=3\kappa = 3 inserted. New capacitance?

C=κC0=3×5=15C = \kappa C_0 = 3 \times 5 = 15 μF ✅

Concept Check 🎯

Dielectrics 🧮

  1. C0=5C_0 = 5 μF, κ=3\kappa = 3. New CC (μF)?

  2. A dielectric triples the capacitance. What is κ\kappa?

  3. C0=10C_0 = 10 μF. A dielectric with κ=2\kappa = 2 is inserted. Capacitor is charged to Q=100Q = 100 μC (battery disconnected). V=Q/CV = Q/C (V)?

Concept Check 🔍

Practice

#ScenarioEffect
1Insert dielectric (battery disconnected)C↑C \uparrow, V↓V \downarrow
2Insert dielectric (battery connected)C↑C \uparrow, Q↑Q \uparrow
3Remove dielectric (battery disconnected)C↓C \downarrow, V↑V \uparrow

Challenge Question 📋

Part 6: Problem-Solving Workshop

⚡ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Capacitor Problem Strategy

  1. Identify the capacitor configuration (series, parallel, or single)
  2. Determine if a battery is connected or disconnected
  3. Apply Q=CVQ = CV and energy formulas
  4. For dielectrics, multiply CC by κ\kappa
  5. Use energy density u=12epsilon0E2u = \frac{1}{2}epsilon_0 E^2 for field energy problems

Worked Example

Three capacitors: 2 μF, 3 μF in series, then that combination in parallel with 5 μF. V=12V = 12 V. Find total charge.

Series: 1/Cs=1/2+1/3=5/61/C_s = 1/2 + 1/3 = 5/6, Cs=1.2C_s = 1.2 μF

Parallel: Ceq=1.2+5=6.2C_{eq} = 1.2 + 5 = 6.2 μF

Q=CeqV=6.2×12=74.4Q = C_{eq}V = 6.2 \times 12 = 74.4 μC ✅

Concept Check 🎯

Problem-Solving Workshop 🧮

  1. 2 μF and 3 μF in series → 1.2 μF. In parallel with 5 μF → CeqC_{eq} (μF)? (Round to nearest integer.)

  2. Ceq=6.2C_{eq} = 6.2 μF, V=12V = 12 V. QQ (μC)? (Round to nearest integer.)

  3. C=10C = 10 μF, V=100V = 100 V. Energy (mJ)?

Concept Check 🔍

Practice

#ConfigurationKey Step
1Series-parallelSimplify step by step
2With dielectricMultiply CC by κ\kappa
3Energy sharingTwo capacitors connected

Challenge Question 📋

Part 7: Review & Applications

⚡ Review & Applications

Part 7 of 7 — Review & Applications

Key Formulas

  • C=Q/VC = Q/V, C=epsilon0A/dC = epsilon_0 A/d, C=κC0C = \kappa C_0
  • Series: 1/Ceq=∑1/Ci1/C_{eq} = \sum 1/C_i
  • Parallel: Ceq=∑CiC_{eq} = \sum C_i
  • Energy: U=12CV2=Q2/(2C)U = \frac{1}{2}CV^2 = Q^2/(2C)
  • Energy density: u=12epsilon0E2u = \frac{1}{2}epsilon_0 E^2

Worked Example

A 20 μF capacitor with κ=5\kappa = 5 dielectric is charged to V=50V = 50 V. Find the stored energy.

C=κC0=5(20)=100C = \kappa C_0 = 5(20) = 100 μF

U=12CV2=12(100×10−6)(2500)=0.125U = \frac{1}{2}CV^2 = \frac{1}{2}(100 \times 10^{-6})(2500) = 0.125 J ✅

Concept Check 🎯

Review & Applications 🧮

  1. C=100C = 100 μF, V=50V = 50 V. Energy (mJ)?

  2. κ=5\kappa = 5, C0=20C_0 = 20 μF. New CC (μF)?

  3. C0=100C_0 = 100 pF, κ=4\kappa = 4. New CC (pF)?

Concept Check 🔍

Practice

#TopicFormula
1CapacitanceC=Q/VC = Q/V
2Parallel plateC=epsilon0A/dC = epsilon_0 A/d
3Energy storageU=12CV2U = \frac{1}{2}CV^2

Challenge Question 📋