electrical units - university of plymouth · electrical units f hamer, ... with the use of c for...

Basic Engineering

Electrical Units

F Hamer, A Khvedelidze & M Lavelle

The aim of this package is to provide a short selfassessment programme for students who want tounderstand some of the units used in electricity.

c 2006 chamer, akhvedelidze, [email protected] Revision Date: October 16, 2006 Version 1.0

http://www.plymouth.ac.uk/mathsmailto:protect egingroup catcode ` active def { }catcode `%active let %%let %%catcode `#active def mailto:[email protected]:protect egingroup catcode ` active def { }catcode `%active let %%let %%catcode `#active def

Table of Contents

1. Introduction2. Force between Charges (Electric Field)3. Electric Potential and Capacitance4. Current and Resistance5. Final Quiz

Solutions to ExercisesSolutions to Quizzes

The full range of these packages and some instructions,should they be required, can be obtained from our webpage Mathematics Support Materials.

http://www.plymouth.ac.uk/mathaid

Section 1: Introduction 3

1. IntroductionIn the package on Units some Systeme International (SI) units suchas the metre, kilogram and second were introduced. Derived unitssuch as newtons (for forces) and joules (for energy) which may beexpressed in terms of the basic units were also discussed.

Example 1 Newtons equation expresses force F = ma in terms ofmass (m) and acceleration (a). So the units of force are

units of force = units of mass units of acceleration= kg m s2 .

For clarity 3 kg m s2 of force is called 3 newtons of force.

Exercise 1. (Click on the green letters for solutions.)(a) Gravitational potential energy, Ep is given in terms of mass, m,

acceleration due to gravity g and height h by Ep = mgh.Express, joules, the units of energy, in terms of basic units.

(b) Thus show that 1 N= 1 Jm1 (a newton is a joule per metre).

Section 2: Force between Charges (Electric Field) 4

2. Force between Charges (Electric Field)Electricity is based on the existence of charges and their interactions.The SI unit of charge is the coulomb (C).

1 coulomb = charge of 6.2 1018 electronsQuiz What is the charge of a single electron (in Coulombs)?

(a) 1 C (b) 1.61019 C (c) 1.61018 C (d) 6.21018 C

N.B. the symbol C for the unit of charge should not be confusedwith the use of C for the physical quantity called capacitance. Thesymbols for quantities and their units are normally different.Here are some examples.

Quantity charge, Q current, I capacitance, C resistance, R

Unit coulomb, C ampere, A farad, F ohm,

All of the quantities above, and more, will be discussed in this package.


By convention electrons have negative charge (and protons or atomicnuclei have positive charge). If two static charges are separated by adistance r

rq1 q2

the size of the force between them is experimentally described by

F = kq1q2r2

where k is a constant. This is an inverse square law (the gravitationalforce between two masses also follows an inverse square law). Charges of the same sign repel each other (e.g., two electrons) Opposite charges attract (e.g., electron and atomic nucleus).

Exercise 2.(a) Rewrite the above equation for the force between static charges

in the form k = . . . .

(b) So find the units of k in terms of newtons, metres and coulombs.


In the SI system of units, we represent the constant k using theGreek symbol as follows

F =1

4q1q2r2

where is called the permittivity of the region between them.

Example 2 The permittivity, , is given from rewriting the aboveequation by

=q1q2

4Fr2,

so the units of permittivity are (4 is a number and has no units)

units of permittivity =C2

N m2= C2 N1 m2 .

This is complicated and does not bring out any physical meaning. It isusual to express permittivity in Fm1 (farads per metre), where thefarad, symbol F, is the (derived) unit of capacitance (see Section 4).


The permittivity of free space is written as 0. Permittivitiesin materials are often written as = r0, where r is the relativepermittivity, which is a dimensionless ratio, i.e., a number which isindependent of the units chosen.

Quiz The permittivity of free space is 8.91012 Fm1. If r in glassis 8, what is the permittivity of glass (in Fm1)?

(a) 7 1011 (b) 9 1011 (c) 9 1012 (d) 8 1013

Quiz The permittivity of water is 7 1010 Fm1. What is r inwater (to one significant figure)?

(a) 8 (b) 70 (c) 0.01 (d) 80


A charge q will experience a force from other charges attracting (orrepelling it). The electric field, E, is the force per unit charge

E =F

q.

The units of E are thus N C1 (newtons per coulomb). Charges inlarge electric fields experience stronger forces.

Exercise 3. (Click on the green letters for solutions.)(a) If a charge of 3 C experiences a force of 6 N in an electric field,

what is the size of the electric field E?

(b) The typical electric field produced by rubbing a plastic comb isof the order of 1, 000 NC1. Calculate the force (in newtons) onan electron (charge 1.6 1019 C) in such a field.

(c) The mass of an electron is 9 1031 kg. Calculate the force dueto gravity on the electron. (Use F = mg where g = 10 ms2 isthe acceleration due to gravity.)

Section 3: Electric Potential and Capacitance 9

3. Electric Potential and CapacitanceWhen an object is lifted up/dropped on Earth (moved through agravitational field) it gains/loses potential energy. Similarly a chargemoved through an electric field changes its electric potential energy.The electric potential is defined as electric potential energy per unitof charge (unit volts, symbol V).This means that if a charge Q moves through a potential V it gainsor loses electrical potential energy, Ep = QV .

Quiz From this definition, express volts in terms of joules and coulombs.

(a) 1V = 1J C (b) 1V = 1J2 C2 (c) 1V = 1J C1 (d) 1V = 1J1 C

Quiz The electron volt, eV, is a widely used unit of energy inmodern physics. It is the energy change when an electron (charge1.61019 C) moves through a potential difference of one volt. Whatis one electron volt in joules?

(a) 6.2 1018 J (b) 1.6 1019 J (c) 106 J (d) 9.1 1031 J


In the Introduction we saw that 1 N = 1Jm1 (a newton is a jouleper metre). The units of electric field E, which from the definitionare N C1, can thus be written as:

units of E = NC1 = J m1 C1 = V m1 ,

where we used that 1V = 1J C1.

This result shows that the electric field is a potential gradient (ifthe potential changes rapidly over a short distance then E is large).

Quiz If the electric potential changes at a constant rate by 12 joulesper coulomb over a distance of 2 metres what is the electric fieldstrength in this region?

(a) 14 V m1 (b) 10 V m1 (c) 24 V m1 (d) 6 V m1

Quiz Using the relation between joules and newtons, select the equiv-alent expression for a volt.

(a) N m1 C (b) N m1 C1 (c) N m C (d) N m C1


A capacitor stores electric charge. This means it is a store of electricpotential energy. Capacitance, variable symbol C, is measured infarads (symbol F).

A capacitor of one farad stores one coulomb of charge when one voltis applied. The general relation is charge = capacitancepotential:

Q = CV

Rewriting this as C = Q/V , we see that

units of capacitance =coulombs

volts= C V1

Thus a farad is a coulomb per volt.

Exercise 4. (Click on the green letters for solutions.)(a) If a charge 2 C is produced by applying a voltage 3 V, what is

the capacitance?

(b) Show that, as stated on page 6, C2 N1 m2 = Fm1.

Section 4: Current and Resistance 12

4. Current and ResistanceWhen charges move past a point, we say that a current, I, flows.

One ampere (unit A) of current corresponds to onecoulomb of charge flowing past a point per second.

Quiz Select the correct expression for an ampere (amp) below.

(a) C s (b) C s2 (c) C s1 (d) s C1

Example 3 If one ampere flows through the copper wire in the picture

a

I

then a coulomb of charge, 6.2 1018 electrons, flows through the areaa every second.


For historical reasons, the flow of current is conventionally taken inthe opposite direction to that which electrons are really moving in.

Current depends upon the voltage applied and upon the material(copper is a better conductor than paper) but it also depends uponthe shape of the material (there is a greater current if the area a islarger).

To remove this dependence upon the area a, we define the currentdensity J by

J =I

a, units Am2 read as amps per metre squared .

Quiz Use the expression above to select an alternative expression forthe unit of current density.

(a) C s1 m2 (b) C s m2 (c) C s1 m2 (d) C s m2


The current density J is related to the electric field strength in theconductor by

J = Ewhere is called the conductivity of the wire.

Conductivity is a property of the material. From the last equationif is large then the current density in a material will be higher.The (derived) unit of is Sm1 (read as siemens per metre). Theconductivity of copper is 5.8 107 Sm1

Quiz The units of conductivity can also be expressed in terms of otherunits from rewriting its definition as

=J

E

Select the correct result from the possibilities below.

(a) A V1 m1 (b) A V m1 (c) A V1m3 (d) A V m3


Example 4 Ohms Law In a resistor with a constant electric field Eand length ` (metres) the potential difference is V = E` (volts).

Similarly the current in the conductor is I = Ja (amps). Hence fromOhms law V = IR we have that resistance R is given by

R =V

I R = E`

Ja=

`

a.

To understand this equation note that:

Doubling the length ` doubles the resistance.

Increasing the cross-sectional area a decreases the resistance.

For two identically shaped resistors, the one made from materialwith a higher conductivity will have the lower resistance.

Quiz The units of resistance R are ohms (symbol ). From the aboveequation select the correct expression for an ohm.

(a) Sm (b) m S1 (c) S1 (d) S1 m3


Exercise 5. (Click on the green letters for solutions.)(a) From Ohms law, express the unit of voltage, the volt in terms of

amperes and ohms.

(b) By rewriting Ohms law, express the unit of resistance, the ohmin terms of volts and amperes.

(c) From

R =`

aand the conductivity of copper stated above, calculate theresistance of a copper wire of length 10 m and radius 2 mm.

Section 5: Final Quiz 17

5. Final Quiz

Begin Quiz

1. Coulomb volts, C V, correspond to which physical quantity?(a) Capacitance (b) Resistance (c) Current (d) Energy .

2. Express the siemens, S, in terms of ohms, ?(a) S = m2 (b) S = A (c) S = V 1 (d) S = 1

3. Using Ohms law, select the equivalent expression for the unit ofcurrent, the ampere.(a) V (b) V1 1 (c) V 1 (d) V1

4. Select the electric field value which produces a force on an electronsimilar to its weight on Earth.(a) 1011 N C1 (b) 1013 N C1 (c) 1048 N C1 (d) 1010 N C1

End Quiz

Solutions to Exercises 18

Solutions to ExercisesExercise 1(a) The SI unit of energy, the joule, may be writtenin terms of basic units from the equation Ep = mgh, where m isthe mass, g is the acceleration due to gravity and h is a height. Thisgives

the joule, J = kg m s2 m ,which simplifies to

J = kg m2 s2 .

Click on the green square to return


Exercise 1(b) To show that a newton is a joule per metre, 1 N =1 Jm1, recall from Example 1 that a newton is

1 N = 1 kg m s2 ,

and that a joule is1 J = 1 kg m2 s2 .

HenceJm

=kg m2 s2

m= kg m s2 .

which is indeed a newton.

This means that if an object is accelerated by a force of one newtonover a distance of one metre, it will acquire one joule of energy.



Exercise 2(a) To find an equation for the constant k, multiply bothsides of

F = kq1q2r2

by r2. This givesFr2 = kq1q2

and then divide both sides by q1q2 to get

Fr2

q1q2= k .



Exercise 2(b) To find the units of k, use the equation derived above

k =Fr2

q1q2.

Since F is a force (unit newtons), r is a distance (unit metres), andq1 & q2 are both charges (units coulombs), this means

units of k =N m2

C2= Nm2 C2 .

This is read as newtons, metres squared per square coulomb.

The constant k measures the strength of the force betweencharges. The larger the value of k, the larger the force betweentwo charges at a fixed separation would be.



Exercise 3(a) If 3 C of charge experiences a force of 6 N from anelectric field, the value of the electric field must be

E =F

Q=

63

= 2N C1 ,

which is read as two newtons per coulomb.Click on the green square to return


Exercise 3(b) If an electron is in an electric field of 1, 000 NC1 =103 N C1, then since the charge on the electron is 1.6 1019 C, theelectric force on it can be found by rearranging E = F/Q to read

F = EQ= 103 N C1 1.6 1019 C= 1.6 1019+3 N ,= 1.6 1016 N .

This force acts on each individual electron in such an electric field.Click on the green square to return


Exercise 3(c) If the mass of an electron is 91031 kg, then the forceon it due to gravity (i.e., its weight) on Earth where the accelerationdue to gravity is g = 10 ms2 is given by

F = mg= 9 1031 kg 10 ms2 ,= 9 1030 kg ms2 ,= 9 1030 N

since one newton is one kilogram metre per second squared.

It is clear that the force due to gravity (the weight of the electron)is much smaller than the electric force of part 3(b). This indicateswhy it is possible to lift small pieces of paper by rubbing a comb(although the weight of the paper is much more than just the weightof the electrons).



Exercise 4(a) If a charge of 2 C (two coulombs) is produced on acapacitor by applying a voltage of 3 V, the capacitance C can befound as follows. From

Q = CVwe have

C =Q

Vso substituting the data

C =2 C3 V

= 0.7 F ,

to one significant figure. In the final step the definition of the farad,F = C V1 was used. Note the distinction between C (the symbol forthe physical quantity capacitance) and C (the symbol for the coulomb,the unit of charge).



Exercise 4(b) To prove that

C2 N1 m2 = Fm1

recall the definition of capacitance

C =Q

V

which means that the units are related by

F =CV

.

However, it was shown in the second quiz on page 10 that

1 V = 1 Nm C1

so that

Fm1 =C m1

N m C1

= C1(1) m11N1 = C2 N1 m2



Exercise 5(a) The unit of voltage, the volt, may be rewritten interms of amperes and ohms by using Ohms law, V = IR. Thus

the units of voltage = amperes ohms = A .We read that a volt is a ampere ohm.

Thus a four ohm resistor with five amps flowing through it must havea potential difference of twenty volts across it.



Exercise 5(b) The unit of resistance, the ohm, may be rewritten interms of volts and amperes by using ohms law, V = IR, as

R =V

I.

Thus

the units of resistance =volts

amperes=

VA

= V A1 .

We read that an ohm is a volt per ampere.

Thus a three ohm resistor needs three volts applied to it to produceone amp of current.



Exercise 5(c) To find the resistance, R, of copper wire of length` = 10m and radius 2 mm (= 2 103 m), use

R =`

a

and the value = 5.8 107 Sm1 for copper. The wires cross-sectional area is a = r2 = (2 103)2 m2 = 4 106 m2 . Thus

R =10 m

5.8 107 Sm1 4 106 m2

=10 m

4 5.8 1076 Sm1+2

=10 m

4 5.8 10 Sm=

14 5.8 S

= 0.01 ,

where in the last step we used that S1 = . The tiny result isbecause copper wire is a good conductor.


Solutions to Quizzes 30

Solutions to QuizzesSolution to Quiz: A coulomb is the charge of 6.2 1018 electrons.

1C = 6.2 1018 electron chargesTherefore

16.2 1018

C = 1 electron charge .

So the charge of a single electron is1

6.2 1018C = 1.6 1019 C .

To check this, note that the charge of 6.2 1018 electrons is indeed6.2 1018 1.6 1019 C = 1 C .

End Quiz


Solution to Quiz: To find the permittivity in glass, , given thatthe relative permittivity is r = 8 use

= r0 ,

where 0 = 8.9 1012 Fm1 is the permittivity of free space. Thus = 8 8.9 1012 Fm1 = 7 1011 Fm1 ,

to one significant figure. End Quiz


Solution to Quiz: To find the relative permittivity of water, r,given that its permittivity is = 7 1010 Fm1 rearrange = r0to read

r =

0where 0 = 8.9 1012 Fm1 is the permittivity of free space. Thus

r =7 1010 Fm1

8.9 1012 Fm1,

which shows that r is a dimensionless ratio (a number) and does nothave any units. Its numerical value is

r =7 1010

8.9 1012

=7 1010(12)

8.9=

7 102

8.9= 80 ,

to one significant figure.End Quiz


Solution to Quiz: To express volts in terms of joules andcoulombs, use

Ep = QV ,which implies that

V =EpQ

.

Thus the units of volts are:

units of volts, V =joules

coulombs=

JC

.

In other words V = J C1 (a volt is a joule per coulomb).

Thus if a coulomb of charge passes through a potential difference ofone volt, it will gain or lose one joule of electrical potential energy.

End Quiz


Solution to Quiz: The charge of an electron is 1.6 1019 C. Thusan electron volt (eV), the change in energy of an electron movingthrough a one volt potential difference is

Ep = QV= 1.6 1019 C 1 V= 1.6 1019J .

In the last step the relation between the units V = J C1 was used.End Quiz


Solution to Quiz: The electric field, E, is a potential gradient

E =change in V

distance.

Thus if the change in V is 12 V over a distance of 2 m

E =12 V2 m

= 6 V m1 .

End Quiz


Solution to Quiz: To find an equivalent expression for a volt, wehave:

N = J m1

andV = J C1 .

From the first equationJ = N m

which may be inserted into the second equation to give

V = N m C1 .

End Quiz


Solution to Quiz: An ampere is related to a coulomb by

one amp = one coulomb per second

which means1 A = 1 C s1 .

End Quiz


Solution to Quiz: To find an expression for the units of currentdensity, J , use the definition

J =I

a.

Since the unit of current, the ampere, can be written as C s1 andthe units of area are m2, the units of J are

C s1

m2= Cs1 m2 .

This is read as coulombs per second per square metre.End Quiz


Solution to Quiz: To find an alternative expression for the units ofconductivity, , recall its definition

=J

E.

The units of current density, J , are A m2 and the units of electricfield, E, may be written as either N C1 or, equivalently, V m1.

From the possible answers given, we see we have to use the expressionV m1. Thus

units of =A m2

V m1

= AV1 m2(1)

= AV1 m1 .

End Quiz


Solution to Quiz: The units ohms and siemens are related asfollows. Resistance is related to conductivity by

R =`

a,

and ` is a length (unit metres), a is an area (unit square metres) andthe units of conductivity, are siemens per metre (Sm1). Thus

the units of resistance R =m

Sm1 m2=

mSm

= S1 .

Thus an ohm is an inverse siemen.End Quiz

Table of Contents1 Introduction2 Force between Charges (Electric Field)3 Electric Potential and Capacitance4 Current and Resistance5 Final Quiz Solutions to Exercises Solutions to Quizzes

eqs@qz:elec_units: 1: 1: 2: 3: 4:

2: 1: 2: 3: 4:

3: 1: 2: 3: 4:

4: 1: 2: 3: 4:

qz:elec_units: Score:eq@Mrk:qz:elec_units:

electrical units - university of plymouth · electrical units f hamer, ... with the use of c for...

Documents