Combination of Capacitors: Series and parallel groupings and energy stored in a capacitor

📘 1. Capacitance

The capacitance of a conductor is defined as the ratio of charge stored on the conductor to its potential difference.

C = Q/V

SI unit of capacitance is farad (F).

⚡ 2. Parallel-Plate Capacitor Without Dielectric

For two large parallel conducting plates having area A and separation d:

C₀ = ε₀A/d

where ε₀ is the permittivity of free space.

Plate Area ↑

C ∝ A

Increasing plate area increases capacitance.

Separation ↑

C ∝ 1/d

Increasing separation decreases capacitance.

Vacuum / Air

K ≈ 1

🟢 3. Capacitor With Dielectric

If a dielectric material of dielectric constant K completely fills the space between the plates:

C = Kε₀A/d

C = KC₀

⭐ Important: A dielectric increases capacitance by a factor K.

🔬 4. Interactive Practical — Dielectric Insertion

Plate Area A 5

Plate Separation d 5

Dielectric Constant K 4

Dielectric

+ + +

− − −

+

−

Vacuum Capacitance: 0 units
With Dielectric: 0 units
Ratio C/C₀: 1
Status: No dielectric

🔋 5. Dielectric With Battery Connected

When the capacitor remains connected to the battery, the potential difference remains constant.

V = constant

C' = KC

Q' = KQ

E' = E

U' = KU

🔋 6. Dielectric With Battery Disconnected

When the battery is disconnected, the charge cannot flow to or from the capacitor.

Q = constant

C' = KC

V' = V/K

E' = E/K

U' = U/K

📐 7. Dielectric Slab of Thickness t

If a dielectric slab of thickness t is inserted between plates separated by distance d:

C = ε₀A / [d − t + t/K]

Equivalent form:

C = Kε₀A / [K(d − t) + t]

📚 8. Important Formula Summary

Without Dielectric

C₀ = ε₀A/d

Complete Dielectric

C = Kε₀A/d

Capacitance Ratio

C/C₀ = K

Energy

U = ½CV²

Charge Form

U = Q²/2C

Electric Field

E = V/d

🧮 9. Solved Numerical

Question:

A parallel-plate capacitor has plate area 2 m² and separation 2 mm. Find its capacitance in vacuum.

C = ε₀A/d

C = (8.85×10⁻¹² × 2)/(2×10⁻³)

C = 8.85 × 10⁻⁹ F

📝 10. MCQ Practice

1. Capacitance of a parallel-plate capacitor is:

A. ε₀d/A

B. ε₀A/d

C. Ad/ε₀

D. ε₀Ad

✔ Answer: B

2. If plate area is doubled, capacitance becomes:

A. C/2

B. C

C. 2C

D. 4C

✔ Answer: C

3. If separation between plates is doubled:

A. C doubles

B. C becomes half

C. C becomes four times

D. C remains same

✔ Answer: B

4. When dielectric completely fills the capacitor:

A. C₀/K

B. KC₀

C. C₀

D. K²C₀

✔ Answer: B

5. Dielectric constant of vacuum is:

A. 0

B. 1

C. 9

D. ∞

✔ Answer: B

6. With battery disconnected, dielectric insertion causes charge to:

A. Increase

B. Remain constant

C. Decrease K times

D. Become zero

✔ Answer: B

7. With battery disconnected, potential difference becomes:

A. KV

B. V

C. V/K

D. K²V

✔ Answer: C

8. Energy stored in a capacitor is:

A. CV²

B. ½CV²

C. 2CV²

D. C/V²

✔ Answer: B

9. SI unit of capacitance is:

A. Coulomb

B. Volt

C. Farad

D. Ohm

✔ Answer: C

10. If dielectric constant K increases, capacitance:

A. Decreases

B. Remains same

C. Increases

D. Becomes zero

✔ Answer: C

🟢 11. 2 Marks — 6 Questions

Q1 Define capacitance and write its SI unit.

Q2 Write the formula for capacitance of a parallel-plate capacitor.

Q3 What happens to capacitance when plate area is doubled?

Q4 Define dielectric constant.

Q5 Write the expression for energy stored in a capacitor.

Q6 What happens to capacitance when a dielectric completely fills the space?

🟡 12. 3 Marks — 6 Questions

Q1 Derive C = ε₀A/d for a parallel-plate capacitor.

Q2 Explain the effect of increasing plate separation on capacitance.

Q3 Explain the effect of inserting a dielectric slab.

Q4 Differentiate between capacitance with and without dielectric.

Q5 Write three expressions for energy stored in a capacitor.

Q6 Explain why capacitance increases when dielectric is inserted.

🟠 13. 4 Marks — 6 Questions

Q1 Derive the capacitance of a parallel-plate capacitor.

Q2 Derive C = Kε₀A/d when a dielectric completely fills the space.

Q3 Explain changes in Q, V, C and U when dielectric is inserted with battery connected.

Q4 Explain changes in Q, V, C and U when dielectric is inserted after disconnecting the battery.

Q5 Derive capacitance when a dielectric slab of thickness t is inserted.

Q6 Explain the effect of plate area and plate separation on capacitance.

🔴 14. 5 Marks — 6 Questions

Q1 Derive the expression for capacitance of a parallel-plate capacitor in vacuum.

Q2 Explain in detail the effect of a dielectric slab on capacitance.

Q3 Derive the capacitance of a capacitor partially filled with dielectric.

Q4 Compare Q, V, E, C and U when dielectric is inserted with battery connected.

Q5 Compare Q, V, E, C and U when dielectric is inserted with battery disconnected.

Q6 Derive different expressions for energy stored in a capacitor.

🔵 15. 6 Marks — 6 Questions

Q1 Derive the capacitance of a parallel-plate capacitor and discuss all factors affecting it.

Q2 Derive the capacitance when a dielectric slab of thickness t is inserted between plates separated by d.

Q3 Explain completely the effect of dielectric insertion when the capacitor remains connected to a battery.

Q4 Explain completely the effect of dielectric insertion when the capacitor is disconnected from the battery.

Q5 Derive the energy stored in a capacitor and discuss the effect of dielectric.

Q6 A charged parallel-plate capacitor is disconnected from the battery. A dielectric slab is inserted completely. Derive changes in C, V, E and U.

🚀 Quick Revision

Capacitance: C = Q/V

Parallel Plate: C₀ = ε₀A/d

Complete Dielectric: C = Kε₀A/d = KC₀

Partial Dielectric: C = ε₀A/[d − t + t/K]

Energy: U = ½CV² = Q²/2C = ½QV

Battery Connected: V constant, Q increases K times, C increases K times

Battery Disconnected: Q constant, V becomes V/K, C increases K times