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Capacitors and Dielectrics

Capacitance calculations, energy storage, and dielectric materials

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Capacitors and Dielectrics

Capacitance

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

where QQ is charge on each plate, VV is potential difference.

Units: 1 farad (F) = 1 coulomb/volt

Parallel Plate Capacitor

Area AA, separation dd:

E=σϵ0=Qϵ0AE = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}

V=Ed=Qdϵ0AV = Ed = \frac{Qd}{\epsilon_0 A}

C=QV=ϵ0AdC = \frac{Q}{V} = \frac{\epsilon_0 A}{d}

Other Geometries

Cylindrical Capacitor

Inner radius aa, outer radius bb, length LL:

C=2πϵ0Lln⁡(b/a)C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}

Spherical Capacitor

Inner radius aa, outer radius bb:

C=4πϵ0abb−aC = 4\pi\epsilon_0\frac{ab}{b-a}

Isolated Sphere

Radius RR (other conductor at infinity):

C=4πϵ0RC = 4\pi\epsilon_0 R

Capacitors in Series

Same charge QQ on each:

1Ceq=1C1+1C2+⋯\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Two capacitors: Ceq=C1C2C1+C2C_{eq} = \frac{C_1C_2}{C_1 + C_2}

Capacitors in Parallel

Same voltage VV across each:

Ceq=C1+C2+⋯C_{eq} = C_1 + C_2 + \cdots

Energy Stored in Capacitor

Work to charge capacitor:

W=∫0QV dq=∫0QqC dq=Q22CW = \int_0^Q V \, dq = \int_0^Q \frac{q}{C} \, dq = \frac{Q^2}{2C}

Energy: U=12Q2C=12CV2=12QVU = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}CV^2 = \frac{1}{2}QV

Energy density (parallel plate): u=UAd=12ϵ0E2u = \frac{U}{Ad} = \frac{1}{2}\epsilon_0 E^2

Dielectrics

Insulating material inserted between plates:

Dielectric constant: κ\kappa (or KK)

With dielectric:

  • Capacitance: C=κC0C = \kappa C_0
  • Electric field: E=E0/κE = E_0/\kappa
  • Potential: V=V0/κV = V_0/\kappa (if charge constant)

Permittivity of material: ϵ=κϵ0\epsilon = \kappa\epsilon_0

Modified equations: C=κϵ0AdC = \frac{\kappa\epsilon_0 A}{d}

u=12κϵ0E2u = \frac{1}{2}\kappa\epsilon_0 E^2

Dielectric Breakdown

Maximum field before dielectric breaks down:

Air: ~3×1063 \times 10^6 V/m

Different materials have different breakdown strengths.

Gauss's Law with Dielectrics

∮E⃗⋅dA⃗=Qfreeκϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{free}}{\kappa\epsilon_0}

or using electric displacement D⃗=κϵ0E⃗\vec{D} = \kappa\epsilon_0\vec{E}:

∮D⃗⋅dA⃗=Qfree\oint \vec{D} \cdot d\vec{A} = Q_{free}

Polarization

Dielectric polarizes in electric field:

Polarization: P⃗=κϵ0(κ−1)E⃗\vec{P} = \kappa\epsilon_0(\kappa - 1)\vec{E}

Bound surface charge: σb=P\sigma_b = P

This reduces net field inside dielectric.

Energy with Dielectric

If dielectric inserted with:

Constant charge: Uf=Ui/κU_f = U_i/\kappa (energy decreases)

Constant voltage: Uf=κUiU_f = \kappa U_i (energy increases)

Force on dielectric:

Dielectric pulled into capacitor (lower energy state).

📚 Practice Problems

1Problem 1medium

❓ Question:

A parallel-plate capacitor has circular plates of radius R = 5.0 cm separated by distance d = 2.0 mm. (a) Find the capacitance in vacuum. (b) If a dielectric material with κ = 3.5 is inserted, what is the new capacitance? (c) If the capacitor is charged to 12 V with the dielectric in place, how much charge is stored?

💡 Show Solution

Given:

  • R = 5.0 cm = 0.05 m
  • d = 2.0 mm = 0.002 m
  • κ = 3.5
  • V = 12 V
  • ε₀ = 8.85 × 10⁻¹² F/m

(a) Capacitance in vacuum:

The area of the circular plates is: A=πR2=π(0.05)2=7.85×10−3 m2A = \pi R^2 = \pi (0.05)^2 = 7.85 \times 10^{-3} \text{ m}^2

The capacitance is: C0=ϵ0Ad=(8.85×10−12)(7.85×10−3)0.002C_0 = \frac{\epsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(7.85 \times 10^{-3})}{0.002}

C0=3.47×10−11 F=34.7 pFC_0 = 3.47 \times 10^{-11} \text{ F} = \boxed{34.7 \text{ pF}}

(b) Capacitance with dielectric:

With a dielectric, the capacitance increases by factor κ: C=κC0=(3.5)(34.7 pF)C = \kappa C_0 = (3.5)(34.7 \text{ pF})

C=121 pFC = \boxed{121 \text{ pF}}

(c) Charge stored:

Using Q = CV: Q=CV=(121×10−12)(12)Q = CV = (121 \times 10^{-12})(12)

Q=1.45×10−9 C=1.45 nCQ = 1.45 \times 10^{-9} \text{ C} = \boxed{1.45 \text{ nC}}

2Problem 2hard

❓ Question:

Three capacitors (C₁ = 4.0 μF, C₂ = 6.0 μF, C₃ = 3.0 μF) are connected with C₁ and C₂ in series, and this combination is in parallel with C₃. The entire network is connected to a 24 V battery. Find: (a) the equivalent capacitance, (b) the charge on each capacitor, and (c) the voltage across each capacitor.

💡 Show Solution

Given:

  • C₁ = 4.0 μF, C₂ = 6.0 μF, C₃ = 3.0 μF
  • V = 24 V

(a) Equivalent capacitance:

C₁ and C₂ in series: 1C12=1C1+1C2=14.0+16.0=512\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4.0} + \frac{1}{6.0} = \frac{5}{12}

C12=2.4 μFC_{12} = 2.4 \text{ μF}

C₁₂ in parallel with C₃: Ceq=C12+C3=2.4+3.0C_{eq} = C_{12} + C_3 = 2.4 + 3.0

Ceq=5.4 μFC_{eq} = \boxed{5.4 \text{ μF}}

(b) Charges:

Total charge from battery: Qtotal=CeqV=(5.4)(24)=129.6 μCQ_{total} = C_{eq}V = (5.4)(24) = 129.6 \text{ μC}

For parallel combination:

  • Q₃ = C₃V = (3.0)(24) = 72 μC
  • Q₁₂ = C₁₂V = (2.4)(24) = 57.6 μC

In series, C₁ and C₂ have same charge:

  • Q₁ = Q₂ = Q₁₂ = 57.6 μC

(c) Voltages:

V1=Q1C1=57.64.0=14.4 VV_1 = \frac{Q_1}{C_1} = \frac{57.6}{4.0} = \boxed{14.4 \text{ V}}

V2=Q2C2=57.66.0=9.6 VV_2 = \frac{Q_2}{C_2} = \frac{57.6}{6.0} = \boxed{9.6 \text{ V}}

V3=24 VV_3 = \boxed{24 \text{ V}}

Note: V₁ + V₂ = 14.4 + 9.6 = 24 V ✓

3Problem 3medium

❓ Question:

A parallel-plate capacitor with capacitance C₀ = 50 pF is charged to voltage V₀ = 100 V and then disconnected from the battery. A dielectric slab with κ = 2.5 is then inserted between the plates. Find: (a) the charge on the capacitor (before and after), (b) the new voltage, and (c) the change in stored energy.

💡 Show Solution

Given:

  • C₀ = 50 pF
  • V₀ = 100 V
  • κ = 2.5
  • Capacitor is isolated (Q constant)

(a) Charge:

Initial charge: Q=C0V0=(50×10−12)(100)=5.0×10−9 CQ = C_0 V_0 = (50 \times 10^{-12})(100) = 5.0 \times 10^{-9} \text{ C}

Since capacitor is isolated, charge remains constant: Qbefore=Qafter=5.0 nCQ_{before} = Q_{after} = \boxed{5.0 \text{ nC}}

(b) New voltage:

With dielectric: C=κC0=(2.5)(50)=125 pFC = \kappa C_0 = (2.5)(50) = 125 \text{ pF}

Voltage decreases: V=QC=5.0×10−9125×10−12V = \frac{Q}{C} = \frac{5.0 \times 10^{-9}}{125 \times 10^{-12}}

V=40 VV = \boxed{40 \text{ V}}

(c) Change in energy:

Initial energy: U0=12C0V02=12(50×10−12)(100)2=2.5×10−7 JU_0 = \frac{1}{2}C_0 V_0^2 = \frac{1}{2}(50 \times 10^{-12})(100)^2 = 2.5 \times 10^{-7} \text{ J}

Final energy: U=12CV2=12(125×10−12)(40)2=1.0×10−7 JU = \frac{1}{2}CV^2 = \frac{1}{2}(125 \times 10^{-12})(40)^2 = 1.0 \times 10^{-7} \text{ J}

Change in energy: ΔU=U−U0=1.0×10−7−2.5×10−7\Delta U = U - U_0 = 1.0 \times 10^{-7} - 2.5 \times 10^{-7}

ΔU=−1.5×10−7 J\Delta U = \boxed{-1.5 \times 10^{-7} \text{ J}}

Energy decreases (absorbed by dielectric being pulled in).

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Capacitance calculations, energy storage, and dielectric materials
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