ElectromagnetismHigh Yield (2-3 Qs in JEE)

Capacitance & Dielectrics Formulas

Parallel plate capacitors, dielectric insertion, stored energy, common potential, and transient RC circuits.

8 Formulas•431+ Questions in Bank•

Concept Diagram

Parallel Plate Capacitor & Dielectric

Parallel plates of area A and separation d, uniform electric field E = σ/ε₀, and dielectric polarization.

Core Formulas & Mathematical Identities

Parallel Plate Capacitance

C = \frac{\varepsilon_0 A}{d}, \quad C_{\text{dielectric}} = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}

Diagram Reference:Parallel Plate Capacitance

Capacitance with vacuum and with dielectric slab of thickness t and constant K.

Examiner Tip: Filling entire gap with dielectric increases capacitance to K*C₀.

Stored Energy & Energy Density

U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV, \quad u_E = \frac{1}{2}\varepsilon_0 E^2

Electrostatic potential energy stored in the electric field between capacitor plates.

Examiner Tip: Energy density u_E has units of Joules per cubic meter (J/m³).

Combinations of Capacitors

C_{\text{parallel}} = \sum C_i, \quad \frac{1}{C_{\text{series}}} = \sum \frac{1}{C_i}

Series combination has same charge on each capacitor; parallel has same potential difference.

Examiner Tip: Notice rules are opposite to resistor combinations.

Charge Sharing & Heat Loss

V_{\text{common}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}, \quad \Delta H = \frac{1}{2}\frac{C_1 C_2}{C_1 + C_2}(V_1 - V_2)^2

Common potential and heat dissipated when two charged capacitors are connected.

Examiner Tip: Heat loss ΔH is always positive (energy is lost, never gained).

Transient RC Circuit (Charging & Discharging)

q(t) = Q_0(1 - e^{-t/RC}) \; (\text{Charging}), \quad q(t) = Q_0 e^{-t/RC} \; (\text{Discharging}), \quad \tau = RC

Exponential charging and discharging with capacitive time constant τ = RC.

Examiner Tip: At t = τ, capacitor charges to ~63.2% of Q₀; at t = 5τ, it is ~99.3% charged.

Attractive Force Between Plates & Electrostatic Pressure

F = \frac{Q^2}{2\varepsilon_0 A} = \frac{1}{2}QE, \quad P_{\text{elec}} = \frac{F}{A} = \frac{\sigma^2}{2\varepsilon_0} = \frac{1}{2}\varepsilon_0 E^2

Mutual electrostatic attraction between oppositely charged plates, and electrostatic outward pulling pressure.

Examiner Tip: The factor of 1/2 arises because the charge on one plate experiences the electric field E/2 produced solely by the other plate.

Spherical & Cylindrical Capacitor Geometries

C_{\text{spherical}} = 4\pi\varepsilon_0 \left(\frac{ab}{b - a}\right), \quad C_{\text{isolated sphere}} = 4\pi\varepsilon_0 R, \quad \frac{C_{\text{cylindrical}}}{L} = \frac{2\pi\varepsilon_0}{\ln(b/a)}

Capacitance formulas for concentric spherical shells and coaxial cylindrical conductors of radii a and b (b > a).

Examiner Tip: For Earth (R ≈ 6400 km), isolated spherical capacitance is C ≈ 711 μF.

Dielectric Slab Partially Inserted (Series & Parallel)

\text{Slab (thickness } t\text{, full area):}\; C = \frac{\varepsilon_0 A}{d - t + t/K} \quad \text{(series stacking)}\\[4pt] \text{Slab (partial area } A_1\text{, rest } A_2\text{):}\; C = \frac{\varepsilon_0 K A_1}{d} + \frac{\varepsilon_0 A_2}{d} = \frac{\varepsilon_0}{d}(K A_1 + A_2)

When dielectric occupies only part of the gap thickness (series): effective gap is reduced by t(1 - 1/K). When dielectric covers part of the plate area (parallel): two sub-capacitors add directly.

Examiner Tip: Battery connected while inserting: V = const → Q increases, C increases, E between plates unchanged. Battery disconnected: Q = const → V decreases, C increases, E decreases. Energy is LOST when battery connected, REDUCED when disconnected.

Top Examiner Traps & Pitfalls

Frequently tested misconceptions in Capacitance & Dielectrics

•Battery connected vs battery disconnected when inserting dielectric: battery connected -> V remains constant; battery disconnected -> Q remains constant.
•Energy lost during charge sharing is dissipated as heat and electromagnetic radiation.

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