Electrical Capacitance

1. Concept Overview

Every conductor can store electric charge. However, the amount of charge it can hold depends on how much electric potential develops on it when that charge is placed.

Electrical Capacitance is a measure of the ability of a conductor to store charge without becoming too highly charged (high potential).

• If a conductor gains charge [Q] and its potential becomes [V], then its capacitance is defined as:

[
\boxed{C = \dfrac{Q}{V}}
]

• Large capacitance means:
  • the conductor can hold a large charge
  • for a small rise in potential
  • i.e., it is a “good charge-storer”

Capacitance depends only on the shape, size, and surroundings of the conductor, not on the charge stored.

This concept is foundational for understanding capacitors, energy storage, filters, and nearly all electronic circuits.

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2. Clear Explanation and Mathematical Derivation

Consider an isolated conductor given a charge [Q]. The potential developed is [V].

Experimentally and from electrostatics:

• [V \propto Q]
• Or: [V = kQ]
  where [k] depends on the geometry and environment.

Thus:

[C = \dfrac{Q}{V} = \dfrac{1}{k}]

Because [k] depends only on shape and medium, capacitance is purely a property of geometry.

Example: Capacitance of an isolated spherical conductor

For a sphere of radius [R]:

 

[V] [= \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}]

Thus:

[C] [= \dfrac{Q}{V}] [= 4\pi\varepsilon_0 R]

This shows capacitance increases with size.

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3. Dimensions and Units

• SI Unit of capacitance: Farad (F)
  [
  1\ \text{F} = 1\ \text{C V}^{-1}
  ]
• Practical units:
  • microfarad: [\mu\text{F} = 10^{-6}\ \text{F}]
  • nanofarad: [\text{nF} = 10^{-9}\ \text{F}]
  • picofarad: [\text{pF} = 10^{-12}\ \text{F}]
• Dimensions of capacitance:
  [[M^{-1} L^{-2} T^{4} A^{2}]]

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4. Key Features

1. Capacitance depends on geometry, not on charge.
2. Directly proportional to the ability to store charge.
3. Medium affects capacitance through permittivity [\varepsilon].
4. Larger conductor → larger capacitance.
5. Capacitance increases when conductors are brought near others (basis of capacitors).
6. Used in energy storage: [U = \dfrac{1}{2}CV^2].
7. Determines how circuits respond to AC signals (filters, timing circuits).

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5. Important Formulas to Remember

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6. Conceptual Questions with Solutions

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7. FAQ / Common Misconceptions

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8. Practice Questions (With Step-by-Step Solutions)

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Q1. A conductor develops a potential of [50\ \text{V}] when it holds a charge [2.5\times10^{-6}\ \text{C}]. What is its capacitance?

Solution:

[C] [= \dfrac{Q}{V}] [= \dfrac{2.5\times10^{-6}}{50}]

[C] [= 5.0\times10^{-8}\ \text{F}] [= 50\ \text{nF}]

Answer: [C = 50\ \text{nF}]

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Q2. An isolated spherical conductor has radius [0.5\ \text{m}]. Find its capacitance.

Solution:

[C] [= 4\pi\varepsilon_0 R] [= 4\pi (8.85\times10^{-12})(0.5)]

[C] [= 4\pi \times 4.425\times10^{-12}] [\approx 55.6\times10^{-12}\ \text{F}]

[C] [\approx 5.6\times10^{-11}\ \text{F}] [= 56\ \text{pF}]

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Q3. A conductor has capacitance [20\ \text{pF}]. What charge is required to raise potential by [200\ \text{V}]?

Solution:

[Q = CV] [= 20\times10^{-12} \times 200]

[Q = 4000\times10^{-12}] [= 4.0\times10^{-9}\ \text{C}]

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Q4. A sphere of radius [R] has capacitance [C_1]. If its radius is doubled, what is new capacitance [C_2]?

Solution:

[
C = 4\pi\varepsilon_0 R
]

So if [R \rightarrow 2R]:

[C_2] [= 4\pi\varepsilon_0 (2R)] [= 2C_1]

Capacitance doubles.

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Q5. A conductor requires [Q=5\times10^{-6}\ \text{C}] to raise its potential from [0\ \text{V}] to [100\ \text{V}]. How much work is done in charging it?

Solution:

Energy stored:

[U] [= \dfrac{1}{2}QV] [= \dfrac{1}{2}(5\times10^{-6})(100)]

[U] [= \dfrac{500\times10^{-6}}{2}] [= 250\times10^{-6}] [= 2.5\times10^{-4}\ \text{J}]