Capacitor and Capacitance

1.0What is a Capacitor?

Capacitors are simple passive devices that store electrostatic energy in an electric field. A capacitor or condenser consists of two conductors separated by an insulator or dielectric. Having opposite charges on which sufficient quantity of charge may be accommodated. It is a device which is used to store energy in the form of an electric field by storing charge.

2.0Circuit Symbol of Capacitor

The capacitor is represented as following:

3.0What is Capacitance?

Capacitance is the ability of a body to store an electrical charge. It is a fundamental concept in electrostatics and is quantified by the ratio of the charge stored on a conductor to the potential difference applied across it. The SI unit for capacitance is the Farad (F), named after Michael Faraday. The capacitance symbol is 'C'.

Capacitance of conductor depends upon shape, size, presence of medium and nearness of other conductor. Electrical capacitance is a Scalar quantity.

General Formula:

$C=Q/V$

This equation shows that capacitance is the amount of charge stored per unit voltage.

Capacitance Unit

• Capacitance Unit: The SI unit of capacitance is the farad (F).
• Capacitance Unit Symbol: F
• Capacitance Symbol: C

Remember: 1 Farad (F): A capacitor has a capacitance of one farad if a charge of one coulomb increases its potential by one volt.

In practice, one farad is very large. Therefore, smaller units are commonly used:

Microfarad (μF = $10^{-6}$ F)
$Nanofarad(nF=10^{-9}F)$
Picofarad (pF = $10^{-12}$ F)

4.0Capacitors in Series and Parallel

Capacitors can be connected in circuits to achieve a desired equivalent capacitance.

Series Combination

When initially uncharged capacitors are connected as shown so that charges do not have any alternative path(s) to flow then the combination is called series combination.

To find equivalent capacitance of this combination lets connect a battery across its terminals.

$C_{eq}=\frac{Q}{V}$

Let's assume that initially, the capacitors were uncharged and after connecting to the battery, Q charge flows through the battery as shown in above figure.

$\frac{-Q}{C_1}+\frac{-Q}{C_2}+\frac{-Q}{C_3}+V=0;\frac{V}{Q}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\Rightarrow \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\text{or in general}\frac{1}{C_{eq}}={\sum }_{n=1}^n\frac{1}{C_n}$

Parallel Combination

When one plate of each capacitor (more than one) is connected together and the other plate of each capacitor is connected together, such a combination is called a parallel combination.

5.0Derivation of Equivalent Capacitance

Q = Q₁ + Q₂ + Q₃
Q = C₁V + C₂V + C₃V = V(C₁ + C₂ + C₃)
Q / V = C₁ + C₂ + C₃
Ceq = C₁ + C₂ + C₃

In General $C_{eq}={\sum }_{n=1}^n{C}_n$

Remember: Half of the energy supplied by the battery is stored in the form of electrostatic energy and half of the energy is converted into heat through resistance. (If all capacitors are initially uncharged) 

6.0Potential Energy of a Charged Conductor

Since the electrostatic field is conservative field and in conservative field work done by an external agent is stored in the form of potential energy.

* Potential energy of the conductor will be stored in the form of an electric field.

* Potential energy of a conductor which is charged by V potential is given by

$P.E.=W_{ext},U=\frac{1}{2}C{V}^2=\frac{1}{2}QV=\frac{Q^2}{2C}$

7.0Potential Energy of Conducting Sphere

U = 1/2 CV² = 1/2 QV = Q²/2C

∴ C = 4πε₀R

U = Q² / (2·4πε₀R)

U = KQ² / (2R)

Self-potential energy of a spherical conductor. This potential energy is stored in the form of an electric field.

8.0Energy of Parallel Plate Capacitor

Energy stored in capacitor = Q² / 2C putting the value of capacitance for parallel plate capacitor

U = 1/2 · Q²/C = (Aσ)² / 2 × d / (ε₀A)

U = 1/2 (Ad) · σ² / ε₀

Energy density

U / V = σ² / 2ε₀ = 1/2 ε₀ E² (here V is volume i.e. A·d and E = σ / ε₀)

Once it is established that a region containing electric field E has energy 1/2 ε₀ E² per unit volume, the result can be used for any electric field whether it is due to a capacitor or otherwise.

9.0Effect of a Dielectric

A dielectric is an insulating material (e.g., mica, glass, oil) that can be polarized by an external electric field. When a dielectric is placed between the plates of a capacitor, it reduces the electric field and therefore the potential difference, increasing the capacitance.

• The insulators in which at microscopic level displacement of charges takes place in presence of electric field are known as dielectrics.

• Dielectrics are non-conductors upto certain value of field depending on its nature. If the field exceeds this limiting value called dielectric strength they lose their insulating property and begin to conduct.

• Dielectric strength is defined as the maximum value of electric field that a dielectric can tolerate without breakdown. Unit is volt/metre. Dimensions M¹ L¹ T⁻³ A⁻¹.

10.0Polarisation Vector ($\vec{P}$)

This is a vector quantity which describes the extent to which molecules of dielectric become polarized by an electric field or oriented in direction of field.

$\vec{P}$ = the dipole moment per unit volume of dielectric = $n\vec{p}$

where n is number of atoms per unit volume of dielectric and p is dipole moment of an atom or molecule.

$\vec{P}=n\vec{p}=\frac{q_{i}d}{Ad}=(q_{i}A)={\sigma }_i$

Polarization in a Dielectric

The alignment of dipole moments of permanent or induced dipoles in the direction applied electric field is called polarisation.

When a dielectric is placed in an external electric field (E0​), its molecules align with the field. This creates an internal induced electric field (Ei​) that opposes the external field. The net electric field (E) inside the dielectric is therefore reduced:

$E=E_0-E_i=\frac{E_0}{K}$

• K is the dielectric constant (or relative permittivity) of the material. K>1.

11.0Capacitance with a Dielectric

(i) In absence of dielectric, E = σ / ε₀

(ii) When a dielectric fills the space between the plates then molecules having dipole moment align themselves in the direction of electric field.

σᵦ = induced (bound) charge density (called bound charge because it is not due to free electrons).
The induced charge also produce electric field.

Let E₀, V₀, C₀ be electric field, potential difference and capacitance in absence of dielectric and E, V, C are electric field, potential difference and capacitance respectively in presence of dielectric.

Electric field in absence of dielectric, E₀ = V₀ / d = σ / ε₀ = Q / (ε₀A)

Electric field in presence of dielectric, E = E₀ − Eᵢ = (σ − σᵦ) / ε₀ = (Q − Qᵦ) / (ε₀A) = V / d

Capacitance in absence of dielectric, C₀ = Q / V₀

Capacitance in presence of dielectric, $C=\frac{Q-Q_b}{V}$

The dielectric constant or relative permittivity K or εᵣ = $\frac{E_0}{E}=\frac{V_0}{V}=\frac{C}{C_0}=\frac{Q}{Q-Q_b}=\frac{\sigma }{\sigma -{\sigma }_b}=\frac{\epsilon }{{\epsilon }_0}$

From $K=\frac{Q}{Q-Q_b},Q_b=Q\left(1-\frac{1}{K}\right)andK=\frac{\sigma }{\sigma -{\sigma }_b}$$\Rightarrow {\sigma }_b=\sigma \left(1-\frac{1}{K}\right)$

12.0Capacitance in the Presence of Dielectric

$C=\frac{\sigma A}{V}=\frac{\sigma A}{\frac{\sigma }{K{\epsilon }_0}d}=\frac{AK{\epsilon }_0}{d}$

Here capacitance is increased by a factor K.

$C=\frac{AK{\epsilon }_0}{d}$

13.0How to Increase Capacitance

Capacitance can be increased when:

• A capacitor's plates (conductors) are positioned closer together.
• Larger plates offer more surface area.
• The dielectric is the best possible insulator for the application.

In electrical circuits, capacitors are frequently used to block direct current (dc) while permitting alternating current (ac) to flow.

14.0Factors Affecting Capacitance

The capacitance of a capacitor depends on:

1. Area of plates (A): Larger plate area increases capacitance.
2. Separation between plates (d): Smaller distance increases capacitance.
3. Dielectric constant (εrεr​): Inserting a dielectric material increases capacitance by a factor of εrεr​.
4. Geometry of conductors: Shape and arrangement (parallel plates, cylindrical, spherical) affect capacitance.

15.0Sample Questions on Capacitor and Capacitance

Question 1: Find out equivalent capacitance between A and B (take each plate Area = A)

Solution:

$C_{eq}=\frac{Q}{V}=\frac{2xA}{V}$

$V=V_2-V_4=(V_2-V_3)+(V_3-V_4)=\frac{xd}{{\epsilon }_0}+\frac{2xd}{{\epsilon }_0}=\frac{3xd}{{\epsilon }_0}$

$\therefore C_{eq}=\frac{2Ax{\epsilon }_0}{3xd}=\frac{2A{\epsilon }_0}{3d}=\frac{2C}{3}.$

Question 2: In the given circuit find out charge on 6 μF and 1 μF capacitor.

Solution:

Solution:

It can be simplified as $C_{eq}=18/9=2\mu F$

Charge flow through the cell = 30 × 2 μC

Q = 60 μC

Now charge on 3 μF = Charge on 6 μF = 60 μC

Potential difference across 3 μF = 60 / 3 = 20 V

∴ Charge on 1 μF = 20 μC

Question 3: The plates of small size of a parallel plate capacitor are charged as shown. The magnitude of force on the charged particle of 'q' at a distance 'l' from the capacitor is: (Assume that the distance between the plates is d << l)

Solution:

Question 4: Three initially uncharged capacitors are connected in series as shown in circuit with a battery of emf 30 V. Find out following:

(i) charge flow through the battery

(ii) potential energy in 3 μF capacitor

(iii) Uₜₒₜₐₗ in capacitors

(iv) heat produced in the circuit

Solutions:

$\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1$

$C_{eq}=1\mu \text{F}$

(i) $Q=C_{eq}V=30\mu C$

(ii) Charge on 3 μF capacitor = 30 μC

Energy = Q² / 2C = (30 × 30) / (2 × 3) = 150 μJ

(iii) U_total = [(30 × 30) / 2] μJ = 450 μJ

(iv) Heat produced = (30 μC)(30) − 450 μJ = 450 μJ