Capacitance

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Capacitance — Key Facts Capacitance: $C = \frac{Q}{V}$; farad (F) = coulomb/volt; $1 \mu F = 10^{-6}$ F, $1 pF = 10^{-12}$ F Parallel plate capacitor (vacuum): $C = \frac{\varepsilon_0 A}{d}$; $A$ = plate area (m²), $d$ = separation (m) Energy stored: $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$ Dielectric: increases capacitance by factor $K$ (dielectric constant); $C’ = KC$; stored energy reduces to $U’ = U/K$ ⚡ Exam tip: Capacitors in parallel add directly ($C_{eq} = C_1 + C_2$); in series, reciprocals add ($\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$)

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of capacitance and dielectric behaviour.

Capacitance — CUET Physics Study Guide

A capacitor consists of two conductors separated by an insulator (dielectric or vacuum). When a voltage $V$ is applied, charge $+Q$ accumulates on one plate and $-Q$ on the other. The capacitance $C$ measures the ability to store charge: $C = Q/V$. A larger capacitance means more charge can be stored at a given voltage.

Parallel Plate Capacitor: With vacuum between plates, $C = \frac{\varepsilon_0 A}{d}$. This shows that capacitance increases with larger plate area and decreases with smaller plate separation — intuitive since larger plates hold more charge and closer plates reduce the potential difference for a given charge. The permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m.

Dielectric Materials: Inserting a dielectric increases capacitance to $C = \frac{K\varepsilon_0 A}{d}$ where $K$ is the dielectric constant. Common dielectrics: air ($K \approx 1$), mica ($K \approx 5$), glass ($K \approx 4-7$), paper ($K \approx 3$), ceramic ($K \approx 100-10000$). The dielectric reduces the electric field to $E = E_0/K$ while increasing capacitance.

Combination of Capacitors:

• Parallel: $C_{eq} = C_1 + C_2 + …$ (voltage same across all, charge adds)
• Series: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$ (charge same on all, voltage adds)

Energy Density: The energy stored per unit volume in the electric field is $u = \frac{1}{2}\varepsilon_0 E^2$ (for vacuum). With dielectric, $u = \frac{1}{2}K\varepsilon_0 E^2$.

Example: Two capacitors $C_1 = 3 \mu F$ and $C_2 = 6 \mu F$ are connected in series across a 100 V supply.

• Equivalent capacitance: $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$, so $C_{eq} = 2 \mu F$
• Charge on each: $Q = C_{eq}V = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4}$ C
• Voltage across $C_1$: $V_1 = Q/C_1 = 66.7$ V; across $C_2$: $V_2 = Q/C_2 = 33.3$ V

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Capacitance — Complete CUET Physics Notes

Force Between Plate Charges: Two parallel plates with opposite charges attract each other. The force on one plate due to the other: $F = \frac{Q^2}{2\varepsilon_0 A}$ (in vacuum). The work needed to increase separation from $d_1$ to $d_2$ equals the change in stored energy.

Capacitor with Dielectric Partially Filled: When a dielectric slab of thickness $t < d$ is inserted between plates, the effective capacitance depends on what fraction is filled. For a slab filling partially (cross-sectional), treat as two capacitors in series: $C = \frac{\varepsilon_0 A}{(d-t) + t/K}$.

Leakage Current and Time Constant: A real capacitor slowly discharges through its dielectric (leakage resistance $R$). The time constant $\tau = RC$ gives the time for voltage to drop to $1/e$ of initial value. For a charging capacitor in an RC circuit: $V(t) = V_0(1 - e^{-t/RC})$; discharge: $V(t) = V_0 e^{-t/RC}$.

Van de Graaff Generator: A large spherical capacitor achieving very high voltages. The sphere of radius $R$ has capacitance $C = 4\pi\varepsilon_0 R$. Maximum potential $V = Q/C = Q/(4\pi\varepsilon_0 R)$. The sphere potential is limited by dielectric breakdown of surrounding air ($E_{max} \approx 3 \times 10^6$ V/m).

Dielectric Breakdown: When electric field exceeds a threshold, dielectric conducts — creating a spark. For air, breakdown field $\approx 3 \times 10^6$ V/m. This limits the maximum charge a capacitor can hold. $V_{max} = E_{max} \times d$.

Cylindrical Capacitor: Two coaxial cylinders of radii $a$ and $b > a$, length $L$: $C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}$. Spherical capacitor (two concentric spheres): $C = 4\pi\varepsilon_0 \frac{ab}{b-a}$.

Effect of Plate Area and Separation on Energy: Energy can be expressed as $U = \frac{Q^2}{2C} = \frac{Q^2 d}{2\varepsilon_0 A}$ — for fixed charge, energy increases with plate separation (useful in capacitor microphone). For fixed voltage, $U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{\varepsilon_0 A}{d}V^2$ — energy decreases as plates are separated (pulling plates apart requires external work).

CUET Exam Patterns (2022–2024):

• Combination of capacitors (series/parallel) is a frequent 2-mark question
• Energy stored in capacitor tested in 2023 with dielectric insertion
• Time constant $\tau = RC$ rarely tested in CUET but appears in JEE
• Common mistakes: confusing series and parallel formulas, forgetting the $1/2$ in energy formula

⚡ Key insight: When capacitors are connected and then disconnected from the battery, charge is conserved (total charge remains fixed). When connected to a battery, voltage is fixed and charge redistributes. Choose the right conservation law based on whether the battery is connected or disconnected.

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