Capacitance

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Capacitance — Key Facts

Capacitance (C) is the ability of a system to store electric charge: C = Q/V. The SI unit is Farad (F), though most practical capacitors are in μF, nF, or pF.

Parallel Plate Capacitor:

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

where A = plate area, d = plate separation, ε₀ = 8.85 × 10⁻¹² F/m

With dielectric: $C’ = KC = \frac{K\varepsilon_0 A}{d}$

where K = dielectric constant (relative permittivity)

Energy Stored in Capacitor:

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

Energy density in electric field: $$u = \frac{1}{2}\varepsilon_0 E^2 \quad \text{(in vacuum)}$$ $$u = \frac{1}{2}K\varepsilon_0 E^2 \quad \text{(with dielectric)}$$

⚡ JEE Exam Tip: The dielectric constant K > 1 always. When a dielectric is inserted, if the capacitor is disconnected from battery, charge stays constant but voltage decreases (C increases, V = Q/C decreases, U decreases — energy is extracted). If connected to battery, voltage stays constant but charge increases.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding…

Capacitor Combinations:

Series Combination: $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + …$$

For two capacitors: $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$

In series: same charge Q on each capacitor; voltage drops add: V = V₁ + V₂

Parallel Combination: $$C_{eq} = C_1 + C_2 + …$$

In parallel: same voltage V across each; charges add: Q = Q₁ + Q₂

Effect of Dielectrics:

Condition	Q	V	C	U
Battery connected (V constant)	increases	constant	increases	changes
Disconnected (Q constant)	constant	decreases	increases	decreases

Dielectric Strength:

Maximum electric field before breakdown:

• Air: ~3 × 10⁶ V/m
• Mica: ~100 × 10⁶ V/m
• Paper: ~16 × 10⁶ V/m

Maximum voltage: V_max = E_breakdown × d

⚡ JEE Exam Tip: For capacitors with different dielectrics filling partial space, calculate as series/parallel combination of partial capacitors. If dielectric fills half the volume, treat as two capacitors in parallel each with area A/2.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Derivation of Capacitance of Parallel Plate:

Consider parallel plate capacitor with plate area A, separation d, charges +Q and -Q.

Surface charge density: $\sigma = \frac{Q}{A}$

Electric field between plates (from one plate): $E_1 = \frac{\sigma}{2\varepsilon_0}$ (directed away)

Electric field between plates (from two plates): $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$

Potential difference: $V = Ed = \frac{Qd}{\varepsilon_0 A}$

Therefore: $C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d}$

Spherical Capacitor:

Conducting sphere in free space: $$C = 4\pi\varepsilon_0 R$$

Concentric spherical shells (inner radius a, outer radius b): $$C = \frac{4\pi\varepsilon_0 ab}{b-a}$$

Special cases:

• When outer sphere is at infinity: C = 4πε₀a (same as isolated sphere)
• When a → b (thin spherical capacitor): $C \approx 4\pi\varepsilon_0 b^2/(b-a) = 4\pi\varepsilon_0 b \cdot \frac{b}{b-a}$ (like parallel plate with curvature)

Cylindrical Capacitor (coaxial cylinders):

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

where L = length, a = inner radius, b = outer radius

Force on Dielectric:

When dielectric is partially inserted into capacitor connected to battery:

• Force is attractive and pulls dielectric inside
• Work done = change in stored energy

$$F = \frac{\varepsilon_0 bV^2}{2d}\left(\frac{1}{\ln(b/a)}\right)^2 \times \text{(for cylindrical)}$$

Charging/Discharging of Capacitor:

Charging: $q = Q_0(1 - e^{-t/RC})$ Discharging: $q = Q_0 e^{-t/RC}$

Time constant: $\tau = RC$

After time τ: capacitor reaches 63% of final charge (charging) or has 37% remaining (discharging)

Energy in Charging Process:

When capacitor charges through resistor, only 50% of energy supplied by battery becomes stored energy in capacitor. The other 50% is dissipated as heat in resistor.

$$P_{avg} = \frac{1}{2}CV^2 \quad \text{(stored)}$$ $$P_{dissipated} = \frac{1}{2}CV^2 \quad \text{(as heat)}$$

Van de Graaff Generator:

Uses corona discharge to charge a hollow sphere. Maximum charge: $Q_{max} = C \cdot V_{breakdown}$ Capacitance of sphere: $C = 4\pi\varepsilon_0 R$ Maximum voltage before air breaks down: $V = E_{max} \cdot R$

⚡ JEE Advanced 2023 Analysis: Questions involving cylindrical capacitors, energy calculations with dielectrics, and RC circuits have appeared. The combination of series/parallel capacitors with multiple dielectrics requires careful application of the dielectric constant. Remember: the energy density formula u = ½ε₀E² applies to any region with an electric field.

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