What is capacitance and what is its SI unit?

Capacitance is the ability of a capacitor to store electric charge per unit potential difference. It is defined as C = Q/V where Q is charge stored and V is potential difference. The SI unit of capacitance is Farad (F), where 1 Farad = 1 Coulomb/Volt. Practical units include microfarad (μF), nanofarad (nF), and picofarad (pF).

What is the formula for capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is C = ε₀A/d, where ε₀ is permittivity of free space (8.85×10⁻¹² F/m), A is the area of each plate, and d is the separation between plates. With a dielectric of constant K, the formula becomes C = Kε₀A/d.

How do capacitors combine in series and parallel?

In series combination: 1/Ceq = 1/C₁ + 1/C₂ + 1/C₃ + ... (charge is same, voltage divides). In parallel combination: Ceq = C₁ + C₂ + C₃ + ... (voltage is same, charge divides). For two capacitors in series: Ceq = C₁C₂/(C₁+C₂).

What are the three formulas for energy stored in a capacitor?

Energy stored in a capacitor can be calculated using three equivalent formulas: U = ½CV² (using capacitance and voltage), U = ½Q²/C (using charge and capacitance), U = ½QV (using charge and voltage). The energy is stored in the electric field between the plates.

What happens when a dielectric is inserted in a capacitor?

When dielectric (K) is inserted: Capacitance increases by factor K (C = KC₀). If battery is connected: V stays same, Q increases K times, Energy increases K times. If battery is disconnected: Q stays same, V decreases K times, Energy decreases K times. Electric field inside dielectric becomes E = E₀/K.

What is the time constant in an RC circuit?

The time constant τ (tau) of an RC circuit is τ = RC, where R is resistance and C is capacitance. It represents the time taken for the capacitor to charge to 63.2% of maximum charge (or discharge to 36.8% of initial charge). After 5τ, the capacitor is considered fully charged/discharged (99.3%).

What is the formula for force between parallel plate capacitor plates?

The force between plates of a parallel plate capacitor is F = Q²/(2ε₀A) = ε₀AV²/(2d²) = CV²/(2d). Alternatively, force per unit area (electrostatic pressure) is P = ε₀E²/2 = σ²/(2ε₀), where E is electric field and σ is surface charge density.

Capacitance for JEE 2025-26 | Complete Notes with Formulas, Derivations & Solved Examples

Capacitance JEE notes, Formulas, PYQs — Capacitance JEE Notes, Formulas, PYQs - Complete Chapter

Capacitance - Fundamentals

A capacitor is a device that stores electrical energy in an electric field. It consists of two conductors (plates) separated by an insulating medium (dielectric). Capacitance is the measure of a capacitor's ability to store electric charge per unit potential difference.

1.1 Definition of Capacitance

Capacitance Definition

C = \frac{Q}{V}

Where:

C = Capacitance
Q = Charge stored on each plate
V = Potential difference between plates

Key Points:

Capacitance is always positive
It depends on geometry only
Independent of Q and V individually
Property of the capacitor system

1.2 Units and Dimensions

Property	Details
SI Unit	Farad (F) = Coulomb/Volt = C/V
Dimensional Formula	[M⁻¹L⁻²T⁴A²]
Practical Units	1 mF (millifarad) = 10⁻³ F 1 μF (microfarad) = 10⁻⁶ F 1 nF (nanofarad) = 10⁻⁹ F 1 pF (picofarad) = 10⁻¹² F
1 Farad means	1 Coulomb of charge stored per 1 Volt of potential difference

1.3 Capacitance of an Isolated Conductor

Isolated Spherical Conductor

For an isolated spherical conductor of radius R:

C = 4\pi\epsilon_0 R

Derivation:

V = \frac{kQ}{R} = \frac{Q}{4\pi\epsilon_0 R}

C = \frac{Q}{V} = 4\pi\epsilon_0 R

Key Insight:

C ∝ R (larger sphere → larger C)
For Earth (R = 6400 km):
C ≈ 711 μF
Even Earth has small capacitance!

💡 JEE Quick Facts

1 Farad is huge! Most practical capacitors are in μF or pF range
Capacitance depends only on geometry (size, shape, separation)
C does NOT depend on: Q or V (doubling Q doubles V, C stays same)
Symbol: ─┤├─ (two parallel lines with gap)

📝 Solved Example 1

Question: A capacitor stores 200 μC of charge when connected to a 50V battery. Find (a) capacitance (b) charge stored at 100V (c) energy stored at 50V.

Solution:

Given: Q = 200 μC = 200 × 10⁻⁶ C, V = 50 V

(a) Capacitance:

C = \frac{Q}{V} = \frac{200 \times 10^{-6}}{50} = 4 \times 10^{-6} \text{ F} = 4 \text{ μF}

(b) Charge at 100V:

Since C is constant (property of capacitor)

Q' = CV' = 4 \times 10^{-6} \times 100 = 400 \text{ μC}

(c) Energy at 50V:

U = \frac{1}{2}CV^2 = \frac{1}{2} \times 4 \times 10^{-6} \times (50)^2 = 5 \times 10^{-3} \text{ J} = 5 \text{ mJ}

\text{Answers: (a) } 4 \text{ μF}, \text{ (b) } 400 \text{ μC}, \text{ (c) } 5 \text{ mJ}

Parallel Plate Capacitor

The parallel plate capacitor is the most fundamental and commonly asked capacitor configuration in JEE. It consists of two large parallel conducting plates separated by a small distance, with opposite charges on each plate.

2.1 Capacitance Formula & Derivation

Parallel Plate Capacitor Formula

C = \frac{\epsilon_0 A}{d}

Where:

C = Capacitance
ε₀ = Permittivity of free space = 8.85 × 10⁻¹² F/m
A = Area of each plate (not total area)
d = Separation between plates

With Dielectric (K):

C = \frac{K\epsilon_0 A}{d} = \frac{\epsilon A}{d}

where ε = Kε₀ = permittivity of dielectric

2.2 Complete Derivation

Step-by-Step Derivation

Step 1: Surface charge density on plates

\sigma = \frac{Q}{A}

Step 2: Electric field between plates (using Gauss law)

E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}

Step 3: Potential difference between plates

V = E \cdot d = \frac{Qd}{\epsilon_0 A}

Step 4: Capacitance

C = \frac{Q}{V} = \frac{Q}{\frac{Qd}{\epsilon_0 A}} = \frac{\epsilon_0 A}{d}

2.3 Important Parameters of Parallel Plate Capacitor

Parameter	Formula	Notes
Capacitance	C = ε₀A/d	Depends only on geometry
Electric Field	E = σ/ε₀ = V/d	Uniform between plates
Surface Charge Density	σ = Q/A = ε₀E	On inner surfaces only
Energy Stored	U = ½CV² = ½QV = ½Q²/C	All three formulas equivalent
Energy Density	u = ½ε₀E²	Energy per unit volume
Force Between Plates	F = Q²/(2ε₀A) = ε₀AE²/2	Always attractive

2.4 Effect of Changing Parameters

Battery Connected (V = constant)

When a parameter changes with battery connected:

If C increases (↑) Q↑, U↑, V same

If C decreases (↓) Q↓, U↓, V same

Insert dielectric (K) C→KC, Q→KQ, U→KU

Battery Disconnected (Q = constant)

When a parameter changes after battery removal:

If C increases (↑) V↓, U↓, Q same

If C decreases (↓) V↑, U↑, Q same

Insert dielectric (K) C→KC, V→V/K, U→U/K

⚠️ Most Common JEE Mistakes

Area confusion: A is the area of ONE plate, not both!
Battery connected vs disconnected: Results are completely opposite - always check this first
Electric field outside: E = 0 outside the plates (ideal case)
Edge effects: Ignored in derivation (assumes infinite plates)

📝 Solved Example 2

Question: A parallel plate capacitor has plate area 100 cm² and separation 2 mm. It is connected to a 100V battery. Find: (a) Capacitance (b) Charge (c) Electric field (d) Energy stored.

Solution:

Given: A = 100 cm² = 100 × 10⁻⁴ m² = 10⁻² m²

d = 2 mm = 2 × 10⁻³ m, V = 100 V

(a) Capacitance:

C = \frac{\epsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 10^{-2}}{2 \times 10^{-3}}

C = 44.25 \times 10^{-12} \text{ F} = 44.25 \text{ pF}

(b) Charge:

Q = CV = 44.25 \times 10^{-12} \times 100 = 4.425 \times 10^{-9} \text{ C} = 4.425 \text{ nC}

(c) Electric field:

E = \frac{V}{d} = \frac{100}{2 \times 10^{-3}} = 5 \times 10^4 \text{ V/m} = 50 \text{ kV/m}

(d) Energy stored:

U = \frac{1}{2}CV^2 = \frac{1}{2} \times 44.25 \times 10^{-12} \times (100)^2 = 2.21 \times 10^{-7} \text{ J}

\text{Answers: (a) 44.25 pF, (b) 4.425 nC, (c) 50 kV/m, (d) 0.221 μJ}

📝 Solved Example 3 (JEE Advanced Pattern)

Question: A parallel plate capacitor of capacitance C₀ is charged to potential V₀. The battery is then disconnected. If the separation between plates is doubled, find the new capacitance, voltage, and energy. Comment on energy change.

Solution:

Initial: C₀, V₀, Q₀ = C₀V₀, U₀ = ½C₀V₀²

Battery disconnected → Charge remains constant (Q = Q₀)

New capacitance (d → 2d):

C' = \frac{\epsilon_0 A}{2d} = \frac{C_0}{2}

New voltage:

V' = \frac{Q}{C'} = \frac{Q_0}{C_0/2} = \frac{2Q_0}{C_0} = 2V_0

New energy:

U' = \frac{1}{2}C'V'^2 = \frac{1}{2} \times \frac{C_0}{2} \times (2V_0)^2 = C_0V_0^2 = 2U_0

Alternative using Q:

U' = \frac{Q^2}{2C'} = \frac{Q_0^2}{2 \times C_0/2} = \frac{Q_0^2}{C_0} = 2U_0

Results:

C → C₀/2 (halved)
V → 2V₀ (doubled)
U → 2U₀ (doubled)

Comment: Energy increased! Work was done by external agent to separate the plates against attractive force.

Types of Capacitors

Different geometries of conductors give rise to different types of capacitors. Each has its own formula for capacitance, which can be derived using basic electrostatic principles.

3.1 Spherical Capacitor

Spherical Capacitor (Two Concentric Spheres)

Inner sphere of radius a, outer sphere of radius b

C = \frac{4\pi\epsilon_0 ab}{b - a}

With dielectric K:

C = \frac{4\pi K\epsilon_0 ab}{b - a}

Special Cases:

If b → ∞ (Isolated sphere):

C = 4\pi\epsilon_0 a

If (b - a) << a (thin shell):

C \approx \frac{4\pi\epsilon_0 a^2}{b-a} = \frac{\epsilon_0 A}{d}

Same as parallel plate!

3.2 Cylindrical Capacitor

Cylindrical Capacitor (Two Coaxial Cylinders)

Inner radius a, outer radius b, length L

C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}

Capacitance per unit length:

\frac{C}{L} = \frac{2\pi\epsilon_0}{\ln(b/a)}

Key Points:

Used in coaxial cables
C ∝ L (longer cable → more C)
C increases if (b-a) decreases
ln is natural logarithm

With dielectric K:

C = \frac{2\pi K\epsilon_0 L}{\ln(b/a)}

3.3 Comparison of All Capacitor Types

Capacitor Type	Capacitance Formula	Electric Field	Geometry
Parallel Plate	C = ε₀A/d	E = σ/ε₀ = V/d	Two flat plates
Spherical	C = 4πε₀ab/(b-a)	E = kQ/r² (a<r<b)	Concentric spheres
Cylindrical	C = 2πε₀L/ln(b/a)	E = λ/(2πε₀r)	Coaxial cylinders
Isolated Sphere	C = 4πε₀R	E = kQ/r² (r>R)	Single sphere

📝 Solved Example 4

Question: A spherical capacitor has inner and outer radii of 10 cm and 12 cm. Find the capacitance. If the outer sphere is earthed, what is the capacitance?

Solution:

Given: a = 10 cm = 0.1 m, b = 12 cm = 0.12 m

Capacitance:

C = \frac{4\pi\epsilon_0 ab}{b-a} = \frac{4\pi \times 8.85 \times 10^{-12} \times 0.1 \times 0.12}{0.12 - 0.1}

C = \frac{4\pi \times 8.85 \times 10^{-12} \times 0.012}{0.02}

C = \frac{1.334 \times 10^{-12}}{0.02} = 66.7 \times 10^{-12} \text{ F} = 66.7 \text{ pF}

When outer sphere is earthed:

The capacitance remains the same because earthing only fixes the potential of outer sphere to zero, but doesn't change the geometry.

\text{Answer: } C = 66.7 \text{ pF}

Series & Parallel Combinations

Capacitors can be connected in series or parallel to achieve desired capacitance values. Understanding these combinations is crucial for JEE as 30-40% of capacitor problems involve finding equivalent capacitance.

4.1 Series Combination

Capacitors in Series

─┤├─┤├─┤├─

C₁ C₂ C₃

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...

For two capacitors:

C_{eq} = \frac{C_1 C_2}{C_1 + C_2}

Key Properties:

✓ Same charge Q on all capacitors
✓ Voltage divides: V = V₁ + V₂ + V₃...
✓ C_eq is smaller than smallest capacitor
✓ Voltage across each: V₁ = Q/C₁

Voltage Distribution:

V_1 : V_2 : V_3 = \frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3}

Smaller C gets larger V!

4.2 Parallel Combination

Capacitors in Parallel

┬─┤├─┬

├─┤├─┤

└─┤├─┘

C_{eq} = C_1 + C_2 + C_3 + ...

Simple algebraic sum!

Key Properties:

✓ Same voltage V across all capacitors
✓ Charge divides: Q = Q₁ + Q₂ + Q₃...
✓ C_eq is larger than largest capacitor
✓ Charge on each: Q₁ = C₁V

Charge Distribution:

\[Q_1 : Q_2 : Q_3 = C_1 : C_2 : C_3\]

Larger C gets larger Q!

4.3 Comparison: Series vs Parallel

Property	Series	Parallel
Equivalent Capacitance	1/C_eq = Σ(1/Cᵢ)	C_eq = ΣCᵢ
Charge	Same on all (Q)	Divides (Q₁ + Q₂ + ...)
Voltage	Divides (V₁ + V₂ + ...)	Same across all (V)
C_eq value	Smaller than smallest	Larger than largest
Similar to Resistors	Like R in parallel	Like R in series
Energy stored	U = Q²/(2C_eq)	U = ½C_eq V²

4.4 Mixed Combinations Strategy

💡 JEE Problem-Solving Strategy

Identify series and parallel groups - Look for common nodes
Start from innermost combination - Simplify step by step
Redraw circuit after each simplification
Use symmetry - If circuit is symmetric, capacitors with same potential can be combined
For complex networks: Use Kirchhoff's laws or star-delta transformation

📝 Solved Example 5

Question: Three capacitors of 2μF, 3μF, and 6μF are connected in (a) series (b) parallel. Find equivalent capacitance in each case.

Solution:

Given: C₁ = 2μF, C₂ = 3μF, C₃ = 6μF

(a) Series combination:

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1

C_{eq} = 1 \text{ μF}

(b) Parallel combination:

C_{eq} = C_1 + C_2 + C_3 = 2 + 3 + 6 = 11 \text{ μF}

\text{Answers: (a) } 1 \text{ μF (series)}, \text{ (b) } 11 \text{ μF (parallel)}

📝 Solved Example 6 (JEE Main Pattern)

Question: In the given circuit, find the equivalent capacitance between points A and B. Each capacitor has capacitance C.

    A ─┬─┤├─┬─ B
       │     │
       ├─┤├─┤
       │     │
       └─┤├─┘

(3 capacitors in parallel)

Solution:

All three capacitors are connected between points A and B directly.

This means they are in parallel.

C_{eq} = C + C + C = 3C

\text{Answer: } C_{eq} = 3C

📝 Solved Example 7 (Mixed Combination)

Question: Find equivalent capacitance: 2μF in series with (3μF parallel with 6μF).

Solution:

Step 1: Solve parallel combination first

C_{parallel} = 3 + 6 = 9 \text{ μF}

Step 2: Now solve series (2μF with 9μF)

\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{9} = \frac{9 + 2}{18} = \frac{11}{18}

C_{eq} = \frac{18}{11} \text{ μF} = 1.636 \text{ μF}

\text{Answer: } C_{eq} = \frac{18}{11} \text{ μF} \approx 1.64 \text{ μF}

Energy Stored & Force Between Plates

A charged capacitor stores energy in the electric field between its plates. This energy can be released when the capacitor discharges. Understanding energy storage is crucial for JEE as it connects to many practical applications and complex problems.

5.1 Energy Stored in a Capacitor

Three Equivalent Energy Formulas

Form 1

U = \frac{1}{2}CV^2

Use when C and V are known

Form 2

U = \frac{1}{2}\frac{Q^2}{C}

Use when Q is constant

Form 3

U = \frac{1}{2}QV

Use when Q and V are known

Derivation:

U = \int_0^Q V \, dq = \int_0^Q \frac{q}{C} \, dq = \frac{1}{C} \cdot \frac{q^2}{2}\Big|_0^Q = \frac{Q^2}{2C}

5.2 Energy Density

Energy Per Unit Volume

u = \frac{U}{Volume} = \frac{1}{2}\epsilon_0 E^2

For parallel plate capacitor:

u = \frac{1}{2}\epsilon_0 \left(\frac{V}{d}\right)^2 = \frac{\epsilon_0 V^2}{2d^2}

Also written as:

u = \frac{\sigma^2}{2\epsilon_0} = \frac{D^2}{2\epsilon_0}

where D = ε₀E (displacement field)

With Dielectric (K):

u = \frac{1}{2}\epsilon_0 K E^2 = \frac{1}{2}\epsilon E^2

5.3 Force Between Capacitor Plates

Electrostatic Force & Pressure

Force on one plate:

F = \frac{Q^2}{2\epsilon_0 A} = \frac{\sigma^2 A}{2\epsilon_0}

Also:

F = \frac{1}{2}QE = \frac{\epsilon_0 A E^2}{2}

Electrostatic Pressure:

P = \frac{F}{A} = \frac{\sigma^2}{2\epsilon_0} = \frac{1}{2}\epsilon_0 E^2

Note: Pressure = Energy density!

Key: Force is always attractive (plates attract each other)

5.4 Energy Changes in Different Scenarios

Scenario	What's Constant	Energy Change	Best Formula
Battery connected	V = constant	U ∝ C (if C↑, U↑)	U = ½CV²
Battery disconnected	Q = constant	U ∝ 1/C (if C↑, U↓)	U = Q²/(2C)
Insert dielectric (battery on)	V = constant	U → KU (increases)	U = ½CV²
Insert dielectric (battery off)	Q = constant	U → U/K (decreases)	U = Q²/(2C)
Increase separation (battery on)	V = constant	C↓, so U↓	U = ½CV²
Increase separation (battery off)	Q = constant	C↓, so U↑	U = Q²/(2C)

⚠️ Energy Lost When Capacitors Share Charge

When two capacitors are connected (charge redistribution), energy is always lost as heat, even if no resistance is present.

Energy loss formula:

\Delta U = U_i - U_f = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}

This is always positive (energy is always lost, never gained)

📝 Solved Example 8 (JEE Advanced Pattern)

Question: A 4μF capacitor is charged to 200V. It is then connected to an uncharged 2μF capacitor. Find: (a) Common potential (b) Charge on each (c) Energy loss.

Solution:

Initial: C₁ = 4μF, V₁ = 200V, Q₁ = 800μC

C₂ = 2μF, V₂ = 0V, Q₂ = 0

(a) Common potential:

Total charge is conserved: Q_total = 800μC

V_{common} = \frac{Q_{total}}{C_1 + C_2} = \frac{800}{4 + 2} = \frac{800}{6} = 133.33 \text{ V}

(b) Charge on each:

Q'_1 = C_1 V_{common} = 4 \times 133.33 = 533.33 \text{ μC}

Q'_2 = C_2 V_{common} = 2 \times 133.33 = 266.67 \text{ μC}

(c) Energy loss:

U_i = \frac{1}{2}C_1V_1^2 = \frac{1}{2} \times 4 \times 10^{-6} \times (200)^2 = 0.08 \text{ J}

U_f = \frac{1}{2}(C_1 + C_2)V_{common}^2 = \frac{1}{2} \times 6 \times 10^{-6} \times (133.33)^2 = 0.0533 \text{ J}

\Delta U = U_i - U_f = 0.08 - 0.0533 = 0.0267 \text{ J}

Using direct formula:

\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)} = \frac{4 \times 2 \times (200-0)^2 \times 10^{-6}}{2 \times 6} = 0.0267 \text{ J}

\text{Answers: (a) 133.33V, (b) 533.33μC \& 266.67μC, (c) 26.7 mJ}

Dielectrics in Capacitors

A dielectric is an insulating material that, when inserted between capacitor plates, increases the capacitance. This is one of the most important and most frequently tested topics in JEE.

6.1 Dielectric Constant (K or κ or εᵣ)

Dielectric Constant Definition

K = \frac{C}{C_0} = \frac{E_0}{E} = \frac{V_0}{V} = \frac{\epsilon}{\epsilon_0}

Definitions:

K = Dielectric constant (dimensionless)
K ≥ 1 always (K = 1 for vacuum/air)
C₀ = Capacitance without dielectric
C = Capacitance with dielectric

With dielectric:

C = KC₀
E = E₀/K (field reduces)
ε = Kε₀ (permittivity increases)

6.2 Effect of Dielectric - Complete Analysis

Quantity	Battery Connected (V constant)	Battery Disconnected (Q constant)
Capacitance C	C → KC (increases)	C → KC (increases)
Charge Q	Q → KQ (increases)	Q → Q (same)
Voltage V	V → V (same)	V → V/K (decreases)
Electric Field E	E → E (same, in free space)	E → E/K (decreases)
Energy U	U → KU (increases)	U → U/K (decreases)

6.3 Partial Dielectric Insertion

Dielectric in Series (layers)

Dielectric fills part of the gap (thickness t)

C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}

Treat as two capacitors in series

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}

where C₁ = ε₀A/(d-t), C₂ = Kε₀A/t

Dielectric in Parallel (side by side)

Dielectric covers part of area (A₁)

C = \frac{\epsilon_0 A_1}{d} + \frac{K\epsilon_0 A_2}{d}

Treat as two capacitors in parallel

\[C = C_1 + C_2\]

where A₁ + A₂ = A (total area)

6.4 Multiple Dielectrics

Multiple Dielectric Layers in Series

For n dielectric layers with thicknesses d₁, d₂, ... dₙ and constants K₁, K₂, ... Kₙ:

\frac{1}{C} = \frac{1}{\epsilon_0 A}\left(\frac{d_1}{K_1} + \frac{d_2}{K_2} + ... + \frac{d_n}{K_n}\right)

For two equal halves (d₁ = d₂ = d/2):

C = \frac{2\epsilon_0 A K_1 K_2}{d(K_1 + K_2)}

Multiple Dielectrics in Parallel

For n dielectrics with areas A₁, A₂, ... Aₙ and constants K₁, K₂, ... Kₙ:

C = \frac{\epsilon_0}{d}(K_1 A_1 + K_2 A_2 + ... + K_n A_n)

For two equal halves (A₁ = A₂ = A/2):

C = \frac{\epsilon_0 A(K_1 + K_2)}{2d}

💡 JEE Quick Identification Rules

Series dielectric: Dielectric divides the gap (perpendicular to plates)
Parallel dielectric: Dielectric divides the area (parallel to plates)
Mixed: First identify series parts, then parallel parts, or vice versa

📝 Solved Example 9 (JEE Main Pattern)

Question: A parallel plate capacitor has plate area 200 cm² and separation 4 mm. A dielectric slab of thickness 2 mm and K = 4 is inserted. Find the capacitance.

Solution:

Given: A = 200 cm² = 200 × 10⁻⁴ m²

d = 4 mm = 4 × 10⁻³ m

t = 2 mm (thickness of dielectric), K = 4

Using the series formula:

C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}

C = \frac{8.85 \times 10^{-12} \times 200 \times 10^{-4}}{4 \times 10^{-3} - 2 \times 10^{-3} + \frac{2 \times 10^{-3}}{4}}

C = \frac{1.77 \times 10^{-13}}{2 \times 10^{-3} + 0.5 \times 10^{-3}} = \frac{1.77 \times 10^{-13}}{2.5 \times 10^{-3}}

C = 7.08 \times 10^{-11} \text{ F} = 70.8 \text{ pF}

\text{Answer: } C = 70.8 \text{ pF}

📝 Solved Example 10 (JEE Advanced Pattern)

Question: A parallel plate capacitor has two dielectrics: one with K₁ = 2 filling the upper half of volume, and K₂ = 4 filling the lower half. If C₀ is capacitance without dielectric, find C.

Solution:

Analysis: Each dielectric fills half the gap (d/2 each)

This is a series combination of two capacitors.

C_1 = \frac{K_1 \epsilon_0 A}{d/2} = \frac{2K_1 \epsilon_0 A}{d} = 2K_1 C_0

C_2 = \frac{K_2 \epsilon_0 A}{d/2} = \frac{2K_2 \epsilon_0 A}{d} = 2K_2 C_0

Series combination:

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2K_1 C_0} + \frac{1}{2K_2 C_0}

\frac{1}{C} = \frac{K_2 + K_1}{2K_1 K_2 C_0}

C = \frac{2K_1 K_2 C_0}{K_1 + K_2} = \frac{2 \times 2 \times 4 \times C_0}{2 + 4} = \frac{16C_0}{6} = \frac{8C_0}{3}

\text{Answer: } C = \frac{8C_0}{3} = 2.67C_0

RC Circuits

When a capacitor is connected with a resistor and a voltage source, the charging and discharging processes are not instantaneous but follow exponential curves. This is the basis of RC circuits - a very important topic for JEE.

7.1 Time Constant (τ)

Time Constant Definition

\tau = RC

Where:

τ (tau) = Time constant
R = Resistance in ohms (Ω)
C = Capacitance in farads (F)
Unit of τ = seconds (Ω × F = s)

Physical Meaning:

Time to charge to 63.2% of max
Time to discharge to 36.8% of initial
After 5τ: 99.3% complete (≈ full)
Larger RC = slower charging/discharging

7.2 Charging of Capacitor

Charging Equations

When capacitor is connected to EMF (ε) through resistance R:

Charge on capacitor:

q(t) = Q_0(1 - e^{-t/RC})

where Q₀ = Cε (maximum charge)

Voltage across capacitor:

V_C(t) = \varepsilon(1 - e^{-t/RC})

Current in circuit:

i(t) = I_0 e^{-t/RC} = \frac{\varepsilon}{R}e^{-t/RC}

where I₀ = ε/R (initial current)

Voltage across resistor:

V_R(t) = \varepsilon \cdot e^{-t/RC}

7.3 Discharging of Capacitor

Discharging Equations

When charged capacitor (initial charge Q₀) discharges through R:

Charge on capacitor:

q(t) = Q_0 e^{-t/RC}

Voltage across capacitor:

V_C(t) = V_0 e^{-t/RC}

Current (magnitude):

|i(t)| = \frac{Q_0}{RC}e^{-t/RC} = \frac{V_0}{R}e^{-t/RC}

Note: Current flows in opposite direction

7.4 Important Time Values

Time	Charging (q/Q₀)	Discharging (q/Q₀)	Percentage
t = 0	0	1	0% / 100%
t = τ = RC	1 - 1/e = 0.632	1/e = 0.368	63.2% / 36.8%
t = 2τ	0.865	0.135	86.5% / 13.5%
t = 3τ	0.950	0.050	95% / 5%
t = 5τ	0.993	0.007	99.3% / 0.7%
t = ∞	1	0	100% / 0%

💡 JEE Important Results

At t = 0 (charging): Capacitor acts as short circuit (V_C = 0)
At t = ∞ (steady state): Capacitor acts as open circuit (i = 0)
Half-life: t₁/₂ = τ ln(2) = 0.693RC
Energy dissipated in R: Equals energy stored in C (when charging from zero)

📝 Solved Example 11

Question: A 2μF capacitor is charged through a 1MΩ resistor by a 10V battery. Find: (a) Time constant (b) Maximum charge (c) Charge at t = 2s (d) Time to reach 5V.

Solution:

Given: C = 2μF = 2×10⁻⁶F, R = 1MΩ = 10⁶Ω, ε = 10V

(a) Time constant:

\tau = RC = 10^6 \times 2 \times 10^{-6} = 2 \text{ s}

(b) Maximum charge:

Q_0 = C\varepsilon = 2 \times 10^{-6} \times 10 = 20 \text{ μC}

(c) Charge at t = 2s:

q = Q_0(1 - e^{-t/\tau}) = 20(1 - e^{-2/2}) = 20(1 - e^{-1})

q = 20(1 - 0.368) = 20 \times 0.632 = 12.64 \text{ μC}

(d) Time to reach 5V (half of max):

V_C = \varepsilon(1 - e^{-t/\tau})

5 = 10(1 - e^{-t/2}) \Rightarrow 0.5 = 1 - e^{-t/2}

e^{-t/2} = 0.5 \Rightarrow -\frac{t}{2} = \ln(0.5) = -0.693

t = 1.386 \text{ s}

\text{Answers: (a) 2s, (b) 20μC, (c) 12.64μC, (d) 1.39s}

Special Cases & Advanced Problems

This section covers special configurations and advanced problem types that frequently appear in JEE Advanced. Mastering these will give you an edge in competitive exams.

8.1 Capacitor with Conducting Slab

Conducting Slab Inside Capacitor

When a conducting slab of thickness t is inserted (not touching plates):

C = \frac{\epsilon_0 A}{d - t}

Explanation:

Electric field inside conductor = 0
Effective gap = (d - t)
For K = ∞ (conductor), formula reduces to this
If slab touches one plate: C = ε₀A/(d-t) (same result)

8.2 Guard Ring Capacitor

Guard Ring Arrangement

Used to eliminate edge effects (fringing) in experiments:

Main plate surrounded by a "guard ring" at same potential
Field lines are perpendicular at edges
Gives uniform field and accurate C = ε₀A/d
Used in precise capacitance measurements

8.3 Variable Capacitor

Overlapping Area Change

When plates overlap by area A (can be varied):

Capacitance:

C = \frac{\epsilon_0 A}{d}

If A → A/2, then C → C/2

With n plates overlapping:

C = \frac{(n-1)\epsilon_0 A}{d}

(n-1) capacitors in parallel

8.4 Force on Dielectric Being Inserted

Force on Dielectric Slab

Battery Connected (V constant):

F = \frac{\epsilon_0(K-1)bV^2}{2d}

Force is inward (pulls dielectric in)

Battery Disconnected (Q constant):

F = \frac{Q^2(K-1)}{2\epsilon_0 b(Kx + L - x)^2}

where x = inserted length, L = plate length

Note: b = width of plate, d = separation between plates

8.5 Redistribution of Charge

When Two Charged Capacitors are Connected

Same polarity connection:

V_{common} = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} = \frac{Q_1 + Q_2}{C_1 + C_2}

Opposite polarity connection:

V_{common} = \frac{|C_1V_1 - C_2V_2|}{C_1 + C_2} = \frac{|Q_1 - Q_2|}{C_1 + C_2}

Energy loss:

\Delta U = \frac{C_1C_2(V_1 - V_2)^2}{2(C_1 + C_2)} = \frac{(Q_1 - Q_2)^2}{2(C_1 + C_2)}

For opposite polarity, use (V₁ + V₂)² instead

📝 Solved Example 12 (JEE Advanced)

Question: A capacitor C₁ = 4μF charged to 100V is connected to another capacitor C₂ = 2μF charged to 50V with opposite polarity. Find common potential and energy loss.

Solution:

Initial charges: Q₁ = 4 × 100 = 400μC, Q₂ = 2 × 50 = 100μC

Connected with opposite polarity

Net charge:

Q_{net} = |Q_1 - Q_2| = |400 - 100| = 300 \text{ μC}

Common potential:

V_{common} = \frac{Q_{net}}{C_1 + C_2} = \frac{300}{4 + 2} = 50 \text{ V}

Initial energy:

U_i = \frac{1}{2}C_1V_1^2 + \frac{1}{2}C_2V_2^2 = \frac{1}{2}(4)(100)^2 + \frac{1}{2}(2)(50)^2

U_i = 20000 + 2500 = 22500 \text{ μJ} = 22.5 \text{ mJ}

Final energy:

U_f = \frac{1}{2}(C_1 + C_2)V_{common}^2 = \frac{1}{2}(6)(50)^2 = 7500 \text{ μJ}

Energy loss:

\Delta U = U_i - U_f = 22500 - 7500 = 15000 \text{ μJ} = 15 \text{ mJ}

\text{Answers: } V_{common} = 50V, \Delta U = 15 \text{ mJ}

📋 Complete Formula Sheet - Capacitance

Basic Formulas

C = Q/V

C (parallel plate) = ε₀A/d

C (spherical) = 4πε₀ab/(b-a)

C (cylindrical) = 2πε₀L/ln(b/a)

C (isolated sphere) = 4πε₀R

Combinations

Series: 1/C_eq = Σ(1/Cᵢ)

Parallel: C_eq = ΣCᵢ

Two in series: C_eq = C₁C₂/(C₁+C₂)

n identical in series: C_eq = C/n

n identical in parallel: C_eq = nC

Energy Formulas

U = ½CV² = ½Q²/C = ½QV

Energy density: u = ½ε₀E²

Force: F = Q²/(2ε₀A)

Pressure: P = ½ε₀E² = σ²/(2ε₀)

Loss: ΔU = C₁C₂(V₁-V₂)²/2(C₁+C₂)

Dielectrics

K = C/C₀ = E₀/E = V₀/V

C (with K) = KC₀ = Kε₀A/d

Partial: C = ε₀A/(d-t+t/K)

Conductor: C = ε₀A/(d-t)

Series layers: 1/C = Σ(dᵢ/Kᵢε₀A)

RC Circuits

τ = RC (time constant)

Charging: q = Q₀(1 - e^(-t/RC))

Discharging: q = Q₀e^(-t/RC)

Current: i = I₀e^(-t/RC)

At t = τ: q = 0.632Q₀ (charging)

At t = τ: q = 0.368Q₀ (discharging)

Half-life: t₁/₂ = 0.693RC

At t = 5τ: ~99.3% complete

Download Formula Sheet PDF

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Capacitor Combinations: 35%
✓ Dielectric Effects: 25%
✓ Energy in Capacitors: 20%
✓ RC Circuits: 15%
✓ Basic Formulas: 5%

JEE Advanced (Last 5 Years)

✓ Complex Networks: 30%
✓ Energy & Force Problems: 25%
✓ Variable Capacitors: 20%
✓ Charge Redistribution: 15%
✓ RC Circuit Analysis: 10%

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)

JEE Advanced: 12-18 marks (3-5 questions)

Difficulty Level: Medium to High

Time Required to Master: 10-15 days

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

A parallel plate capacitor has plates of area 200 cm² separated by 2 mm. Find its capacitance.
Three capacitors 3μF, 6μF, and 9μF are connected in series. Find equivalent capacitance.
A 5μF capacitor is charged to 100V. Calculate the energy stored.
Find the capacitance of a spherical capacitor with inner radius 5 cm and outer radius 10 cm.
A capacitor C = 2μF is charged through R = 1MΩ. Find the time constant.
What is the capacitance if a dielectric of K = 5 is inserted in a 10pF capacitor?
Two capacitors 4μF and 6μF in parallel are connected to 20V. Find charge on each.
Find energy density in a region where electric field is 1000 V/m.

Level 2: Intermediate (JEE Main/Advanced)

A 4μF capacitor charged to 200V is connected to an uncharged 2μF capacitor. Find common potential and energy loss.
A parallel plate capacitor has a dielectric slab (K=4) of thickness d/2 inserted. Find the new capacitance in terms of C₀.
In an RC circuit with R = 2MΩ and C = 0.5μF, find time for charge to reach 80% of maximum.
A capacitor of 100pF is charged to 50V, then battery removed. If plate separation is doubled, find new voltage and energy.
Find equivalent capacitance between A and B: C₁ = 2μF in series with parallel combination of C₂ = 3μF and C₃ = 6μF.
A conducting slab of thickness 2 mm is inserted in a capacitor with separation 5 mm. Find new capacitance if original was 10pF.
Two capacitors 3μF (charged to 300V) and 6μF (charged to 150V) are connected with opposite polarity. Find final charge distribution.
Calculate force per unit area between plates of a capacitor if surface charge density is 10 μC/m².

Level 3: Advanced (JEE Advanced/Olympiad)

A capacitor has two dielectric layers: K₁ = 2 (thickness d/3) and K₂ = 6 (thickness 2d/3). Find equivalent K.
A capacitor is charged to V₀, then disconnected. If a dielectric slab (K=3) of thickness d/2 is inserted, find new potential and the work done by external agent.
In an infinite ladder network of identical capacitors C, find equivalent capacitance between input terminals.
A parallel plate capacitor with battery attached is dipped in oil (K=2) to half its height. Find change in charge and energy.
Derive the expression for force on a dielectric slab being inserted in a capacitor (battery connected).
Three capacitors of capacitances C₁, C₂, C₃ are charged to potentials V₁, V₂, V₃. They are then connected in parallel. Find final common potential and total energy loss.
A spherical capacitor with inner radius a and outer radius b has a dielectric of K = br/a filled inside. Find capacitance.
In an RC circuit, if resistance is doubled and capacitance is halved, how does the time to charge to 90% change?

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Most Asked Topic
2024	2 Questions (8 marks)	3 Questions (12 marks)	Dielectric insertion, Energy
2023	3 Questions (12 marks)	4 Questions (15 marks)	Combinations, RC circuits
2022	2 Questions (8 marks)	3 Questions (11 marks)	Charge redistribution, Force
2021	3 Questions (12 marks)	4 Questions (14 marks)	Multiple dielectrics, Networks

📑 Quick Navigation - Capacitance

Capacitance JEE Main & Advanced 2025-26

Capacitance - Fundamentals

1.1 Definition of Capacitance

Capacitance Definition

1.2 Units and Dimensions

1.3 Capacitance of an Isolated Conductor

Isolated Spherical Conductor

💡 JEE Quick Facts

📝 Solved Example 1

Parallel Plate Capacitor

2.1 Capacitance Formula & Derivation

Parallel Plate Capacitor Formula

2.2 Complete Derivation

Step-by-Step Derivation

2.3 Important Parameters of Parallel Plate Capacitor

2.4 Effect of Changing Parameters

Battery Connected (V = constant)

Battery Disconnected (Q = constant)

⚠️ Most Common JEE Mistakes

📝 Solved Example 2

📝 Solved Example 3 (JEE Advanced Pattern)

Types of Capacitors

3.1 Spherical Capacitor

Spherical Capacitor (Two Concentric Spheres)

3.2 Cylindrical Capacitor

Cylindrical Capacitor (Two Coaxial Cylinders)

3.3 Comparison of All Capacitor Types

📝 Solved Example 4

Series & Parallel Combinations

4.1 Series Combination

Capacitors in Series

4.2 Parallel Combination

Capacitors in Parallel

4.3 Comparison: Series vs Parallel

4.4 Mixed Combinations Strategy

💡 JEE Problem-Solving Strategy

📝 Solved Example 5

📝 Solved Example 6 (JEE Main Pattern)

📝 Solved Example 7 (Mixed Combination)

Energy Stored & Force Between Plates

5.1 Energy Stored in a Capacitor

Three Equivalent Energy Formulas

5.2 Energy Density

Energy Per Unit Volume

5.3 Force Between Capacitor Plates

Electrostatic Force & Pressure

5.4 Energy Changes in Different Scenarios

⚠️ Energy Lost When Capacitors Share Charge

📝 Solved Example 8 (JEE Advanced Pattern)

Dielectrics in Capacitors

6.1 Dielectric Constant (K or κ or εᵣ)

Dielectric Constant Definition

6.2 Effect of Dielectric - Complete Analysis

6.3 Partial Dielectric Insertion

Dielectric in Series (layers)

Dielectric in Parallel (side by side)

6.4 Multiple Dielectrics

Multiple Dielectric Layers in Series

Multiple Dielectrics in Parallel

💡 JEE Quick Identification Rules

📝 Solved Example 9 (JEE Main Pattern)

📝 Solved Example 10 (JEE Advanced Pattern)

RC Circuits

7.1 Time Constant (τ)

Time Constant Definition

7.2 Charging of Capacitor

Charging Equations

7.3 Discharging of Capacitor

Discharging Equations

7.4 Important Time Values

💡 JEE Important Results

📝 Solved Example 11

Special Cases & Advanced Problems

8.1 Capacitor with Conducting Slab

Conducting Slab Inside Capacitor

8.2 Guard Ring Capacitor

Guard Ring Arrangement

8.3 Variable Capacitor

Overlapping Area Change