#FocusOnExams : Engineering Science GCE Tecnical – 16th May 2020

ENGINEERING SCIENCE

 

Continuation of electrostatic

Exercise

 

 

In the diagram above, A and B form an ironic dipole with charges of +1.6 x 10-19c and -1.6 x10-19c respectively and at a separation of 2 x10-10m. calculate;

 

 

 

 

1. What is the electric field strength due to the Find the electric field of the dipoles at Y
2. Find the ratio of these two field strength given that 1/4= 9×10⁹ Nm² c^-2

 

dipole at X midway between A and B

 

 

 

    B                X             A                  Y

                       .                                   .

 

           

        2 x10-10m              50 x 10-10m

In the diagram above, A and B form an ironic dipole with charges of +1.6 x 10-19c and -1.6 x10-19c respectively and at a separation of 2 x10-10m. calculate;

       What is the electric field strength due to the dipole at X midway between A and B

       Find the electric field of the dipoles at Y

       Find the ratio of these two field strength given that 1/4πεo = 9×109 Nm2 c-2

 

Correction of assignment

Solution

data

 rAx= 1 x 10^-10m, QA = +1.6 x 10^-19c

rBx = 1×10^-10m, QB= -1.6x 10^-19c

1. Calculation of electric field strength (E)

E_Ax=

1. =

   

 E_Ax = +1.44 x 10¹¹vm^-1

 

E_Bx ⁼

1. =

   

E_Bx = -1.44 x 10¹¹vm^-1

 

Therefore;

EX = E_Ax – E_Bx

= +1.44 x 10¹¹vm^-1 – -1.44 x 10¹¹vm^-1

 

    EX= 2.88 x 10¹¹vm^-1

 

1. E_AY =
2. =

              E_AY = 5.76 x 10⁷vm^-1

 

E_{BY =}

1. =

 

E_{BY =}-5.33 x10⁷vm^-1

 

Then Ey= E_AY + E_BY

= 4.3 x 10⁶vm^-1

Since they are in opposite directions

1. Ratio = =

 

= 0.67 x 10⁵

 

 

Chapter 8: Capacitors and Inductors

Objectives    

At the end of this lesson students should be able to:

• Explain how a capacitor stores energy
• Appreciate Q It, E , D , Q  CV and D E  0r and perform calculations using these formulae
• State the unit of capacitance
• Define a dielectric
• state the value of 0 and appreciate typical values of r
• Perform calculations involving the parallel plate capacitor
• Define dielectric strength
• Determine the energy stored in a capacitor and state its unit
• Describes practical types of capacitor
• Define inductance and state its unit
• State the factors which affect the inductance of an inductor
• Describe examples of practical inductors and their circuit diagrams
• Calculate the energy stored in the magnetic field of an inductor and state its units

 

8.1 Capacitors and capacitance

A capacitor is a device capable of storing electrical energy.

Figure 8.1 shows a capacitor consisting of a pair of parallel

 

Metal plates X and Y are separated by an insulator, which could be air. Since the plates are electrical conductors each will contain a large number of mobile electrons. Because the plates are connected to a d.c. supply the electrons on plate X, which have a small negative charge, will be attracted to the positive pole of the supply and will be repelled from the negative pole of the supply on to plate Y. X will become positively charged due to its shortage of electrons whereas Y will have a negative charge due to its surplus of electrons.

The difference in charge between the plates results in a p.d. existing between them, the flow of electrons dying away and ceasing when the p.d. between the plates equals the supply voltage. The plates are then said to be charged and there exists an electric field between them. Figure 8.2 shows a side view of the plates with the field represented by ‘lines of electrical flux’.

 

 

If the plates are disconnected from the supply and connected together through a resistor the surplus of electrons on the negative plate will flow through the resistor to the positive plate. This is called discharging. The current flow decreases to zero as the charges on the plates reduce. The current flowing in the resistor causes it to liberate heat showing that energy is stored in the electric field

 

8.1.1   Summary of important formulae and definitions

From the above lesson charge Q is given by:

Q = I x t coulombs

 

where I is the current in amperes, and t the time in seconds.

A dielectric is an insulating medium separating charged

surfaces.

Electric field strength, electric force, or voltage gradient,

 

E =

E = volts/m

 

 

 

 

Electric flux density,

D =C/m2

 

Charge Q on a capacitor is proportional the applied voltage,

V, i.e. Q  V

Q = CV

Where the constant of proportionality, C, is the capacitance

Capacitance,

 

C =

 

The unit of capacitance is the farad, F (or more usually F =10-⁶F or pF = 10-¹² F), which is defined as the capacitance of a capacitor when a p.d. of one volt appears across the plates when charged with one coulomb. Every system of electrical conductors possesses capacitance. For example, there is capacitance between the conductors of overhead transmission lines and also between the wires of a telephone cable. In these examples the capacitance is undesirable but has to be accepted, minimised or compensated for. There are other situations, such as in capacitors, where capacitance is a desirable property. The ratio of electric flux density, D, to electric field strength, E, is called absolute permittivity, , of a dielectric.

 

Thus            =

Permittivity of free space is a constant, given by

0 = 8.85 x 10^-12 F/m.

Relative permittivity,

 

(r has no units). Examples of the values of r include: air =1, polythene = 2.3, mica =3–7, glass =5–10, ceramics = 6–1000.

Absolute permittivity, =0r, thus  = 0r

8.1.2 The parallel plate capacitor

 

For a parallel-plate capacitor, experiments show that capacitance C is proportional to the area A of a plate, inversely proportional to the plate spacing d (i.e. the dielectric thickness) and depends on the nature of the dielectric:

Capacitance,

C =  farads

 

Where 0 = 8.85 x 10^_12 F/m (constant),

r = relative permittivity,

A = area of one of the plates, in m²,

d = thickness of dielectric in m

and n = number of plates

 

Example 1. (a) A ceramic capacitor has an effective

plate area of 4 cm² separated by 0.1mm of ceramic of relative permittivity 100. Calculate the capacitance of the capacitor in picofarads. (b) If the capacitor in part

(a) is given a charge of 1.2C what will be the p.d. between the plates?

Solution

(a) Area A = 4 cm² =4 x 10^_4 m²,

d = 0.1mm = 0.1 x 10^_3 m, 0 =8.85 x 10^_12 F/m,

r = 100 and n = 2

1. Capacitance, C = farads
2. =  F

C= 3540pF

1. Q = CV thus V =

NA . V= v

 

V = 339v

 

Exercise1. A waxed paper capacitor has two parallel plates, each of effective area 800 cm². If the capacitance of the capacitor is 4425 pF determine the effective thickness of the paper if its relative permittivity is 2.5

Exercise2

1. Define
2. Capacitance
3. Electrical field strength

(2marks)

1. State two practical uses of capacitors (2marks)
2. A 300v battery is connected across capacitors of 3µf and 6 µf
3. In parallel
4. In series

Calculate the charge and energy stored in each capacitor in (i) and (ii)         (4marks)

1. A capacitor of capacitance 4µf is charge to a potential of 100v and another of capacitance 6µf Is charge to a potential of 200v. these capacitors are now join with plates of like charges connected together. Calculate
2. The potential across each after joining (3marks)
3. The total energy stored before joining (3marks)

• The total energy stored after joining (3marks)