Class 11 Physics Capacitor Notes

Unit 7

Electricity and Magnetism

Class 11 Physics

Chapter 22

Capacitor

Class 11 Physics – Capacitor Notes PDF

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Chapter Overview

A capacitor is a device for storing separated electric charge and electrical energy. Its basic form consists of two conductors separated by an insulating region. Capacitors are used in timing, filtering, smoothing, energy storage, camera flashes, tuning, sensing and many other circuits.

This chapter develops capacitance, the parallel-plate capacitor, series and parallel combinations, energy stored in a charged capacitor and the effect of dielectrics.

22.1 Capacitance and Capacitor

Capacitance

For a capacitor, the magnitude of charge Q stored on either conductor is proportional to the potential difference V between the conductors:

C = Q/V

The SI unit of capacitance is the farad (F).

1 F = 1 C V−1

Common practical units are μF, nF and pF.

Charge–Potential Graph

Because Q = CV, a graph of Q against V is a straight line through the origin with gradient C. If V is plotted against Q, the gradient is 1/C.

Diagram 1 — Basic Capacitor and Q–V Graph

+Q−Q two conductors separated by insulator VQ slope = C

Capacitance is the ratio Q/V; on a Q–V graph, the slope equals C.

Uses of Capacitors

  • Temporary energy storage.
  • Smoothing and filtering in power supplies.
  • Timing circuits with resistors.
  • Tuning circuits and oscillators.
  • Blocking steady DC while allowing changing signals in suitable circuits.

22.2 Parallel-Plate Capacitor

Consider two large parallel plates of area A separated by distance d in vacuum or air. Neglect edge effects.

Surface charge density is:

σ = Q/A

Gauss’s law gives the electric field between oppositely charged plates:

E = σ/ε₀ = Q/(ε₀A)

The potential difference is:

V = Ed = Qd/(ε₀A)

Therefore:

C = Q/V = ε₀A/d

Diagram 2 — Parallel-Plate Capacitor

+Q−Q uniform E d C = ε₀A/d

Capacitance increases with plate area and decreases as plate separation increases.

ChangeEffect on C
Increase ACapacitance increases proportionally
Increase dCapacitance decreases inversely
Insert dielectric of relative permittivity κ filling the gapC becomes κ times larger

22.3 Combination of Capacitors

Capacitors in Parallel

All capacitors have the same potential difference V. Charges add:

Q = Q₁ + Q₂ + … = (C₁ + C₂ + …)V
Ceq = C₁ + C₂ + C₃ + …

Diagram 3 — Capacitors in Parallel

C₁C₂C₃ Ceq = C₁ + C₂ + C₃

Parallel capacitors share the same voltage, while their stored charges add.

Capacitors in Series

In a series chain, each capacitor carries the same magnitude of charge Q, while the total potential difference is the sum of individual potential differences:

V = V₁ + V₂ + … = Q(1/C₁ + 1/C₂ + …)
1/Ceq = 1/C₁ + 1/C₂ + 1/C₃ + …

For two capacitors:

Ceq = C₁C₂/(C₁+C₂)

Diagram 4 — Capacitors in Series

C₁C₂C₃ 1/Ceq = 1/C₁ + 1/C₂ + 1/C₃

Series capacitors carry equal charge magnitude; the applied voltage divides among them.

Quick check: A parallel equivalent capacitance is larger than any individual branch capacitance. A series equivalent capacitance is smaller than the smallest individual capacitance.

22.4 Energy Stored in a Charged Capacitor

Charging a capacitor requires work because additional charge is moved against an increasing potential difference.

At an intermediate charge q:

V = q/C

The small work is:

dU = V dq = (q/C)dq

Integrating from 0 to Q:

U = ∫₀Q(q/C)dq = Q²/(2C)

Using Q = CV:

U = ½QV = ½CV² = Q²/(2C)

Diagram 5 — Energy from the V–Q Graph

qV QV Area = ½QV = U

On a V–Q graph, the triangular area under the charging line equals the energy stored.

Energy Density

For a parallel-plate capacitor in vacuum, U = ½CV² leads to energy per unit volume:

u = ½ε₀E²

This expresses the idea that electrical energy is stored in the electric field.

22.5 Effect of a Dielectric

A dielectric is an insulating material that becomes polarized in an electric field. Bound positive and negative charges shift slightly in opposite directions, creating polarization that opposes part of the applied field.

If a dielectric of relative permittivity κ completely fills the gap of a parallel-plate capacitor:

C = κε₀A/d = κC₀

Diagram 6 — Polarization of a Dielectric

+ plate− plate dielectric − +− +− +− +− +− + external E

Polarization produces bound charges whose field partially opposes the applied field, increasing capacitance.

Situation after dielectric insertionCharge QVoltage VCapacitance C
Capacitor disconnected from sourceConstantDecreases to V/κIncreases to κC
Capacitor remains connected to ideal voltage sourceIncreases to κQConstantIncreases to κC

Formula Summary

ConceptFormula
CapacitanceC = Q/V
Parallel-plate capacitorC = ε₀A/d
With dielectricC = κε₀A/d
Parallel combinationCeq = C₁ + C₂ + …
Series combination1/Ceq = 1/C₁ + 1/C₂ + …
Stored energyU = ½QV = ½CV² = Q²/(2C)
Vacuum electric-field energy densityu = ½ε₀E²

Solved Numerical Examples

Example 1 — Basic Capacitance

Question: A capacitor stores 20 μC at 10 V. Find C.

C = Q/V = 20 μC / 10 V = 2.0 μF

Answer: 2.0 μF.

Example 2 — Parallel-Plate Capacitor

Question: A parallel-plate capacitor has A = 0.020 m² and d = 1.0 mm in air. Find C.

C = ε₀A/d = (8.854×10⁻¹²)(0.020)/(1.0×10⁻³) ≈ 1.77×10⁻¹⁰ F

Answer: approximately 177 pF.

Example 3 — Series Combination

Question: Find the equivalent capacitance of 6 μF and 3 μF in series.

Ceq = (6×3)/(6+3) μF = 2 μF

Answer: 2 μF.

Example 4 — Parallel Combination

Question: Find the equivalent capacitance of 2 μF, 4 μF and 6 μF in parallel.

Ceq = 2 + 4 + 6 = 12 μF

Answer: 12 μF.

Example 5 — Stored Energy

Question: A 5 μF capacitor is charged to 200 V. Find the energy stored.

U = ½CV² = ½(5×10⁻⁶)(200)² = 0.10 J

Answer: 0.10 J.

Example 6 — Dielectric

Question: A 100 pF capacitor is fully filled with dielectric of κ = 4. Find the new capacitance.

C = κC₀ = 4×100 pF = 400 pF

Answer: 400 pF.

Important Exam Questions

Short-Answer Questions

  1. Define capacitor and capacitance.
  2. State the SI unit of capacitance.
  3. What does the slope of a Q–V graph represent?
  4. List common uses of capacitors.
  5. How do plate area and separation affect capacitance?
  6. State the formula for a parallel-plate capacitor.
  7. State the equivalent capacitance formulas for series and parallel connections.
  8. Why is series equivalent capacitance smaller than the smallest individual capacitance?
  9. Write three equivalent formulas for capacitor energy.
  10. What is a dielectric? What is polarization?
  11. How does a dielectric change capacitance?

Long-Answer / Derivation Questions

  1. Derive C = ε₀A/d for a parallel-plate capacitor using the field between parallel plates.
  2. Derive the equivalent capacitance of capacitors connected in series.
  3. Derive the equivalent capacitance of capacitors connected in parallel.
  4. Using the V–Q graph, derive U = ½QV and hence U = ½CV².
  5. Explain dielectric polarization and its effect on a parallel-plate capacitor.

Numerical Questions

  1. A 4 μF capacitor carries 12 μC. Find its potential difference.
  2. Two capacitors 4 μF and 12 μF are connected in series. Find Ceq.
  3. The same capacitors are connected in parallel. Find Ceq.
  4. A 10 μF capacitor is charged to 100 V. Find its stored energy.
  5. A parallel-plate capacitor has area 0.01 m² and gap 0.5 mm. Estimate C in air.
  6. A dielectric of κ = 5 is inserted into a 40 pF capacitor. Find its new capacitance.

Diagram Questions

  1. Draw a basic capacitor and its Q–V graph.
  2. Draw a parallel-plate capacitor with field lines.
  3. Draw three capacitors in parallel.
  4. Draw three capacitors in series.
  5. Draw the V–Q energy graph and shade the stored-energy area.
  6. Draw a polarized dielectric between capacitor plates.

One-Minute Revision

  • Capacitance is C = Q/V.
  • The SI unit of capacitance is farad.
  • On a Q–V graph, slope = C.
  • For parallel plates, C = ε₀A/d.
  • Larger plate area gives larger capacitance.
  • Larger plate separation gives smaller capacitance.
  • Parallel capacitors have the same voltage.
  • For parallel capacitors, Ceq = ΣC.
  • Series capacitors carry the same charge magnitude.
  • For series capacitors, 1/Ceq = Σ(1/C).
  • Stored energy is U = ½QV.
  • Also U = ½CV² = Q²/(2C).
  • A dielectric polarizes in an electric field.
  • A dielectric filling the gap increases capacitance by factor κ.
  • Electrical energy is associated with the electric field between the plates.

Diagram Practice

  1. Draw a two-plate capacitor and label +Q, −Q and V.
  2. Draw the Q–V graph and state its slope.
  3. Draw a parallel-plate capacitor and uniform field.
  4. Draw capacitors in parallel and write Ceq.
  5. Draw capacitors in series and write 1/Ceq.
  6. Draw a V–Q graph and shade the triangular energy area.
  7. Draw a dielectric showing polarization between plates.

Syllabus Coverage Checklist

NEB/CDC Chapter 22 scopeCovered
22.1 Capacitance and capacitor; uses; C = Q/V; graph relationYes
22.2 Parallel-plate capacitor; derivation; effects of A, d and dielectricYes
22.3 Combination of capacitors in series and parallelYes — derivations and numericals
22.4 Energy stored in a charged capacitor from potential–charge graphYes
22.5 Effect of dielectric and polarizationYes

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