JEE Main Physics · Chapter 13

Capacitance

Parallel-plate capacitors, series and parallel networks, dielectrics, energy stored and charge sharing.

NCERT Class 12JEE Main: 1 Q per shift 9 min read

1. Chapter at a Glance

A capacitor stores electric charge and energy in the electric field between two conductors. This chapter covers capacitance, parallel-plate and spherical capacitors, combinations in series and parallel, the effect of dielectrics, energy stored and charge sharing. In the JEE syllabus it is part of the electrostatics unit; capacitor networks and dielectric questions appear almost every year.

What JEE asks most
  • Equivalent capacitance of networks; charge and voltage on each capacitor.
  • Effect of inserting a dielectric with the battery connected or disconnected.
  • Energy stored and energy lost when capacitors share charge.
  • Capacitor with a partial dielectric slab or a conducting slab.

2. Capacitance

  • C = Q/V, unit farad (F). It depends only on the geometry and the medium — not on Q or V.
  • Isolated sphere: C = 4πε₀R (Earth ≈ 711 μF).
  • Parallel-plate capacitor: C = ε₀A/d; with dielectric filling the gap, C = Kε₀A/d.
  • Spherical capacitor (radii a < b): C = 4πε₀ab/(b − a). Cylindrical (length L): C = 2πε₀L/ln(b/a).
  • The field between plates is E = σ/ε₀ = V/d; outside an ideal parallel-plate capacitor E = 0.
K +Q−Q d t E₀ in airE₀
A dielectric slab of thickness t (constant K) reduces the field inside it to E₀/K, so C = ε₀A/(d − t + t/K).

3. Combinations of Capacitors

FeatureSeriesParallel
Equivalent1/C = 1/C₁ + 1/C₂ + …C = C₁ + C₂ + …
Same for allCharge QVoltage V
DividesVoltage (V ∝ 1/C)Charge (Q ∝ C)
Equivalent valueLess than the smallestMore than the largest

4. Dielectrics

  • A dielectric gets polarised in a field; its induced charges reduce the field inside it by a factor K (dielectric constant, K ≥ 1).
  • Filling the gap multiplies C by K.
  • Partial slab of thickness t: C = ε₀A/(d − t + t/K). A metal slab (K → ∞) gives C = ε₀A/(d − t).
  • Plates divided side by side (area split) act as capacitors in parallel; layers stacked across the gap act in series.
After inserting dielectric KBattery stays connected (V fixed)Battery removed (Q fixed)
CapacitanceK timesK times
ChargeK timesSame
VoltageSame1/K times
Field between platesSame1/K times
EnergyK times1/K times

5. Energy Stored and Charge Sharing

U = ½CV² = Q²/2C = ½QV    energy density u = ½ε₀E²
  • When a charged capacitor C₁ (at V₁) is joined to C₂ (at V₂), charge flows until they reach a common potential V = (C₁V₁ + C₂V₂)/(C₁ + C₂).
  • Energy lost as heat = C₁C₂(V₁ − V₂)²/[2(C₁ + C₂)] — always positive.
  • Force of attraction between the plates: F = Q²/(2ε₀A).
  • When a battery charges a capacitor, it supplies QV of energy but only ½QV is stored; the rest is lost as heat in the wires.

6. Formula Master Table

QuantityFormula
DefinitionC = Q/V
Parallel plateC = Kε₀A/d
Isolated sphereC = 4πε₀R
Partial slabC = ε₀A/(d − t + t/K)
Series / parallel1/C = Σ1/Ci; C = ΣCi
EnergyU = ½CV² = Q²/2C
Common potentialV = (C₁V₁ + C₂V₂)/(C₁ + C₂)
Energy lossC₁C₂(V₁ − V₂)²/2(C₁ + C₂)

7. Mind Map

Capacitance
Basics
  • C = Q/V
  • Plate, sphere, cylinder
Networks
  • Series: same Q
  • Parallel: same V
Dielectrics
  • C × K
  • Battery on/off table
  • Partial slabs
Energy
  • ½CV²
  • Charge sharing loss
  • Energy density

8. Tips and Tricks

Shortcuts
  • Two capacitors in series: C = C₁C₂/(C₁ + C₂); n equal ones: C/n.
  • In series, the smaller capacitor gets the larger voltage.
  • First decide what stays constant (V if battery connected, Q if isolated), then use the table.
  • For symmetric bridge networks, look for points at equal potential and remove the capacitor between them.
Mistakes students make
  • Using the resistor rules for capacitors (they are reversed).
  • Assuming energy is conserved when capacitors share charge.
  • Forgetting that capacitance does not depend on the charge or voltage.

9. Important One-liners

  • A capacitor blocks steady DC once charged but allows AC.
  • Inserting a dielectric always increases capacitance.
  • Series combination has the same charge on each capacitor.
  • Half the energy supplied by a battery while charging a capacitor is lost as heat.
  • Energy of a capacitor is stored in its electric field.
  • The capacitance of Earth is less than 1 mF.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹.

Average level

N1Average

Find the capacitance of a parallel-plate capacitor with plate area 0.02 m² and separation 1 mm in air.

Show answer & solution

Concept: C = ε₀A/d.

  1. C = 8.85 × 10⁻¹² × 0.02/10⁻³.
  2. = 1.77 × 10⁻¹⁰ F = 177 pF.

Answer: 177 pF

N2Average

Capacitors of 2 μF, 3 μF and 6 μF are connected (a) in series and (b) in parallel. Find the equivalent capacitance each time.

Show answer & solution

Concept: series adds reciprocals; parallel adds directly.

  1. (a) 1/C = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 1 ⇒ C = 1 μF.
  2. (b) C = 2 + 3 + 6 = 11 μF.

Answer: 1 μF; 11 μF

N3Average

A 10 μF capacitor is charged to 100 V. Find the energy stored.

Show answer & solution

Concept: U = ½CV².

  1. U = ½ × 10 × 10⁻⁶ × 100².
  2. = 0.05 J.

Answer: 0.05 J

N4Average

A 4 μF capacitor is connected across a 50 V battery. Find its charge. If another identical capacitor is connected in series with it across the same battery, what is the charge on each?

Show answer & solution

Concept: Q = CV; in series the charge is the same on both.

  1. Alone: Q = 4 × 10⁻⁶ × 50 = 200 μC.
  2. In series: Ceq = 2 μF, so Q = 2 × 50 = 100 μC on each (25 V across each).

Answer: 200 μC; 100 μC each

N5Average

A dielectric of K = 5 fills a capacitor completely. What happens to C, Q, V and U if (a) the battery stays connected, (b) the battery was removed first?

Show answer & solution

Concept: C always becomes KC; then keep V fixed (a) or Q fixed (b).

  1. (a) V fixed: C × 5, Q = CV × 5, V same, U = ½CV² × 5.
  2. (b) Q fixed: C × 5, Q same, V = Q/C ÷ 5, U = Q²/2C ÷ 5.

Answer: (a) C, Q, U × 5, V same; (b) C × 5, Q same, V and U ÷ 5

Advanced level

N6Advanced

A slab of dielectric constant 4 and thickness d/2 is placed between the plates of a parallel-plate capacitor (plate separation d, capacitance C₀ in air). Find the new capacitance.

Show answer & solution

Concept: C = ε₀A/(d − t + t/K).

  1. d − t + t/K = d − d/2 + d/8 = 5d/8.
  2. C = ε₀A/(5d/8) = (8/5)ε₀A/d.
  3. C = 1.6 C₀.

Answer: 1.6 C₀

N7Advanced

A 2 μF capacitor charged to 100 V is connected to an uncharged 3 μF capacitor. Find the common potential and the energy lost.

Show answer & solution

Concept: charge is conserved; energy is not.

  1. Total charge = 2 × 100 = 200 μC; total C = 5 μF ⇒ V = 40 V.
  2. Energy before = ½ × 2 × 10⁻⁶ × 100² = 0.01 J.
  3. Energy after = ½ × 5 × 10⁻⁶ × 40² = 0.004 J.
  4. Loss = 0.006 J (check: C₁C₂(ΔV)²/2(C₁ + C₂) = 6 × 10⁻¹² × 10⁴/10⁻⁵ = 0.006 J).

Answer: 40 V; 6 mJ

N8Advanced

A 2 μF capacitor is in series with a parallel pair of 3 μF capacitors, and the combination is connected to 12 V. Find the equivalent capacitance, the voltage across the 2 μF capacitor and the charge on each 3 μF capacitor.

Show answer & solution

Concept: reduce step by step, then share charge and voltage.

  1. Parallel pair = 6 μF; in series with 2 μF: C = 2 × 6/8 = 1.5 μF.
  2. Total charge Q = 1.5 × 12 = 18 μC (same on 2 μF and on the pair).
  3. V across 2 μF = 18/2 = 9 V; across the pair = 18/6 = 3 V.
  4. Each 3 μF: Q = 3 × 3 = 9 μC.

Answer: 1.5 μF; 9 V; 9 μC each

N9Advanced

The field between the plates of an air capacitor is 10⁶ V m⁻¹. Find the energy density. If the plates are 1 mm apart with area 0.01 m², find the total energy stored.

Show answer & solution

Concept: u = ½ε₀E²; U = u × volume.

  1. u = ½ × 8.85 × 10⁻¹² × 10¹² ≈ 4.43 J m⁻³.
  2. Volume = 0.01 × 10⁻³ = 10⁻⁵ m³.
  3. U = 4.43 × 10⁻⁵ J (same as ½CV² with V = 1000 V, C = 88.5 pF).

Answer: about 4.43 J m⁻³; about 4.4 × 10⁻⁵ J

N10Advanced

An isolated parallel-plate capacitor with plate area 0.01 m² carries 1 μC. Find the force of attraction between its plates.

Show answer & solution

Concept: each plate feels the field of the other plate only, σ/2ε₀, so F = Q × Q/(2ε₀A).

  1. F = Q²/(2ε₀A) = (10⁻⁶)²/(2 × 8.85 × 10⁻¹² × 0.01).
  2. = 10⁻¹²/(1.77 × 10⁻¹³) ≈ 5.65 N.

Answer: about 5.65 N

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

The capacitance of a parallel-plate capacitor does NOT depend on:

  1. Plate area
  2. Plate separation
  3. Medium between plates
  4. Charge on the plates
Show solution

C = Kε₀A/d depends only on geometry and medium.

Answer: (D)

Q2Numerical

Three 3 μF capacitors are connected in series. Find the equivalent capacitance in μF.

Show solution

C = 3/3 = 1 μF.

Answer: 1

Q3Medium

A charged capacitor is disconnected from its battery and a dielectric slab is inserted. The stored energy:

  1. Increases
  2. Decreases
  3. Stays the same
  4. Becomes zero
Show solution

Q is fixed and C increases, so U = Q²/2C decreases (the slab is pulled in).

Answer: (B)

Q4Hard

Treating Earth as a conducting sphere of radius 6400 km, its capacitance is about:

  1. 711 pF
  2. 711 nF
  3. 711 μF
  4. 711 F
Show solution

C = 4πε₀R = 6.4 × 10⁶/(9 × 10⁹) ≈ 7.1 × 10⁻⁴ F = 711 μF.

Answer: (C)

Q5Numerical

Find the energy (in μJ) stored in a 2 μF capacitor charged to 10 V.

Show solution

U = ½ × 2 × 10⁻⁶ × 100 = 10⁻⁴ J = 100 μJ.

Answer: 100

Q6Medium

In a series combination of two unequal capacitors connected to a battery, the larger voltage appears across:

  1. The larger capacitor
  2. The smaller capacitor
  3. Both equally
  4. Depends on the battery
Show solution

Same Q on both, V = Q/C, so the smaller C has the larger V.

Answer: (B)

12. Quick Revision Summary

Remember before the exam
  • C = Q/V = Kε₀A/d; sphere 4πε₀R.
  • Series: same Q, 1/C adds; parallel: same V, C adds.
  • Dielectric: C × K; battery on → V fixed; battery off → Q fixed.
  • U = ½CV² = Q²/2C; u = ½ε₀E².
  • Charge sharing: common V = ΣQ/ΣC; energy is lost.
All Physics chaptersJEE Main Mock Tests