Capacitance

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Key Formulas:

• Definition: C = Q/V (1 Farad = 1 C/V)
• Parallel plate capacitor: C = ε₀A/d (no dielectric)
• With dielectric: C = KC₀ where K is dielectric constant
• Energy stored: U = ½CV² = ½QV² = Q²/(2C)
• Series combination: 1/C = 1/C₁ + 1/C₂
• Parallel combination: C = C₁ + C₂
• Electric field in parallel plate: E = σ/ε₀ = V/d
• Force between capacitor plates: F = Q²/(2ε₀A)

Exam Tips:

• Capacitors store energy in the electric field between plates
• Adding a dielectric increases capacitance and reduces the field
• In series, voltage divides; in parallel, charge divides
• The Van de Graaff generator uses the principle of charge accumulation on a hollow conductor
• When two capacitors connect, charge redistributes until potentials are equal
• Energy is NOT conserved in redistribution — some energy is lost as heat/spark
• Capacitor blocks DC steady state (open circuit) but allows AC (impedance Z = 1/ωC)

🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of capacitance and its applications.

Types of Capacitors:

Parallel plate capacitors consist of two parallel conducting plates separated by a vacuum or dielectric. The capacitance depends on plate area (larger area → more charge stored per volt), plate separation (closer plates → higher capacitance), and the dielectric material between plates.

Cylindrical capacitors consist of two concentric cylinders with a dielectric in between. The capacitance per unit length is C/L = 2πε₀/ln(b/a), where a and b are the inner and outer radii. This configuration minimises field at edges and is used in coaxial cables.

Spherical capacitors consist of two concentric spherical shells. The capacitance is C = 4πε₀(ab)/(b-a), where a and b are the inner and outer radii. An isolated sphere can be treated as a spherical capacitor with the outer sphere at infinity, giving C = 4πε₀R.

Derivation of Parallel Plate Capacitance:

Consider two parallel plates each with area A, separated by distance d, with vacuum between them. The electric field between the plates (ignoring edge effects) is uniform: E = σ/ε₀, where σ = Q/A is the surface charge density. The potential difference between the plates is V = Ed = σd/ε₀ = (Qd)/(ε₀A). From C = Q/V, we get C = ε₀A/d. This derivation assumes the field is uniform and fringing effects are negligible — valid when d << √A.

Dielectrics at Molecular Level:

A dielectric placed between capacitor plates polarises: its molecules develop induced dipole moments aligning opposite to the existing field. This reduces the effective field inside the dielectric to E = E₀/K, where K is the dielectric constant (always K ≥ 1). Polar dielectrics (water, acetone) have permanent molecular dipoles that align with the field; non-polar dielectrics (air, PTFE) develop induced dipoles only.

The dielectric constant K = ε_r = ε/ε₀, where ε is the permittivity of the dielectric. For vacuum, K = 1; for air, K ≈ 1.0006 (often taken as 1); for water, K ≈ 80; for glass, K ≈ 4-8; for mica, K ≈ 6.

Dielectric Strength and Breakdown Voltage:

Every dielectric has a maximum electric field it can withstand before electrical breakdown occurs. For air, breakdown field ≈ 3 × 10⁶ V/m. When this is exceeded, ionisation occurs and the material conducts — producing a spark. Dielectric strength is the maximum field before breakdown: E_max = V_breakdown/d. High-voltage capacitors use dielectrics with high breakdown strengths (plastic films, ceramic, mica).

Energy Stored in a Capacitor:

Charging a capacitor requires work. The incremental work dW to add charge dq to a capacitor already at potential V(q) = q/C is dW = V dq = (q/C) dq. Integrating from 0 to Q: W = ∫[0 to Q] (q/C) dq = Q²/(2C). Since V = Q/C, alternative forms are U = ½QV = ½CV² = Q²/(2C). The energy density (energy per unit volume) in the electric field is u = ½ε₀E² for vacuum, or u = ½ε₀K E² = ½εE² with a dielectric.

Force Between Capacitor Plates:

The two plates of a charged capacitor attract each other. The force can be calculated by considering the work done when the plates move an infinitesimal distance dx apart: dW = F dx = energy change. Since C = ε₀A/x depends on separation x, the energy U = Q²/(2C) = Q²x/(2ε₀A). Therefore F = dU/dx = Q²/(2ε₀A) = ½ε₀E²A. This force is used in capacitor-based sensors and actuators.

Charge Redistribution Between Capacitors:

When two capacitors at different potentials are connected together (positive to positive, negative to negative), charge redistributes until the potentials are equal. The total charge is conserved: Q_total = C₁V₁ + C₂V₂ = (C₁ + C₂)V. The common potential is V = (C₁V₁ + C₂V₂)/(C₁ + C₂). The energy after redistribution is ½(C₁ + C₂)V², which is less than the initial energy ½C₁V₁² + ½C₂V₂². The difference appears as heat in the connecting wires or as a spark. This is a common NEET trap question.

Van de Graaff Generator:

The Van de Graaff generator is a device that accumulates charge on a hollow conducting sphere using a moving belt. A motor drives an insulating belt past a comb electrode at the bottom, which sprays charge onto the belt. At the top, another comb collects charge and transfers it to the outer shell. Charge accumulates on the outer surface because it resides on conductors to minimise repulsion. The process continues until the surface field equals the dielectric strength of air, limiting further accumulation. Applications include particle accelerators, X-ray machines, and demonstrations of high voltage.

Conversion to Ammeter and Voltmeter:

A moving coil galvanometer can be converted to an ammeter by connecting a low resistance shunt in parallel: the shunt carries most of the current while only a small fraction goes through the galvanometer. The shunt resistance S = G/(n - 1), where n = full-scale current/I_g.

For a voltmeter, a high resistance multiplier is connected in series with the galvanometer: total resistance = G + R, and the scale is multiplied by (1 + R/G). The voltmeter consumes some current from the circuit being measured, causing a loading effect — high-quality voltmeters have very high resistance to minimise this.

Worked NEET Example: Two capacitors, C₁ = 2 μF at V₁ = 100 V and C₂ = 4 μF at V₂ = 50 V, are connected in parallel (positive plates together). Find final voltage and charge redistribution.

• Initial charges: Q₁ = 200 μC, Q₂ = 200 μC, total Q = 400 μC
• Total capacitance: C = 6 μF
• Common voltage: V = 400/6 ≈ 66.7 V
• Final charges: Q₁’ = 133.3 μC, Q₂’ = 266.7 μC
• Energy before: ½(2×10⁻⁶)(100)² + ½(4×10⁻⁶)(50)² = 10 mJ + 5 mJ = 15 mJ
• Energy after: ½(6×10⁻⁶)(66.7)² ≈ 13.3 mJ
• Energy lost: 1.7 mJ (appears as heat)

Grading of Capacitors for High Voltage:

When a high voltage is applied across a capacitor, the field distribution matters. Using multiple series-connected sections (“grading”) equalises the voltage distribution across the dielectric and prevents local breakdown. A single thick dielectric may have voids or weaknesses; grading distributes stress more evenly.

🔴 Extended — Deep Study (3mo+)

Comprehensive derivation-based theory for complete mastery of capacitance.

Derivation of Spherical Capacitance:

Consider a conducting sphere of radius R surrounded by a concentric spherical shell of radius r > R. The inner sphere carries charge +Q and the outer shell carries charge -Q. The field in the region R < r < r₀ (between the spheres) is E = Q/(4πε₀r²), directed radially outward. The potential difference is V = ∫[R to r₀] E dr = Q/(4πε₀) ∫[R to r₀] (1/r²) dr = Q/(4πε₀)(1/R - 1/r₀). For an isolated sphere (outer shell at infinity, r₀ → ∞), V = Q/(4πε₀R). Hence C = Q/V = 4πε₀R. For finite outer radius r₀, C = 4πε₀R × (r₀/(r₀ - R)).

Derivation of Cylindrical Capacitance:

Consider two coaxial cylinders of radii a and b (b > a), length L >> a, b. The field at distance r from axis (a < r < b) due to inner cylinder charge +λ per unit length is E = λ/(2πε₀r). The potential difference is V = ∫[a to b] E dr = λ/(2πε₀) ln(b/a). Therefore C/L = λ/V = 2πε₀/ln(b/a). For a complete cylinder of length L: C = 2πε₀L/ln(b/a). This is the basis for the capacitance of coaxial cables and cylindrical capacitors.

Redistribution Energy Loss Derivation:

Consider two capacitors C₁ and C₂ at potentials V₁ and V₂. Initial energy: U_i = ½C₁V₁² + ½C₂V₂². After connection, common potential V = (C₁V₁ + C₂V₂)/(C₁ + C₂). Final energy: U_f = ½(C₁ + C₂)V². Substituting V and simplifying: U_i - U_f = ½C₁C₂(V₁ - V₂)²/(C₁ + C₂).

This is always positive (energy lost), being zero only if V₁ = V₂ (already at same potential). The energy appears as heat in the connecting wires, electromagnetic radiation (spark), or mechanical vibration. This is crucial for NEET questions — students often assume charge and energy are both conserved in redistribution, but only charge is conserved.

Van de Graaff: Continuous Charge Accumulation:

In a Van de Graaff generator, charge is transported by the moving belt to the inner surface of the hollow conductor. The outer surface charge density σ and surface field E = σ/ε₀ increase as more charge is added. When E reaches the breakdown strength of air (~3 × 10⁶ V/m), ionisation occurs and charge leaks into the surrounding air. This sets the maximum achievable voltage: V_max = E_max × R, where R is the radius of the sphere. Larger spheres can reach higher voltages before breakdown.

Force on Dielectric Partially Inserted:

When a dielectric slab is partially pulled out of a capacitor connected to a constant voltage source V, the force pulling it back in can be derived. The capacitance with partially inserted dielectric of length x is C(x) = C₀(K - 1)(x/L) + C₀. The stored energy U = ½C(x)V². The force is F = dU/dx = ½V² × dC/dx. For a parallel plate with dielectric constant K and insertion fraction x/L: F = ½V² ε₀W(K-1)/d, where W is plate width and d is separation. This is positive (force tends to pull dielectric back in) when V is held constant.

Dielectric Behaviour at High Frequencies:

At low frequencies, polar molecules can follow the alternating field, maintaining high polarisation and thus high K. At high frequencies, molecular relaxation prevents dipoles from following the field, reducing effective K. This frequency dependence is described by the complex permittivity ε(ω) = ε’(ω) - iε”(ω). The loss factor ε” peaks at the relaxation frequency, important for AC applications and microwave heating.

Series and Parallel Combination Rules:

For n capacitors in series: 1/C_eq = Σ(1/C_i). The voltage across each is V_i = Q/C_i (same charge Q for all since current is same in series). The largest voltage appears across the smallest capacitance — important for voltage grading.

For n capacitors in parallel: C_eq = Σ C_i. The voltage across all is the same; charge divides: Q_i = C_iV. Larger capacitance stores more charge.

Common Trap Questions in NEET:

1. “Two capacitors are connected and then disconnected — what is the final charge?” Answer: Charge is conserved, voltage equalises on connection but after disconnection, charge stays on each capacitor.

2. “Two capacitors are connected while remaining connected to battery?” Answer: The battery holds voltage constant, so the common voltage equals battery voltage. The total capacitance determines total charge drawn from the battery.

3. “Charge conserved but voltage equalises on connection” — students often forget voltage equalises.

4. “Energy is not conserved in redistribution” — this is counter-intuitive and frequently tested.

5. “Earthing a conductor discharges it completely” — the grounded conductor can exchange charge with Earth until potential is zero.

6. “Capacitor blocks DC steady state” — after transients die out, capacitor allows no current through DC, acting as an open circuit.

7. “Capacitor allows AC” — capacitive reactance X_C = 1/ωC decreases with increasing frequency, approaching zero at very high frequencies.

Solving NEET Previous Year Questions:

Strategy: Identify knowns and unknowns, check whether capacitors are connected in series or parallel, determine if the connection is to a battery or isolated, write charge conservation equations, solve for unknowns, calculate final energy if asked, check for energy loss.

Common question pattern: Three capacitors in a network, find equivalent capacitance, then energy stored or charge on a specific capacitor. Always draw the circuit clearly, identify series and parallel groups, combine systematically from innermost groups outward.