Capacitance

Capacitance is a fundamental concept in electronics and electrical engineering. At its core, capacitance is the ability of a system to store electrical energy in an electric field. It’s most commonly associated with capacitors, components used in countless electronic devices—from smartphones and computers to electric vehicles and power grids.Measured in farads (F), capacitance plays a key role in controlling voltage, filtering signals, and maintaining power supply stability. 

1.0Definition of Capacitance

• Capacitance of conductor is defined as charge required to increase the potential of conductor by one unit.

• When a conductor is charged ,its electric potential increases.
• For an isolated conductor with finite size the potential at infinity is considered zero.
• The potential V of the conductor is directly proportional to the charge q on it.

$q\propto V\Rightarrow q=CV$

• The constant C is called the capacitance of the conductor, it measures the conductor ability to store charge per unit potential.

Graph between Q and V

Understanding Capacitance: Units & Dependencies

• It is a scalar quantity.
• Unit of capacitance is farad in SI units and its dimensional formula is$\left[M^{-1}{L}^{-2}{I}^2{T}^4\right]$.
• 1 Farad: 1 Farad is the capacitance of a conductor for which 1 coulomb charge increases potential by 1 volt.

$1\text{ Farad }=\frac{1\text{ Coulomb }}{1\text{ Volt }}$

$1\mu F=10^{-6}F,1nF=10^{-9}F\text{ or }1pF=10^{-12}F$

$1F=9\times 10^{11}\text{ state }-F$

Capacitance of an isolated conductor depends on following factors:

(a) Shape and size of the conductor: On increasing the size, capacitance increases.

(b) On the surrounding medium: With an increase in dielectric constantK, capacitance increases.

(c) Presence of other conductors: When a neutral conductor is placed near a charged conductor, capacitance of conductors increases.

Capacitance of a conductor does not depend on

(a) Charge on the conductor

(b) Potential of the conductor

2.0Capacitance of Parallel Plate Capacitor

• It consists of two metallic plates M and N each of area A at separation d. Plate M is positively charged and plate N is earthed. If ${\epsilon }_r$ is the dielectric constant of the material medium and E is the field at a point P that exists between the two plates then

Step-1: Finding Electric Field

$E=E_{+}+E_{-}=\frac{\sigma }{2E}+\frac{\sigma }{2E}=\frac{\sigma }{E}=\frac{\sigma }{{ϵ}_0{ϵ}_R}\left[ϵ={ϵ}_0{ϵ}_R\right]$

Step-2: Finding Potential Difference

$V=Ed=\frac{\sigma }{{ϵ}_0{ϵ}_R}d=\frac{qd}{A{ϵ}_0{ϵ}_R}\left(\because E=\frac{V}{d}\text{ and }\sigma =\frac{q}{A}\right)$

Step-3: Finding Capacitance

$C=\frac{q}{V}=\frac{A{ϵ}_0{ϵ}_R}{d}$

Capacitance of a capacitor depends on

(a) Area of plates. 

(b) Distance between the plates. 

(c) Dielectric medium between the plates.

3.0Capacitance of Spherical Capacitor

Capacitance of an Isolated Spherical Conductor 

Let there is charge Q on sphere of radius R

Potential $V=\frac{KQ}{R}$

Hence by formula,

$Q=CV\to Q=\frac{CKQ}{R}\to C=4\pi {ϵ}_{0}R$

(a) If the medium around the conductor is a vacuum or air

$C_{\text{vacuum }}=4\pi {ϵ}_{0}R$

R=Radius of spherical conductor may be solid or hollow

If the medium around the conductor is a dielectric of constant K from surface of sphere to infinity then,

$C_{\text{Medium }}=4\pi {ϵ}_{0}KR$

(b) $\frac{C_{\text{Medium }}}{C_{\text{Air/vacuum }}}=K=DielectricConstant$

Capacitance of a spherical capacitor

1. Outer sphere is earthed

When a charge Q is given to inner sphere it is uniformly distributed on its surface,A charge -Q is induced induced on inner surface of outer sphere. The charge +Q induced on outer surface of outer sphere. The charge +Q induced on the outer surface of the outer sphere flows to earth as it is grounded.

E=0 for r<a

E=0 for r>b

This arrangement is known as a spherical capacitor.

Potential of inner sphere,

$V_1=\frac{Q}{4\pi {ϵ}_{0}a}+\frac{-Q}{4\pi {ϵ}_{0}b}\Rightarrow \frac{Q}{4\pi {ϵ}_0}\left(\frac{b-a}{ab}\right)$

As outer surface is earthed so potential V₂=0

$\begin{array}{rl}& V_1-V_2=\left[\frac{KQ}{a}-\frac{KQ}{b}\right]-\left[\frac{KQ}{b}-\frac{KQ}{b}\right]=\frac{KQ}{a}-\frac{KQ}{b}\\ & C=\frac{Q}{V_1-V_2}=\frac{Q}{\frac{KQ}{a}-\frac{KQ}{b}}=\frac{ab}{K(b-a)}=\frac{4\pi {ϵ}_{0}ab}{b-a}\\ & C=\frac{4\pi {ϵ}_{0}ab}{b-a}\end{array}$

If b>>a then $C=4\pi {ϵ}_{0}a$ (Like an isolated spherical capacitor)

If dielectric mediums are filled as shown then $C=\frac{4\pi {ϵ}_0{ϵ}_{r2}ab}{b-a}$

(b) Inner Sphere is Earthed

Here the system is equivalent to a spherical capacitor of inner and outer radii a and b respectively and a spherical conductor of radius b in parallel. This is because charge Q given to outer sphere distributes in such a way that for the outer sphere charge on the inner sides is $\frac{a}{b}Q$ and charge on the outer side is

$Q-\frac{a}{b}Q=\frac{(b-a)}{b}Q$

So total capacitance of the system

$C=4\pi {ϵ}_0\frac{ab}{b-a}+4\pi {ϵ}_{0}b$

$C=\frac{4\pi {ϵ}_0{b}^2}{b-a}$

4.0Capacitance of Cylindrical Capacitor

There are two co-axial conducting cylindrical surfaces where l>>a and l>>b, where a and b are the radius of cylinders.

When a charge Q is given to inner cylinder it is uniformly distributed on its surface.A charge -Q is induced on inner surface of outer cylinder.The charge +Q induced on the outer surface of the outer cylinder flows to Earth as it is grounded. 

Electrical Field between cylinder, $E=\frac{\lambda }{2\pi {ϵ}_{0}r}=\frac{Q/l}{2\pi {ϵ}_{0}r}$

Potential Difference between plates,

$V={\int }_a^b\frac{Q}{2\pi {ϵ}_{0}rl}dr=\frac{Q}{2\pi {ϵ}_{0}l}\ln \left(\frac{b}{a}\right)$

Capacitance per unit length

$C=\frac{\lambda }{V}=\frac{\lambda }{2K\lambda \ln \frac{b}{a}}=\frac{4\pi {ϵ}_0}{2\ln \frac{b}{a}}=\frac{2\pi {ϵ}_0}{\ln \frac{b}{a}}$

Capacitance per unit length $=\frac{2\pi {ϵ}_0}{\ln \frac{b}{a}}F/m$