ELECTRIC FIELD

Concept of Electric Field

Capacitors may be defined as a two metal plates separated by an insulator the thin insulated materials are paper, mica, air, glass, oil etc. capacitor stores electric charge.

Capacitance is the property exhibited by a capacitor when two metal plates are connected to a source voltage. The unit of capacitance is the Farad (F) or microfarad (µF) – 10^-6F. Farad is the capacitance of a capacitor which has a pd of 1V when maintain a charge of 1C.

 

Electric Field Strength (E)

Electric field strength is defined as the potential drop per unit length. Electric field strength is also called electric field intensity or potential gradient.

E = V v/m  

       D

Electric Flux Density (D)

Electric flux density is the ratio of charge to the cross – sectional area of the plate. 

 If the dielectric has a relative permittivity, then

Electric Charge (Q)

Electric Charge is the current that flows through the plates of a capacitor.

Q = CV

Coulombs (C)

Dielectric

Dielectric is the medium separating the plates of an insulator.

 

Permittivity of free space (𝜀_𝑜)

(Electric Consonants)

It is the ratio of ratio of the electric flux density in vacuum (air) to the electric field strength.

𝜀_𝑜 = electric flux density      = D

       electric field strength      E

 

D = Q/A = Q x d = Q x d

E     v/d      A    v     v   A

 

But Q = C

         v

 

𝜀_𝑜 = C x 𝑑

              A

𝜀_𝑜 = 8.55x10^{-12 F}/m

𝜀_𝑜 = 8.85p F/m

For a given value of applied voltage a material with a higher value of permittivity produces a greater electric flux than a material with a lower value of permittivity.

 

Relative Permittivity (er )

It is defined as the ratio of absolute permittivity to the permittivity of free space.

𝜀_𝑟 =      𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑒𝑟𝑚𝑖𝑡t𝑖𝑣𝑖𝑡y        =  𝜀

         𝑝𝑒𝑟𝑚𝑖𝑡𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑓𝑟𝑒𝑒 𝑠𝑝𝑎𝑐y      𝜀_𝑜 

 

(Dimensionless)

 

Terminologies

1)      Capacitance is the measure of the ability to store electric charge.

2)      Charging is the process of increasing the charge held in a capacitor

3)      Discharging is the process of reducing the charge held in a capacitor.

Leakage current is the role of movement of charge through a dielectric

TYPES OF CAPACITORS

Capacitors are classified by the type of dielectric material used in their construction.

1.   Paper capacitors: They consist of thin aluminium foils separated by a layer of waxed paper. Paper capacitors are used in household appliances (e.g. fluorescent lighting fittings) and power circuits (e.g. motors).  

2.      Electrolytic capacitors: They consist of two aluminium foils, one with an oxide film/skin and one without. Electrolytic capacitors must be connected to the correct voltage polarity to prevent damage to the oxide film/skin. The main disadvantages of this type of capacitor are (a) the insulation resistance is comparatively low (b) it is only suitable for circuits where the voltage applied to the capacitor never reverses its direction. They are used where a large capacitance is required and where the terminal voltage never reverses polarity.

3.      Mica capacitors: They are constructed of thin aluminium foils separated by a layer of mica. With low dielectric loss. The main disadvantage of this type of capacitor is that it is relatively expensive. They are often used in high-frequency electronic circuits.

4.      Ceramic capacitors: A thin ceramic dielectric is coated on both sides with a metal. The capacitor is made up by making a stack of these ceramic. Ceramic capacitors are used in high-temperature situations.

5.      Air capacitors: They are usually constructed of multiple aluminium vanes of which one section moves to make the capacitance variable. They are often used for radio tuning circuits.

6.   Polyester capacitors: They are manufactured in a very thin film of thickness as little as 2μm and metallized on one side. They can operate at high voltage.

 

Application

•      Power circuit of household appliances

•      Reduce ripple in the voltage wave

•      High frequency circuits

•      High temperature situations

 

Parallel – Plate Capacitors

The capacitance of a parallel – plate capacitor (neglecting the effect of electrostatic fringing) is given by

C =  𝜀_𝑜𝜀_𝑟𝐴  (F)

            𝑑  

In case of an n – plate capacitor (multi-plate) capacitor the capacitor is (n – 1) times greater than between one pair of plates.

 

Energy stored in a capacitor

W = ½ CV² 

Multi-plate – plate capacitor

 C =  𝜀_𝑜𝜀_𝑟𝐴_(n-1)  (F)

            𝑑   

 

Numerical Examples

1) A parallel plate capacitor consists of 5 metal plates each of area 40cm² separated by a dielectric of thickness 0.25mm, whose relative permittivity is 5. Calculate the capacitance of the capacitor [𝜀_𝑜  = 8.85 x 10^-12F/m]

 

Solution

Number of plate (n) = 5

Area  (A) = 40cm² = 40x10^-2

            A = 40 (10^-2)² = 40 x 10^-4

Area (A) = 40 x 10^-4 =     40        = 0.004𝑚²

                                     10000  

Relative Permittivity (e_r) = 5

Dielectric thickness (d) = 0.25mm = 0.25/1000

                                 (d) = 0.00025m

Permittivity of free space = (e_o) = 8.85x10^-12F/m

C =  𝜀_𝑜𝜀_𝑟𝐴_(n-1)  (F)

            𝑑   

 C  =  85 x 10^-12 x 5 x 0.004 x (5- 1) (F)

                      0.00025

C =  85 x 10^-12 x 5 x 0.004 x 4 (F)

                      0.00025

C =   7.08 x 10^-12

         0.00025

   = 28320 x 10^-12 F = 28320pF

   = 28.32 x 10^-9F 

   = 28.32nF 

2) A capacitor is made with 7metal and separated by sheets of mica having a thickness of 0.3mm and a relative permittivity of 6. The area of one side of each plate is 500cm² and a Pd of 400V is maintained across its terminals. Calculate

a.       The capacitance in micro farads

b.      The charge

c.       The potential gradient

d.      The energy stored

[𝜀_𝑜 = 8.85x10^-12F/m]

Solution

Number of plates         (n) = 7

Thickness   (d) 0.3mm = 0.3/1000 = 0.3mm   

Relative permittivity (𝜀_𝑟) = 6  

Area (A) = 500cm² = 500(10^-2)² = 5000cm

Voltage (V) = 400V

(a)   C =  𝜀_𝑜𝜀_𝑟𝐴_(n-1)  (F)

                   𝑑   

        C  =  8.85 x 10^-12 x 6 x 500x10^-4 x (7 - 1) (F)

                             0.3 x 10^-3

        C  =  15.93 x 10^-12   = 53100x10^-12

                       0.003

= 0.0531x10^-6

= 0.0531µF

(b) Q = CV

         = 0.0531x10^-6 X 400  = 21.24µC.

 

(c) E = v =   400

            d     0.0003

E = 1333333.3V/m

E = 1333.3KV/m

 

(d)  W = ½ CV²  

     = ½ x 0.0531x10^-6 x (400)²

     = ½ x 0.0531x10^-6 x160000

     = 0.0531x10^-6 x80000

     = 4248x10^-6J

     = 4248µJ

 

3) A capacitor has seven parallel plates separated by sheets of mica, 0.5mm thick and a relative permittivity of 6. The capacitor whose area of each plate is 500cm² is connected to a 400V supply. Calculate the 

a.       Electric field strength

b.      Capacitance of the capacitor

c.       Electric flux density.

[e_o =8.85x10^-12F/m]

Solution

Number of plates         (n) = 7

Dielectric thickness (d) 0.5mm = 0.5𝑥10²𝑚𝑚

Relative permittivity (e_r) = 6

Area (A) = 500cm² = 500(10^-2)² = 5000cm

Voltage (V) = 400V

a. Electric field strength (E)

E = v =        400v        =   800000v/m

    d       0.5 x 10^-3m

E = 800000v/m 

c. Capacitance of the capacitor

C =  𝜀_𝑜𝜀_𝑟𝐴_(n-1)  (F)

               𝑑   

C  =  8.85 x 10^-12 x 6 x 500 x10^-4 x (7 - 1) (F)

                                 0.5 x 10^-3 

C  =  15.93 x 10^-12   = 31860x10^-12 F 

              0.0005

C = 31860pF

C = 0.03186µF

d.      Electric flux density (D)

D =  Q   =

        𝜀_𝑟𝐴

 

But Q = CV

                 = 31860x10^-12 X 400

       = 12744000 x10^-12 C

                 = 12.744µC

 

D =  Q   =  12.7 x 10^-6   

       𝜀_𝑟𝐴       6 x 500 x10^-6 

    = 42.48 x10^-6 c/m²

 

Type of Capacitor Connections

Series Connection

If V₁ and V₂ are the corresponding p.ds across C₁ and C₂ respectively

Q = C₁V₁                                 Q = C₂V₂

V₁ = Q/C₁                               V2 = Q/C₂

But Q = CV    or  V = Q/C 

V = V₁ + V₂

V₁ + V₂ = 𝑄/𝐶 

𝑄/𝐶 = V₁ + V₂

 𝑄/𝐶 = 𝑄/𝐶₁ + 𝑄/𝐶₂

Q (1/𝐶) = Q(1/𝐶₁ + 1/𝐶₂ )

Dividing through by Q

𝟏/𝑪_𝑻 = 𝟏/𝑪_𝟏 + 𝟏 𝑪_𝟐 + ⋯ + 𝟏/𝑪_𝒏          

 

Numerical Example

1) If two capacitors having capacitance of 6µF and 12µF respectively are connected in series across a 200V supply, find the 

a.       Total capacitance of the capacitor

b.      The pd (voltage) across each capacitor

c.       The charge on each capacitor

Solution

For series connection

a)      Total capacitance

𝟏/𝑪_𝑻 = 𝟏/𝑪_𝟏 + 𝟏 𝑪_𝟐 = 1/6 + 1/10

       𝟏/𝑪_𝑻 = 10 +  6  = 16

           60          60

 

C_T = 60/16 = 3.75µF

C_T = 3.75µF

 

b)     The charge on each capacitor

Q = CV

    = 3.75x10^-6 X 200

   = 750 x10^-6 C

    = 750µC

Charge is common to all series capacitors

c)      Pd across each capacitor

V₁ = 𝑄 = 750𝑥10 ⁻⁶

         𝐶₁     6𝑥10⁻⁶

 

 V₁ = 750 = 125V

            6

 

V₂ = 𝑄 = 750𝑥10 ⁻⁶

         𝐶₂     10𝑥10⁻⁶

 

V₂ = 750 = 75V

            10

Note

In series connection

V₁ = 𝑄 =    V₂ = 𝑄      

        𝐶₁              𝐶₂

Parallel Connection

If Q₁ and Q₂ are the corresponding charges in capacitance C₁ and C₂ respectively

Q₁ = C₁V             Q₂ = C₂V

Q = CV 

But Q = Q₁ + Q₂

Q₁ + Q₂= CV

CV = Q₁ + Q₂

CV= C₁V + C₂V 

CV= C₁V + C₂V

V(C) = V (C₁ + C₂)

Dividing through by V

𝐶𝑉 = 𝑉(𝐶₁+𝐶₂) 

𝑉             𝑉

C = C₁ + C₂

CT =C1 + C2

In parallel connection

𝑉₁ = 𝑉_T x 𝐶₂

       C₁ + C₂

𝑉₂ = 𝑉_T x 𝐶₁

       C₁ + C₂ 

Numerical Example

(1) Three capacitors have capacitances 2µF, 4µF and 8µF respectively are connected in parallel across a 40V supply. Calculate the 

a)      Total capacitance of the capacitor

b)      Charge on each capacitor

c)      Total energy stored

Solution

a)      Total capacitance of the capacitor

For parallel connection

C_T = C₁ + C₂ + C₃ 

     = 2 + 4 + 8

    = 14µF

(b)      Charge on each capacitor 

Q₁ = C₁V 

     = 2x10^-6 X 40

Q₁ = 80x10^-6C Q₁ = 80µC 

Q₂ = C₂V

    = 4x10^-6 X 40 

    = 160µC

Q₃  =  C₃V

    = 8x10^-6 X 40  = 320µC

c)      Total energy stored

W = ½ CV²

     = ½ X 14x10^-6 X (40)²

     = 7x10^-6 X 1600

= 11200 x10^-6 J

     = 11200µJ 

In the diagram above, calculate the 

(i) Voltage V₁ across the 3F capacitor

(ii) Every stored in the 6F capacitor

 

Solution

i) V₁ = 𝑄 = Q   = 1Q volts

     𝐶₁     3      3

W = ½ C₂V₂² = ½ x 6 x V₂² = 3V₂²J

 

Trial Test

a.       The diagram above is an arrangement of capacitors in a circuit. Calculate the 

(a)      Equivalent capacitance of the two capacitors in series 

(b)      Effective capacitance in the 

(c)      Charge on the 40µF

(d)      Capacitor in the 

(e)      Voltage across C₂.

b.      Show that the total capacitance of two capacitors having capacitances C₁ and C₂ connected in series is  C₁C₂/(C₁+C₂)

c.       What factors affect the capacitance that exists between two plates insulated from each other

d.      A capacitor consists of two metal plates, each having an area of 900cm², spaced 3.0mm apart. The whole of the space between the plates is filled with a dielectric having a relative permittivity of 6.A p.d. of 500V is maintained between the two plates. Calculate the (a) capacitance

(b)  charge

(c)   electric field strength

(d)  electric flux density

e.       What are the factors which determine the capacitance of a parallel-plate capacitor?