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Capacitance and Dielectrics — Notes

6 sections 3 min readBeta

1. Conductors and capacitance

  • In electrostatic equilibrium: E = 0 inside a conductor; excess charge resides on the outer surface; the conductor is an equipotential; just outside, E is normal to the surface. Free charges move through conductors; bound charges in insulators only shift within molecules.
  • Capacitance C=Q/VC = Q/V; unit farad (C V⁻¹); μF, nF, pF in practice. Depends on geometry and medium, not on Q.
  • Isolated sphere: C=4πε0RC = 4\pi\varepsilon_0R (R = 9 cm gives 10 pF; the Earth about 711 μF).

Formulas

  • Capacitance
    C=QVC = \dfrac{Q}{V}

Key terms

Capacitance

Ratio of the charge on a conductor to its potential (or on a capacitor’s plates to their potential difference).

Farad

Capacitance that holds one coulomb per volt.

2. Parallel-plate capacitor

Plates of area A, separation d, charges ±Q: uniform field E=σ/ε0=Q/(ε0A)E = \sigma/\varepsilon_0 = Q/(\varepsilon_0A); V=Ed=Qd/(ε0A)V = Ed = Qd/(\varepsilon_0A); so C=ε0AdC = \dfrac{\varepsilon_0A}{d} (edge effects neglected).

Formulas

  • Parallel-plate capacitor
    C=ε0AdC = \dfrac{\varepsilon_0A}{d}

3. Combinations

  • Series: same Q on each; V=∑ViV = \sum V_i; 1Cs=∑1Ci\dfrac{1}{C_s} = \sum \dfrac{1}{C_i}; smaller than the smallest; the smallest C takes the largest V.
  • Parallel: same V; Q=∑QiQ = \sum Q_i; Cp=∑CiC_p = \sum C_i; larger than the largest.
  • Mixed networks: reduce step by step; find Q on series branches and V across parallel ones.

Formulas

  • Series and parallel
    1Cs=∑1Ci,Cp=∑Ci\dfrac{1}{C_s} = \sum\dfrac{1}{C_i},\qquad C_p = \sum C_i

4. Energy stored

  • U=12CV2=12QV=Q22CU = \dfrac12CV^2 = \dfrac12QV = \dfrac{Q^2}{2C} (area under the V–Q line; formulae used, not derived, in the ISC scope).
  • Energy density of the field: u=12ε0E2u = \dfrac12\varepsilon_0E^2 (J m⁻³).
  • Connecting a charged C1C_1 (at V1V_1) to an uncharged C2C_2: charge conserved, common V=C1V1/(C1+C2)V = C_1V_1/(C_1 + C_2); energy is lost as heat.
  • V = Q/C (C = 2.0 μF)
Area under the V–Q line = work done in charging = ½QV.

Formulas

  • Energy stored
    U=12CV2=Q22CU = \dfrac12CV^2 = \dfrac{Q^2}{2C}
  • Energy density
    u=12ε0E2u = \dfrac12\varepsilon_0E^2

Key terms

Energy density

Energy stored per unit volume of the electric field, ½ε₀E².

5. Dielectrics

  • In a field a dielectric polarises (polar molecules align; non-polar molecules become induced dipoles); bound surface charges appear (− facing the + plate) and reduce the field inside to E0/KE_0/K.
  • Dielectric constant K=C′/C0=ε/ε0=εrK = C'/C_0 = \varepsilon/\varepsilon_0 = \varepsilon_r; filled capacitor C′=Kε0A/dC' = K\varepsilon_0A/d.
  • Battery disconnected (Q fixed): C × K; V, E, U ÷ K. Battery connected (V fixed): C, Q, U × K; E unchanged.

Formulas

  • Filled capacitor
    C′=KC0C' = KC_0

Key terms

Polarisation

Alignment or induction of molecular dipoles in a dielectric placed in a field.

Dielectric constant

K = C with dielectric ÷ C in vacuum = ε/ε₀.

6. Capacitor partly filled with a slab

Slab of thickness t (t < d), constant K: V=E0(d−t)+E0KtV = E_0(d - t) + \dfrac{E_0}{K}t, so C=ε0Ad−t+t/KC = \dfrac{\varepsilon_0A}{d - t + t/K}. t = 0 gives C0C_0; t = d gives KC0KC_0; a metal slab (K → ∞) gives ε0A/(d−t)\varepsilon_0A/(d - t).

Drawn to scale: d = 6.0 mm, t = 3.0 mm.
  • slab with K = 4
Half filled (t = d/2) with K = 4 gives C = 1.6 C₀.

Formulas

  • Partial slab
    C=ε0Ad−t+t/KC = \dfrac{\varepsilon_0A}{d - t + t/K}

Exam tips

  • Derive C = ε₀A/d from E = σ/ε₀ and V = Ed.
  • Series: common Q; parallel: common V.
  • Battery on or off decides whether V or Q is fixed.
  • Charge sharing: conserve Q; energy is lost.