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Capacitance

Part of the JEE Main study roadmap. Physics topic phy-016 of Physics.

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Capacitance

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Capacitance measures how much charge a conductor system stores per unit potential difference: C = Q/V, where Q is in coulomb, V in volt, and C in farad (1 F = 1 C/V). The geometry of a parallel-plate capacitor gives C = ε₀A/d, with ε₀ = 8.854 × 10⁻¹² F/m, A the plate area in m², and d the plate separation in m. Inserting a dielectric of relative permittivity K multiplies capacitance by K: C = Kε₀A/d.

Energy stored is U = ½CV² = Q²/(2C) = ½QV in joules, equivalent to field energy density u = ½ε₀E² J/m³. The RC time constant governs transient charging and discharging through τ = RC seconds.

Exam pointers:

  • Series: 1/C_eq = Σ 1/C_i; charge identical, voltage divides.
  • Parallel: C_eq = Σ C_i; voltage identical, charge divides.
  • Dielectric inserted, battery connected → V fixed, Q rises by K.
  • Dielectric inserted, battery disconnected → Q fixed, V drops by K.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Definition and the Farad

A capacitor is a two-conductor system that stores electric energy in the field between its plates. Capacitance is the proportionality constant between charge on one plate and the potential difference across them:

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

The SI unit is the farad. Most practical capacitors sit in the picofarad to microfarad range; 1 F would require plates of order 10⁷ m² at 1 mm separation.

Parallel-Plate Geometry and the Dielectric

For two parallel plates of area A separated by distance d in vacuum, the field between them is uniform with E = σ/ε₀ = Q/(ε₀A). The potential difference V = Ed = Qd/(ε₀A), giving:

$$C_0 = \frac{\varepsilon_0 A}{d}$$

Filling the gap with a dielectric of relative permittivity K (also written ε_r) reduces the internal field by polarization, and the capacitance becomes:

$$C = \frac{K\varepsilon_0 A}{d}$$

Dielectric constants: air ≈ 1.0006, water ≈ 80, mica ≈ 6, ceramic ≈ 10–10⁴.

Combinations

ConfigurationEquivalent CCharge on eachVoltage across each
Series (n identical, each C₀)C₀/nSame Q on allV divides equally if equal
Parallel (n identical, each C₀)nC₀Divides equallySame V across all

In series, the equivalent is always smaller than the smallest individual capacitor; in parallel, always larger.

Energy Storage

Work done by the battery during charging accumulates in the field between the plates. Integrating V from 0 to the final voltage gives:

$$U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$$

This equals the volume integral of energy density u = ½ε₀E² over the gap region — an explicit confirmation that energy resides in the field, not on the metal surfaces.

RC Transients

When a capacitor charges through resistor R from source V₀:

$$Q(t) = Q_0!\left(1 - e^{-t/RC}\right), \qquad I(t) = I_0 e^{-t/RC}$$

Discharging follows Q(t) = Q₀e^(−t/RC). At t = τ = RC, the charge reaches about 63.2% of its final value during charging or drops to 36.8% during discharging.

Common Exam Traps

  • Confusing ε₀ with K; the formula needs the product Kε₀.
  • Forgetting that charge is conserved (not voltage) when a dielectric is inserted on an isolated capacitor.
  • Treating a capacitor as a wire in steady DC — in steady state it blocks current.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Geometries and Other Capacitor Types

Beyond parallel plates, JEE occasionally tests cylindrical and spherical capacitors. For a spherical capacitor with inner radius a and outer radius b in vacuum:

$$C = \frac{4\pi\varepsilon_0 ab}{b-a}$$

For a long coaxial cylinder of length L with inner radius a and outer radius b:

$$C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}$$

The Van de Graaff generator is essentially a spherical capacitor: a high-voltage terminal sphere of radius R stores charge Q with C = 4πε₀R, and the achievable voltage is limited by dielectric breakdown of the surrounding air (~3 × 10⁶ V/m).

Sharing of Charge Between Connected Capacitors

When two charged capacitors (C₁ at V₁, C₂ at V₂) are connected positive-to-positive, charge redistributes until they reach a common potential V. By conservation of total charge:

$$V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$$

Energy is lost in the redistribution as heat in the connecting wires — a direct analogue of inelastic collisions.

Force on a Dielectric Slab

A partially inserted dielectric slab is pulled into a charged isolated capacitor because doing so increases C, lowering energy U = Q²/(2C) at constant Q. The attractive force per unit area on the slab equals ½ε₀(K − 1)E² where E is the original field without the slab. Similarly, two parallel plates carrying opposite charges attract each other with:

$$F = \frac{Q^2}{2\varepsilon_0 A}$$

Common Mistakes in JEE Problems

  1. Computing C_eq of a network without redrawing the symmetry — Wheatstone-bridge-style capacitor networks have a balanced condition under which the middle branch carries no charge.
  2. Using U = ½CV² with the post-dielectric C while the charge Q stays the same — energy must be recomputed as Q²/(2C) instead.
  3. Ignoring the role of fringing fields in non-ideal geometries; the parallel-plate formula assumes d ≪ √A.

Practice Prompts

  • A 10 μF capacitor charged to 100 V is connected across an uncharged 30 μF capacitor. Find the common voltage, charge on each, and energy dissipated in the process.
  • A parallel-plate capacitor with plate area 100 cm² and separation 2 mm is half-filled with a dielectric of K = 4. Determine the equivalent capacitance as a function of insertion depth and identify the configuration that gives the minimum value.

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