Capacitors and Capacitance 

1.0What is a Capacitor?

A capacitor is a passive two-terminal electrical component used to store electric charge and electrical energy in an electric field. It typically consists of two conducting plates separated by a non-conducting material called a dielectric. When a capacitor is connected to a voltage source, positive charge builds up on one plate and an equal amount of negative charge builds up on the other.

Principle of Capacitor

When uncharged conductor is placed nearer to the charged conductor and uncharged conductor is connected to earth, then capacitance of charged conductor is increased.

Circuit Symbol of Capacitor

The capacitor is represented as following:

2.0What is Capacitance?

Capacitance (C) is a measure of a capacitor's ability to store charge. It is defined as the ratio of the magnitude of the charge (Q) on either conductor to the potential difference (V) between them.

$C=\frac{Q}{V}$

The SI unit of capacitance is the farad (F), which is equal to one coulomb per volt $(1F=1C/V)$. Since the farad is a very large unit, capacitance is often expressed in microfarads

$(\mu \mathrm{F})$, nanofarads $(nF)$, or picofarads $(pF)$.

How Does a Capacitor Work?

When a capacitor is connected to a battery or power supply, electrons are drawn from one plate and deposited on the other. This creates a potential difference across the plates. The plate that loses electrons becomes positively charged, while the plate that gains electrons becomes negatively charged. The electric field established between the plates stores the electrical energy.

Capacitance of a Parallel Plate Capacitor

The simplest and most common type of capacitor is the parallel plate capacitor. Its capacitance depends on its physical dimensions and the material between the plates.

Parallel Plate Capacitor

It consists of two large plane parallel conducting plates separated by a small distance.

Capacitance of Parallel Plate Capacitor

$\sigma =\frac{Q}{A}$

Surface charge density:

$E_0=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{{\epsilon }_{0}A}$

Electric field intensity between plates:

$V=E_0\times d=\frac{Qd}{{\epsilon }_{0}A}$

Potential difference between the plates:

$C_0=\frac{Q}{V}=\frac{Q}{\frac{Qd}{{\epsilon }_{0}A}}=\frac{{\epsilon }_{0}A}{d}$

Capacitance of parallel plate capacitor:

For a parallel plate capacitor with plate area A and distance d between the plates (filled with vacuum or air), the capacitance is given by:

$C_0=\frac{{ϵ}_{0}A}{d}$

where ${ϵ}_0$ is the permittivity of free space $\left(\approx 8.85\times 10^{-12}\mathrm{F}/\mathrm{m}\right)$

3.0Capacitors in Series and Parallel

Capacitors can be connected in a circuit in two primary ways:

Capacitors in Series

When capacitors are connected in series, they are joined end-to-end. The charge (Q) on each capacitor is the same, but the total voltage is the sum of the individual voltages.

The equivalent capacitance (Ceq​) for series connection is given by:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\dots$

Capacitors in Parallel

When capacitors are connected in parallel, their terminals are connected to the same two points. The voltage (V) across each capacitor is the same, but the total charge is the sum of the charges on individual capacitors.

The equivalent capacitance (Ceq​) for parallel connection is given by:

$C_{eq}=C_1+C_2+C_3+\dots$

4.0Energy Stored in a Capacitor

A charged capacitor stores electrical potential energy. The energy (U) stored in a capacitor can be calculated using one of the following equivalent formulas:

$U=\frac{1}{2}QV=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}$

This energy is stored in the electric field between the plates.

5.0Dielectrics and Capacitance

A dielectric is an insulating material placed between the plates of a capacitor. When a dielectric is inserted, it gets polarized by the electric field, which reduces the effective electric field between the plates. This, in turn, reduces the potential difference (V) and increases the capacitance.

The new capacitance (C) is given by:

$C=K{C}_0$

where K is the dielectric constant of the material (K>1). The dielectric constant is a dimensionless quantity that depends on the material.

How Do You Determine the Value of Capacitance?

The capacitance (C) of a capacitor is a measure of its ability to store charge. It is defined as the ratio of the magnitude of the charge (Q) stored on one plate to the potential difference (V) between the plates.

$C=\frac{Q}{V}$

This formula is fundamental for determining a capacitor's value. While it appears that capacitance depends on charge and voltage, it's actually an intrinsic property of the capacitor's physical characteristics, such as the shape, size, and material of the plates and the dielectric

6.0Solved Problems

Problem 1: A parallel plate capacitor has an area of 50 cm2 and a plate separation of 1 mm. A voltage of 100 V is applied across it. Calculate the capacitance and the charge stored.

Solution:

$\text{ Area, }A=50{\mathrm{cm}}^2=50\times 10^{-4}{\mathrm{m}}^2$

$\text{ Distance, }d=1\mathrm{mm}=1\times 10^{-3}\mathrm{m}$

$\text{ Voltage, }V=100\mathrm{V}$

Capacitance: 

$C=\frac{{ϵ}_{0}A}{d}=\frac{\left(8.85\times 10^{-12}\right)\left(50\times 10^{-4}\right)}{1\times 10^{-3}}=4.425\times 10^{-11}\mathrm{F}$

Charge stored: 

$Q=CV=\left(4.425\times 10^{-11}\right)(100)=4.425\times 10^{-9}\mathrm{C}$

Problem 2: Two capacitors, $C_1=2\mu F\text{ and }{C}_2=4\mu F$, are connected in series to a 12 V battery. Find the equivalent capacitance and the voltage across each capacitor.

Solution: Equivalent capacitance (series):

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

$C_{eq}=\frac{4}{3}\mu F$

Total charge stored: 

$Q=C_{eq}\bar{V}=\left(\frac{4}{3}\times 10^{-6}\right)(12)=16\times 10^{-6}\mathrm{C}=16\mu C$

Voltage across each capacitor: 

$V_1=\frac{Q}{C_1}=\frac{16\mu C}{2\mu F}=8\mathrm{V}$

$V_2=\frac{Q}{C_2}=\frac{16\mu C}{4\mu F}=4\mathrm{V}$

Note that $V_1+V_2=8+4=12\mathrm{V},$ which matches the total voltage.

Problem 3

A capacitor has two circular plates whose radius are 8 cm and distance between them is 1 mm.
When mica (dielectric constant k=6k=6) is placed between the plates, calculate the capacitance of this capacitor and the energy stored when it is given a potential of 150 V.

Solution:
Area of plate $A=\pi r^2=\pi \times {\left(8\times 10^{-2}\right)}^2=0.0201m^{2}andd=1\mathrm{mm}=1\times 10^{-3}\mathrm{m}$ Capacitance of capacitor,

$C=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}=\frac{8.85\times 10^{-12}\times 6\times 0.0201}{1\times 10^{-3}}=1.068\times 10^{-9}F$

Potential difference,

V=150 V

Energy stored,

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\times \left(1.068\times 10^{-9}\right)\times (150{)}^2=1.2\times 10^{-5}J$

Problem 4

An infinite number of capacitors of capacitance C, 4C, 16C, $\dots \dots \infty$

are connected in series.
Find their resultant capacitance.

Solution:
Let the equivalent capacitance of the combination be $C_{eq}$ .

$\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}}+\frac{1}{4\mathrm{C}}+\frac{1}{16\mathrm{C}}+\dots \dots \infty =\left[1+\frac{1}{4}+\frac{1}{16}+\dots \dots \infty \right]\frac{1}{\mathrm{C}}\text{ (this is G.P. series) }$

$\Rightarrow S_{\infty }=\frac{a}{1-r};\text{ first term }a=1\text{, common ratio }r=\frac{1}{4}$

$\Rightarrow \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{1-\frac{1}{4}}\times \frac{1}{\mathrm{C}}\Rightarrow {\mathrm{C}}_{\mathrm{eq}}=\frac{3}{4}\mathrm{Ce}$

Problem 5

The plates of small size of a parallel plate capacitor are charged as shown.
The magnitude of force on the charged particle of charge ‘q’ at a distance $‘l’$ from the capacitor is:
(Assume that the distance between the plates is $d\ll l,$).

Solution:
The two plates act as a dipole.

The magnitude of force on charge q:

$|\vec{F}|=|\vec{E}|q=\left(\frac{2kQd}{l^3}\right)q=\frac{Qqd}{2\pi {\epsilon }_0{l}^3}$

Problem 6

An infinite number of identical capacitors, each of capacitance $1\mu \mathrm{F}1,$ are connected as in adjoining figure. Then the equivalent capacitance between A and B.

Solution:

$\mathrm{C}=1\mu \mathrm{F}$

$C_{\text{eq. }}=C+\frac{C}{2}+\frac{C}{4}+\frac{C}{8}+\frac{C}{16}+\dots \dots \infty \Rightarrow C\left[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots .\infty \right]$

This is a G.P. series with first term a=1, common ratio $r=\frac{1}{2}$.

$S_{\infty }=\frac{a}{1-r}Herea=firstterm=1,r=commonratio=\frac{1}{2}.$

$=\mathrm{C}\left[\frac{1}{\left(1-\frac{1}{2}\right)}\right]=2\mathrm{C}=2\mu \mathrm{F}$