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Capacitor formulas

Master Capacitor through 32 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

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Capacitor, every formula

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Charge–capacitance relation
$q = C V$
Parallel plate capacitor
$C = \frac{ε _{0} A}{d}$
Q1MCQParallel plate
The capacitance of a parallel plate capacitor is:
- A
  $\dfrac{\varepsilon_0 A}{d}$
- B
  $\dfrac{\varepsilon_0 d}{A}$
- C
  $\varepsilon_0 A d$
- D
  $\dfrac{A}{d}$
Capacitor with dielectric
$C = \frac{K ε _{0} A}{d} = K C_{0}$
Q1MCQDielectric
Inserting a dielectric of constant $K$ changes the capacitance to:
- A
  $KC_0$
- B
  $\dfrac{C_0}{K}$
- C
  $C_0$
- D
  $K^{2}C_0$
Isolated spherical conductor
$C = 4 π ε_{0} R$
Q1MCQIsolated sphere
The capacitance of an isolated sphere of radius $R$ is:
- A
  $4\pi\varepsilon_0 R$
- B
  $\dfrac{4\pi\varepsilon_0}{R}$
- C
  $\dfrac{\varepsilon_0}{R}$
- D
  $\varepsilon_0 R^{2}$
Spherical capacitor
$C = \frac{4 π ε _{0} ab}{b - a}$
Q1MCQSpherical capacitor
The capacitance of a spherical capacitor (radii $a<b$) is:
- A
  $\dfrac{4\pi\varepsilon_0 ab}{b-a}$
- B
  $4\pi\varepsilon_0(b-a)$
- C
  $\dfrac{4\pi\varepsilon_0}{ab}$
- D
  $4\pi\varepsilon_0 ab$
Cylindrical capacitor
$C = \frac{2 π ε _{0} L}{ln ( b / a )}$
Q1MCQCylindrical
The capacitance of a cylindrical capacitor is:
- A
  $\dfrac{2\pi\varepsilon_0 L}{\ln(b/a)}$
- B
  $\dfrac{\varepsilon_0 L}{b-a}$
- C
  $2\pi\varepsilon_0 L\ln(b/a)$
- D
  $\dfrac{2\pi\varepsilon_0}{L}$
Energy stored
$U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$
Energy density
$u = \frac{1}{2} ε_{0} E^{2}$
Q1MCQEnergy density
The energy density of an electric field is:
- A
  $\tfrac12\varepsilon_0 E^{2}$
- B
  $\varepsilon_0 E^{2}$
- C
  $\tfrac12\varepsilon_0 E$
- D
  $\dfrac{E^{2}}{2\varepsilon_0}$
Field between plates
$E = \frac{σ}{ε _{0}} = \frac{V}{d}$
Q1MCQField between plates
The field between capacitor plates is:
- A
  $\dfrac{V}{d}$
- B
  $Vd$
- C
  $\dfrac{d}{V}$
- D
  $\varepsilon_0 V$
Force on a plate
$F = \frac{q ^{2}}{2 A ε _{0}}$
Q1MCQForce on plate
The force on one plate of a capacitor is:
- A
  $\dfrac{q^{2}}{2A\varepsilon_0}$
- B
  $\dfrac{q^{2}}{A\varepsilon_0}$
- C
  $qE$
- D
  $\dfrac{q}{2A\varepsilon_0}$
Series combination
$\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \dots$
Parallel combination
$C_{e q} = C_{1} + C_{2} + \dots$
n identical in series
$C_{e q} = \frac{C}{n}, V^{'} = \frac{V}{n}$
Q1MCQn series
Three identical $C$ capacitors in series give:
- A
  $\dfrac{C}{3}$
- B
  $3C$
- C
  $C$
- D
  $\dfrac{3}{C}$
n identical in parallel
$C_{e q} = n C$
Q1MCQn parallel
Three identical $C$ capacitors in parallel give:
- A
  $3C$
- B
  $\dfrac{C}{3}$
- C
  $C$
- D
  $9C$
Voltage division (series)
$V_{1} : V_{2} = \frac{1}{C _{1}} : \frac{1}{C _{2}}$
Q1MCQVoltage division
In series, the voltages divide as:
- A
  $V_1:V_2=\dfrac{1}{C_1}:\dfrac{1}{C_2}$
- B
  $V_1:V_2=C_1:C_2$
- C
  $V_1=V_2$
- D
  $V_1:V_2=C_2:C_1$... same
Charge division (parallel)
$Q_{1} : Q_{2} = C_{1} : C_{2}$
Q1MCQCharge division
In parallel, the charges divide as:
- A
  $Q_1:Q_2=C_1:C_2$
- B
  $Q_1:Q_2=\dfrac{1}{C_1}:\dfrac{1}{C_2}$
- C
  $Q_1=Q_2$
- D
  $Q_1:Q_2=C_2:C_1$
Common potential
$V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$
Heat loss on sharing
$Δ U = \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}$
Q1MCQHeat loss
The energy lost when two capacitors share charge is:
- A
  $\tfrac12\dfrac{C_1C_2}{C_1+C_2}(V_1-V_2)^{2}$
- B
  $0$
- C
  $\tfrac12(C_1+C_2)V^{2}$
- D
  $C_1V_1$
Charging: charge
$q = q_{0} (1 - e^{- t / R C})$
Q1MCQCharging
During charging, the charge on a capacitor is:
- A
  $q_0(1-e^{-t/RC})$
- B
  $q_0 e^{-t/RC}$
- C
  $q_0 t$
- D
  $q_0$
Discharging: charge
$q = q_{0} e^{- t / R C}$
Q1MCQDischarging
During discharging, the charge is:
- A
  $q_0 e^{-t/RC}$
- B
  $q_0(1-e^{-t/RC})$
- C
  $q_0$
- D
  $0$
Time constant
$τ = R C$
Charging current
$i = \frac{V}{R} e^{- t / R C}$
Q1MCQCharging current
The charging current in an RC circuit is:
- A
  $\dfrac{V}{R}e^{-t/RC}$
- B
  $\dfrac{V}{R}(1-e^{-t/RC})$
- C
  $\dfrac{V}{R}$
- D
  $0$
63% rule (charging)
$t = τ \Rightarrow q = 0.63 q_{0}$
Q1MCQ63% rule
After one time constant of charging, the charge is:
- A
  $63\%$ of maximum
- B
  $37\%$ of maximum
- C
  $100\%$
- D
  $50\%$
37% rule (discharging)
$t = τ \Rightarrow q = 0.37 q_{0}$
Q1MCQ37% rule
After one time constant of discharging, the charge is:
- A
  $37\%$ of initial
- B
  $63\%$ of initial
- C
  $0$
- D
  $50\%$
Dielectric constant
$K = \frac{ε}{ε _{0}}$
Q1MCQDielectric constant
The dielectric constant is:
- A
  $\dfrac{\varepsilon}{\varepsilon_0}$
- B
  $\varepsilon\varepsilon_0$
- C
  $\dfrac{\varepsilon_0}{\varepsilon}$
- D
  $\varepsilon+\varepsilon_0$
Bound (induced) charge density
$σ_{b} = σ (1 - \frac{1}{K})$
Q1MCQBound charge
The bound surface charge density in a dielectric is:
- A
  $\sigma\left(1-\dfrac{1}{K}\right)$
- B
  $\dfrac{\sigma}{K}$
- C
  $\sigma K$
- D
  $\sigma$
Field inside dielectric
$E_{in} = \frac{σ}{K ε _{0}} = \frac{E}{K}$
Q1MCQField in dielectric
The field inside a dielectric-filled capacitor is:
- A
  $\dfrac{E}{K}$
- B
  $KE$
- C
  $E$
- D
  $\dfrac{E}{K^{2}}$
Partially filled dielectric
$C = \frac{ε _{0} A}{d - t + \frac{t}{K}}$
Q1MCQPartial dielectric
With a dielectric slab of thickness $t$ inside, the capacitance is:
- A
  $\dfrac{\varepsilon_0 A}{d-t+\tfrac{t}{K}}$
- B
  $\dfrac{K\varepsilon_0 A}{d}$
- C
  $\dfrac{\varepsilon_0 A}{d-t}$
- D
  $\dfrac{\varepsilon_0 A}{d}$
Metal slab inserted
$C = \frac{ε _{0} A}{d - t}$
Q1MCQMetal slab
With a metal slab of thickness $t$ inserted, the capacitance is:
- A
  $\dfrac{\varepsilon_0 A}{d-t}$
- B
  $\dfrac{\varepsilon_0 A}{d}$
- C
  $\dfrac{\varepsilon_0 A}{d+t}$
- D
  $\infty$
Battery disconnected (dielectric in)
$Q const, V = \frac{V _{0}}{K}, U = \frac{U _{0}}{K}$
Q1MCQBattery disconnected
When a dielectric is inserted with the battery disconnected:
- A
  $Q$ stays, $V$ and $U$ decrease
- B
  $V$ stays, $Q$ increases
- C
  everything increases
- D
  nothing changes
Battery connected (dielectric in)
$V const, Q = K Q_{0}, U = K U_{0}$
Q1MCQBattery connected
When a dielectric is inserted with the battery connected:
- A
  $V$ stays, $Q$ and $U$ increase
- B
  $Q$ stays, $V$ decreases
- C
  everything decreases
- D
  nothing changes
Force on dielectric (battery on)
$F = \frac{ε _{0} b ( K - 1 ) V ^{2}}{2 d}$
Q1MCQForce on dielectric
With the battery connected, the force pulling a dielectric in is:
- A
  $\dfrac{\varepsilon_0 b(K-1)V^{2}}{2d}$
- B
  $\dfrac{Q^{2}}{2C}$
- C
  $0$
- D
  $\dfrac{V}{d}$

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Capacitor formulas

Capacitor, every formula

Charge–capacitance relation

Parallel plate capacitor

Capacitor with dielectric

Isolated spherical conductor

Spherical capacitor

Cylindrical capacitor

Energy stored

Energy density

Field between plates

Force on a plate

Series combination

Parallel combination

n identical in series

n identical in parallel

Voltage division (series)

Charge division (parallel)

Common potential

Heat loss on sharing

Charging: charge

Discharging: charge

Time constant

Charging current

63% rule (charging)

37% rule (discharging)

Dielectric constant

Bound (induced) charge density

Field inside dielectric

Partially filled dielectric

Metal slab inserted

Battery disconnected (dielectric in)

Battery connected (dielectric in)

Force on dielectric (battery on)

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