Capacitors: Charge Storage, Types, and Applications

What is a Capacitor?

A capacitor is an electronic device that stores electrical energy in an electric field. It consists of two conductors (called plates) separated by an insulating material (called a dielectric).

• When connected to a battery, one plate gains +Q charge, and the other gains -Q charge.
• The potential difference (V) between the plates depends on the charge (Q) stored and the capacitance (C) of the capacitor.

Capacitance (C)

Capacitance is the ability of a capacitor to store charge. It is defined as: $$C=\frac{Q}{V}$$

• Unit: Farad (F), where 1 F = 1 C/V.
• 1 Farad is a very large unit, so we often use microfarad (µF) or picofarad (pF).

Factors Affecting Capacitance

The capacitance of a parallel-plate capacitor depends on:

1. Area of the plates (A): Larger plates → more charge stored → higher C.
2. Distance between plates (d): Smaller distance → stronger electric field → higher C.
3. Dielectric constant (K): The dielectric material between plates increases C by a factor of K (e.g., air has K=1, mica has K=6).

The formula is: $$C=\frac{K{ϵ}_{0}A}{d}$$ where:

• ${ϵ}_0=8.85\times 10^{-12}\text{F/m}$ (permittivity of free space).

Types of Capacitors

1. Parallel-Plate Capacitor

• Two parallel metal plates separated by a dielectric.
• Used in circuits, filters, and tuning devices.

2. Cylindrical Capacitor

• Two coaxial cylinders (one inside the other).
• Used in cables and radio frequency applications.

• Formula: $$C=\frac{2\pi {ϵ}_{0}Kl}{\ln (r_2/r_1)}$$ where $r_1$ = inner radius, $r_2$ = outer radius, $l$ = length.

3. Spherical Capacitor

• Two concentric spherical shells.
• Used in high-voltage applications.

• Formula: $$C=\frac{4\pi {ϵ}_{0}K{r}_1{r}_2}{r_2-r_1}$$ where $r_1$ = inner radius, $r_2$ = outer radius.

Capacitors in Series and Parallel

1. Capacitors in Parallel

• Total capacitance (C_eq) increases.
• Formula: $$C_{eq}=C_1+C_2+C_3+\dots$$
• Voltage across each capacitor is the same (V).

graph LR
    A["Battery"] --> B["C1"]
    A --> C["C2"]
    B --> D["Common Point"]
    C --> D
    D --> E["Ground"]

2. Capacitors in Series

• Total capacitance (C_eq) decreases.
• Formula: $$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\dots$$
• Charge on each capacitor is the same (Q).

graph LR
    A["Battery"] --> B["C1"]
    B --> C["C2"]
    C --> D["Ground"]

Energy Stored in a Capacitor

A charged capacitor stores electrical potential energy (U) given by: $$U=\frac{1}{2}C{V}^2=\frac{1}{2}\frac{Q^2}{C}$$

• Used in flash cameras, defibrillators, and energy storage devices.

Applications of Capacitors

Solved Example 1: Calculating Capacitance

Problem: A parallel-plate capacitor has plates of area 100 cm² separated by 2 mm of mica (K=6). Find its capacitance.

Solution:

1. Convert area to m²: $$A=100{\text{cm}}^2=100\times 10^{-4}{\text{m}}^2=0.01{\text{m}}^2$$
2. Use the formula: $$C=\frac{K{ϵ}_{0}A}{d}=\frac{6\times 8.85\times 10^{-12}\times 0.01}{0.002}$$
3. Calculate: $$C=\frac{53.1\times 10^{-12}}{0.002}=26.55\times 10^{-9}\text{F}=26.55\text{nF}$$

Answer: 26.55 nF

Solved Example 2: Capacitors in Series

Problem: Two capacitors, C₁ = 2 µF and C₂ = 3 µF, are connected in series to a 12 V battery. Find:

1. Equivalent capacitance.
2. Charge on each capacitor.
3. Voltage across each capacitor.

Solution:

1. Equivalent capacitance (C_eq): $$\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}⟹C_{eq}=\frac{6}{5}=1.2\mu \text{F}$$
2. Total charge (Q): $$Q=C_{eq}V=1.2\times 10^{-6}\times 12=14.4\times 10^{-6}\text{C}=14.4\mu \text{C}$$ (Same for both capacitors in series.)
3. Voltage across C₁ (V₁) and C₂ (V₂): $$V_1=\frac{Q}{C_1}=\frac{14.4}{2}=7.2\text{V}$$ $$V_2=\frac{Q}{C_2}=\frac{14.4}{3}=4.8\text{V}$$ (Check: $V_1+V_2=12\text{V}$, which matches the battery voltage.)

Answer:

1. 1.2 µF
2. 14.4 µC on each
3. 7.2 V (C₁), 4.8 V (C₂)

NEB Board-Style Questions

Short Answer (3 marks)

1. Define capacitance. Write the formula for capacitance of a parallel-plate capacitor and state the factors affecting it.
2. Two capacitors of 4 µF and 6 µF are connected in parallel to a 12 V battery. Find the total charge stored.

Long Answer (5 marks)

3. Derive the formula for capacitance of a cylindrical capacitor. A coaxial cable has inner radius 1 mm, outer radius 4 mm, and length 1 m. If the space between them is filled with a dielectric of K=2, find its capacitance.
4. Explain how capacitors are used in timing circuits. Draw a simple circuit diagram showing a capacitor and resistor in series with a battery.

Exam Tip

✅ Memorize formulas for parallel-plate, cylindrical, and spherical capacitors. ✅ Practice series-parallel combinations—NEB often tests equivalent capacitance. ✅ Understand energy storage—questions may ask for energy in terms of Q or V. ✅ Draw diagrams—label plates, dielectrics, and connections clearly. ✅ Unit conversions—always convert cm² to m² and mm to m before calculations.

Capacitor symbol and real-world example (e.g., in a smartphone) (Image: Giovanna 27, CC BY 4.0, via Wikimedia Commons)