Capacitors

A capacitor is the primary electronic component used to store electrical energy. Capacitor having two metallic plates separated by a certain distance and filled with dielectric materials like air, ceramic, electrolytes, etc. Capacitors can store energy when charging and release it when discharging, making them essential components in many electronic devices and systems. Their ability to store and manipulate electrical energy makes them indispensable in modern technology.

1.0Concept of Capacitors

• A capacitor or condenser consists of two conductors separated by an insulator or dielectric. Having equal and 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 Electric field by storing charge. Conductors are used to form capacitors.

2.0Electrical Capacitance

• It demonstrates a conductor's ability to store electrical energy through an electric field. If charge(Q) is given to an isolated conducting body and it's potential increases by V, then $\Rightarrow Q\propto V,Q=CV$      $\Rightarrow \mathrm{C}=\frac{Q}{V}$ (C= Capacitance of the capacitor)
• Electrical capacitance is a Scalar quantity.
• Capacitance of conductor depends upon shape, size, presence of medium and nearness of other conductor.

Graph Between Q and V

• S I Unit- Farad
• CGS Unit-Stat Farad
• Dimensional Formula-[M^-1L^-2T⁴A²]

3.0Types of Capacitors

Based on shape and arrangement of capacitor plates there are various types of capacitors

1.  Parallel plate capacitor 
2.  Spherical capacitor
3.  Cylindrical capacitor

4.0Circuit Symbols of Capacitor

5.0Capacitors Applications

Capacitors find extensive applications across diverse fields because of their efficient capability to store and discharge electrical energy.

1. Filtering and Smoothing
2. Energy Storage
3. Timing Circuits
4. Motor Starters
5. Power Factor Correction
6. Memory Backup

6.0Work done by External agent to charge a conductor

• Work done by external agent to increase the potential of conductor from $V_1$ to $V_2$

$W=\frac{1}{2}C\left(V_2^2-V_1^2\right)$

7.0Energy Stored in a Capacitor

• Potential energy of conductor will be stored in the form of electric field

$\Rightarrow P.E=W_{\text{ext }}$

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

$\Rightarrow$ Potential energy of conducting sphere

$\bigcup =\frac{K{Q}^2}{2R}=\frac{1}{2}\frac{Q^2}{4\pi {ϵ}_{0}R}$

Capacitance of Isolated Spherical Conductor:

• If the medium around the conductor is vacuum or air than capacitance is given by,

$\mathrm{C}=4\pi {ϵ}_{0}R$ (∴R = Radius of Spherical Conductor(Solid or Hollow)

• If the medium around the conductor is a dielectric of constant K from surface of sphere to infinity then $C_{\text{medium }}=4\pi {ϵ}_{0}KR$
• $\frac{C_{\text{medium }}}{C_{\text{air/vocuum }}}=K=\text{ Diekectric Constant }$

Capacitance of Spherical Capacitor:

Case 1: Outer Sphere is Earthed

• When a charge Q is given to the inner sphere it is uniformly distributed on its surface A charge –Q is induced on the inner surface of the 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 and E=0 for r>b ,this arrangement is known as spherical capacitor
• Capacitance of spherical capacitor is given by

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

• If dielectric medium is filled then

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

Case 2: Inner Sphere is earthed 

Capacitance is given by,

$\mathrm{C}=\frac{4\pi {ϵ}_0{b}^2}{b-a}$

Cylindrical Capacitor

• There are two coaxial 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 the inner cylinder it is uniformly distributed on its surface. A charge –Q is induced on the inner surface of the outer cylinder. The charge +Q induced on outer surface of outer cylinder flows to earth as it is grounded. Hence Capacitance is given by $\mathrm{C}=\frac{2\pi {ϵ}_{0}l}{\ln \left(\frac{b}{a}\right)}$

8.0Parallel Plate Capacitor(PPC)

• It consist of two large plane parallel conducting plates separated by small distance
• $\mathrm{C}=\frac{{ϵ}_{0}A}{d}$
• Capacitance of parallel plate capacitor with dielectric

$\mathrm{C}=\frac{{ϵ}_0{ϵ}_{r}A}{d}={ϵ}_r{C}_0\left(\because C_0=\frac{{ϵ}_{0}A}{d}\right)$

Capacitance of Parallel Plate Capacitor Depends on

1. Area $\Rightarrow C\propto A$
2. Distance between the plates $\Rightarrow \mathrm{C}\propto \frac{1}{d}$
3. Medium between the Plates $\Rightarrow C\propto {ϵ}_r$

9.0Fringing of Electric Field

• For the plates of finite area the electric field between the two plates will not be uniform and the field lines bend outward at the edges. This is called "fringing of electric field"

Capacitance of Parallel plate Capacitor when dielectric is partially filled:

• Capacitance is given by,

$\mathbf{C}=\frac{{ϵ}_{0}A}{(d-t)+\frac{t}{{ϵ}_r}}$

Force between Plates of Parallel plate Capacitor

$\Rightarrow \mathrm{F}=\frac{1C{V}^2}{2d}$

• Force between two plates means force on a plate.

Pressure on each plate of a capacitor

• P=$\frac{{\sigma }^2}{2{ϵ}_0}$
• Electrostatic pressure always acts perpendicular to the surface and outwards.

Sharing of Charges

• When two charged conductors are connected by a conducting wire then charge flows from a conductor at higher potential to that at lower potential. This flow of charge stops when the potential of both conductors become equal. This potential is called Common Potential.
• By using conservation of charge ,charge before sharing is equal to charge after sharing.
• Common Potential after connection is given by, $\mathrm{C}=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
• Charges after Connection
• $Q_1^{\prime }=\left(\frac{C_1}{C_1+C_2}\right)Q$ (∴ Q = Total charge on the system)
• $Q_2^{\prime }=\left(\frac{C_2}{C_1+C_2}\right)Q$
• In case of spherical conductors ,ratio of the charges after redistribution

$\frac{Q_1^{\prime }}{Q_2^{\prime }}=\frac{C_1}{C_2}=\frac{R_1}{R_2}$

Energy Density $\left(U_E\right)$

• Potential Energy stored in the form of electric field ,the space between two plates and volume of this space is (A ✖ d)

$U_E=\frac{1}{2}{ϵ}_0{E}^2=\frac{1}{2}\frac{{\sigma }^2}{{ϵ}_0}$

Charging of Capacitor by Battery

• When an uncharged capacitor is connected to a battery, half of the energy supplied by the battery is stored in the capacitor, while the other half is dissipated.
• Energy loss does not depend on the resistance of the circuit.
• By using given concept we calculate the heat loss

$W_{\text{by Battery }}=\Delta U$ + Heat Loss

• Heat Loss =$\frac{1}{2}C{V}^2$
  

For Parallel Plate Capacitor

1. If Like plates Connected

• Common Potential $(\mathrm{V})=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
• Heat $=\frac{1}{2}\frac{C_1{C}_2}{\left(C_1+C_2\right)}{\left(V_1-V_2\right)}^2$

2. If Unlike plates are connected

• Common Potential $(\mathrm{V})=\frac{C_1{V}_1-C_2{V}_2}{C_1+C_2}$
• Heat $=\frac{1}{2}\frac{C_1{C}_2}{\left(C_1+C_2\right)}{\left(V_1+V_2\right)}^2$

10.0Combinations of Capacitor(Series and Parallel Combination)

1. Series Combination                 

• Capacitors are connected end-to-end so that the same current flows through each Capacitor. The total Capacitance in series is less than any individual capacitor's Capacitance. The Charge on each Capacitor connected in series is the same.
•  When two capacitors   are connected in series, than effective capacitance is given by

$\Rightarrow \frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}$

$\Rightarrow \frac{1}{C}=\frac{C_1+C_2}{C_1{C}_2}$

$\Rightarrow \mathrm{C}=\frac{C_1{C}_2}{C_1+C_2}$

• The effective capacitance of capacitors in series connection is lower than the capacitance of each capacitor individually.
• The Charge for each capacitor in the series is the same.

2. Parallel combination of Capacitor

• Capacitors are connected across each other's terminals and share the same voltage. When a potential difference V is applied across the terminals all capacitors have equal potential difference. The equivalent Capacitance of parallel combination is more significant than any of the capacitances in the combination.
• Effective Capacitance is given by, C=C₁+C₂+C₃
• Effective Capacitance of parallel combination is greater than any of the capacitance.
• In Parallel combination, voltage across each Capacitor is the same.

11.0Dielectrics and its Types

• The insulators in which microscopic local displacement of charges takes place in presence of electric fields 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-It 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]

1. Polar Dielectrics

• In absence of an external field the centres of positive and negative charge do not coincide-due to asymmetric shape of molecules. Each molecule has permanent dipole moment. The dipole are randomly oriented so average dipole moment per unit volume of polar dielectric in absence of external field is nearly zero. In presence of external field dipoles tends to align in direction of field.
• Example- Water, Alcohol CO₂, NH₃
•  Dipole moment of polar molecules depends on temperature

2. Non Polar Dielectric

• In the absence of an external field the centre of positive and negative charge coincides in these atoms or molecules because they are symmetric. The dipole moment is zero in normal state. In the presence of an external field they acquire induced dipole moments.
• Example: Nitrogen, Oxygen, Benzene, Methane
• Induced electric dipole moment of non-polar molecules is independent of temperature

Polarisation:

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

Polarisation 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 the direction of the field.
• $\vec{P}={\sigma }_i$(Induced surface charge density) Hence Polarisation is equal to induced surface charge density.

Capacitors with Dielectric

1. Electric Field in the absence of Dielectric, $\mathrm{E}=\frac{\sigma }{{ϵ}_0}$
2. Electric Field in absence of Dielectric, $E_0=\frac{Q}{{ϵ}_{0}A}$
3. Electric Field in absence of Dielectric, $\mathrm{E}=\frac{V}{d}$
4. Capacitance in absence of Dielectric, $C_0=\frac{Q}{V_0}$
5. Capacitance in presence of Dielectric, $\mathrm{C}=\frac{Q-Q_b}{V}$
6. Dielectric Constant or Relative Permittivity $\left(\mathrm{K}\text{ or }{ϵ}_r\right)$

$\mathrm{K}=\frac{ϵ}{{ϵ}_0}=\frac{E_0}{E}=\frac{V_0}{V}=\frac{C}{C_0}=\frac{Q}{Q-Q_b}=\frac{\sigma }{\sigma -{\sigma }_b}$

• In presence of dielectric, capacitance is increased by a factor K

$\mathrm{C}=\frac{{ϵ}_{0}AK}{d}=\mathrm{K}{C}_0$

12.0Effects of Dielectrics in Capacitor

1. Distance Division

• Distance is Divided and area remains same
• Capacitors are in Series
• Individual Capacitance are

$C_1=\frac{{ϵ}_0{ϵ}_{r11}A}{d_1}$

$C_2=\frac{{ϵ}_0{ϵ}_{r2}A}{d_2}$

•  Effective Capacitance is given by,  $\mathbf{C}={ϵ}_{0}A\left[\frac{{ϵ}_{r1}{ϵ}_{r2}}{d_1{ϵ}_{r2}+d_2{ϵ}_{r1}}\right]$
•      Special Case: If $d_1=d_2=\frac{d}{2}\Rightarrow C=\frac{{ϵ}_{0}A}{d}\left[\frac{2{ϵ}_{r1}{ϵ}_{r2}}{{ϵ}_{r1}+{ϵ}_{r2}}\right]$

2. Area Division

• Area is divided and distance remains same
• Capacitors are in Parallel
• Individual capacitors are

$C_1=\frac{{ϵ}_0{ϵ}_{r1}{A}_1}{d}$

$C_2=\frac{{ϵ}_0{ϵ}_{r2}{A}_2}{d}$

• Effective Capacitance is given by $\mathbf{C}=\frac{{ϵ}_0{ϵ}_{r1}{A}_1}{d}+\frac{{ϵ}_0{ϵ}_{r2}{A}_2}{d}$
• Special Case: If $A_1=A_2=\frac{A}{2}\Rightarrow \mathrm{C}=\frac{{ϵ}_{0}A}{d}\left(\frac{{ϵ}_{r1}+{ϵ}_{r2}}{2}\right)$

13.0RC Circuits

• An electric circuit consisting of resistor(s) and capacitor(s) only is known as RC circuit.

Charging of Capacitor:

• Quantity of charge at any instant of time t is given by $\mathrm{q}=q_0\left[1-e^{-\left(\frac{t}{RC}\right)}\right]$
• The quantity of charge increases exponentially with increase in time.
• If t=RC= then

$\mathrm{q}=q_0\left[1-e^{-\left(\frac{t}{RC}\right)}\right]=q_0\left[1-\frac{1}{e}\right]=0.63q_0=63\%\text{ of }{q}_0$

• t=RC is known as time constant, this is defined as the time during which the charge rises on the capacitor plate to 63% of its maximum value.
• Potential difference across the capacitor plate at any instant is given by

$V_c=V_0\left[1-e^{-\left(\frac{t}{Rc}\right)}\right]$

• Current at any time t is given by, $\mathrm{i}=\frac{E}{R}{e}^{-\frac{t}{RC}}$
• t=RC==Time constant

$I=I_0\left[1-e^{-\left(\frac{RC}{RC}\right)}\right]=\frac{I_0}{e}=0.37I_0=37\%Of{I}_0$, time constant is that time during which current in the circuit falls to 37% of its maximum value.

• Potential difference across resistor at any time t is given by,

$V_R=\mathrm{iR}=\mathrm{E}{e}^{-\frac{t}{RC}}$

•  Heat dissipated by using energy conservation,

Heat dissipated=Work done by battery- $\Delta {\cup }_{\text{capacitor }}$

$\text{ Heat dissipated }=\mathrm{CE}(\mathrm{E})-\left(\frac{1}{2}C{E}^2-0\right)=\frac{1}{2}\mathrm{C}{E}^2$

14.0Discharging of a Capacitor

• Quantity of charge at any time t is given by, $\mathrm{q}=q_0{e}^{-\frac{t}{RC}}$
• Current at any time, $\mathrm{i}=-\frac{dq}{dt}=\frac{q_0}{RC}{e}^{-\frac{t}{RC}}$
• Potential difference across the capacitor at any time, $V_C=\frac{q}{C}=\frac{q_0}{C}{e}^{-\frac{t}{RC}}$
• Potential difference across the resistor at any time, $V_R=\mathrm{iR}=\frac{q_0}{C}{e}^{-\frac{t}{RC}}$
• Dimension of RC are those of time [M⁰L⁰T¹]
• Dimension of $\frac{1}{RC}$ are those of frequency [M⁰L⁰T^-1] 

15.0Sample Questions On Capacitors

Q1. The stratosphere acts as a conducting layer for the earth. If the stratosphere extends beyond 50 km from the surface of earth, then calculate the capacitance of the spherical capacitor formed between stratosphere and earth's surface. Take the radius of earth as 6400 km.

Solution:

Capacitance of spherical Capacitor $\mathrm{C}=4\pi {ϵ}_0\left(\frac{ab}{b-a}\right)$

 b = radius of the top of stratosphere layer =6400 + 50 =6450 km =6.45 ✕ 106 m

a = radius of Earth =6400 km = $6.4\times 10^6\mathrm{m}$

$\mathrm{C}=\frac{1}{9\times 10^9}\times \frac{6.45\times 10^6\times 6.4\times 10^6}{6.45\times 10^6-6.4\times 10^6}=0.092\mathrm{F}$

Q2. If the distance between the plates of a capacitor of capacitance C₁ is halved and the area of plates is doubled then what will be the capacitance?

Solution:

$\mathbf{C}=\frac{{ϵ}_{0}A}{d}\Rightarrow \frac{C_1}{C_2}=\frac{A_1}{A_2}\times \frac{d_2}{d_1}=\frac{A_1}{2A_1}\times \frac{1}{d_1}\times \frac{d_1}{2}=\frac{1}{4}$        

$C_2=4C_1$

Q3. A slab of material dielectric constant r has the same plate area as that of a   parallel plate capacitor but a different  thickness (2d/3), where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

Solution:

$C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{{ϵ}_r}}=\frac{{ϵ}_{0}A}{d-\frac{2d}{3}+\frac{2d}{{ϵ}_r}}=\frac{{ϵ}_{0}A}{\frac{d}{3}\left(1+\frac{2}{{ϵ}_r}\right)}=\frac{3{ϵ}_{0}A{ϵ}_r}{d\left({ϵ}_r+2\right)}$

Q4. A capacitor of capacitance 1F is connected in a closed series circuit with a resistance of 10⁷ ohms, an open key and a cell of 2 V with negligible internal resistance:

(i) When the key is switched on at time t = 0, find;

       (a) The time constant for the circuit.

       (b) The charge on the capacitor at steady state.

(ii) If after completely charging the capacitor, the cell is shorted by zero resistance at time t = 0,Determine the charge stored in  the capacitor at t = 50 s.    (Given: e^–5 = 6.73 × 10^–3, ln2 = 0.693)  

Solution:

(i) a. Time constant $\text{ ( }\tau \text{ ) }=RC=10^7\times 1\times 10^{-6}=10S$

b. $q_0=\mathrm{CV}=1\times 10^{-6}\times 2=2\mu \mathrm{C}$

(ii) $q=q_0{e}^{-\frac{t}{RC}}=2\times 10^{-6}{e}^{\frac{-50}{10}}=2\times 10^{-6}\times e^{-5}=1.348\times 10^{-8}\mathrm{C}$