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2. Electric Potential Energy, Potential and Capacitance
EPF 0024 PHY II 1
OutlineOutline
2.1 Electric Potential Energy and the Electric Potential2.2 Energy Conservation2.3 The Electron Volt (eV)2.4 Electric Potential of a Point Charge2 5 E i t ti l Li d S f2.5 Equipotential Lines and Surfaces 2.6 Capacitor2 7 Dielectric2.7 Dielectric2.8 Storage of Electrical Energy2.9 Electric Energy and Power in Current Electricity
EPF 0024 PHY II By Dr. John Ojur Dennis 2
gy y
Topics for today's lecture:Topics for today s lecture:
Electric Potential Energy and the Electric Potential.
Energy Conservation.
The Electron Volt (eV).
EPF 0024 PHY II By Dr. John Ojur Dennis 3
Objectives of this lecture:Objectives of this lecture:
Explain the concepts of Electric Potential Energy,Electric Potential and Electric Potential Difference.
Discuss energy conservation in electric fields
Introduce a new unit of energy called the Electron Volt(eV) and its relation to the joule (J).
Solve Problems related to these concepts.
EPF 0024 PHY II By Dr. John Ojur Dennis 4
2.1 Electric Potential Energy and the Electric Potential
In Physics I the concepts of work ,W, & energy,U, have been introduced in a mechanical context.In this section these concepts are used in theIn this section these concepts are used in thecontext of electric fields.
A charged particle moves in an electric field asa result of the field doing work on it. A chargedparticle in an electric field therefore has electricparticle in an electric field therefore has electricpotential energy , U.
EPF 0024 PHY II By Dr. John Ojur Dennis 5
Fig. 2.1: Electric & Gravitational Potential Energy
Fi 2 1 h if A t t h i l dFig. 2.1 shows a uniform E. A +ve test charge q0 is placedat A and experiences a downward electric force ofmagnitude F = q0E which displaces it to point B.
Electric field
+ ++ +Gravitational field
mA
+ q0 A
yA
AA mgyU =AA EyqU 0=
y
x
BE B
yB
yA
BB mgyU =BB EyqU 0=g
F = mgF = q0E
yB
EPF 0024 PHY II By Dr. John Ojur Dennis 6
Ground level = zero potential
2.1.1 Electric Potential Energy
The work done WAB by the electric force F:
2.1.1 Electric Potential Energy
The work done WAB by the electric force F:
While the change in potential energy U is -ve:( )( ) ( )( ) yEqyFyyFW ABAB === 0 (2.1)
While the change in potential energy U is ve:
yEqEyqEyqUUU ABAB 000 === (2.2) Therefore
UWAB = (2.3)Where U is set to zero is arbitrary. The absolute
U of a test charge q0 is set to zero with respect toh h th i fi it l f t
EPF 0024 PHY II By Dr. John Ojur Dennis 7
a charge q when they are infinitely far apart.
2.1.2 Electric Potential2.1.2 Electric Potential
Electric potential energy U is associated withelectric potential that is known as voltage V inelectric circuits. The electric potential V isdefined asdefined as
UV =or
0q (2.4)VqU 0=
qVU =
V is a scalar quantity and has the SI unit of thej l / l b lt b l V
q0
EPF 0024 PHY II By Dr. John Ojur Dennis 8
joule/coulomb = volt, symbol V.
2.1.3 The Electron Volt (eV)2.1.3 The Electron Volt (eV)
W i ti (2 4) th t th l t iWe see in equation (2.4) that the electricpotential energy is given by .qVU =
This equation suggests, at the atomic level, aconvenient and commonly used unit of energyy gycalled the Electron Volt (eV) defined as
( )( ) ( )( ) J10601V1C10601V1e1eV1 1919 === (2 5)( )( ) ( )( ) J10601V1C10601V1e1eV 1 === .. (2.5)
EPF 0024 PHY II By Dr. John Ojur Dennis 9
2.2 Electric potential difference (p.d.)
The electric p.d. V between the
2.2 Electric potential difference (p.d.)
ptwo points yA and yB in Fig. 2.1 is:
EyEqWUV AB 0 (2 5)
+ ++ ++ q0 A AA EyqU 0=
From which we obtain the
yEq
yqqq
V AB 0
0
00
==== (2.5)E B
yA
BB EyqU 0=From which we obtain theimportant relation between E andV, that is
VF = q0E
yB
yVE = (2.6)
(The SI Unit for E can therefore
EPF 0024 PHY II By Dr. John Ojur Dennis 10
(The SI Unit for E can therefore also be volts/meter, V/m).
2.3 Energy Conservationgy
Gravitational potential energy U of a mass mdropped in gravitational field decreases whiledropped in gravitational field decreases whileits kinetic energy K increases. The total energy,however, remains the same (i.e. energy ishowever, remains the same (i.e. energy isconserved).
The same applies to electric fields, i.e.:BBAA UKUK +=+
( )BBAA
BBAA
VVqUUvvm
UmvUmvUKUK
+=+++
)( 221
2212
21 (2.7)
EPF 0024 PHY II By Dr. John Ojur Dennis 11
( )BABAAB VVqUUvvm == )(21
Example 1Example 1
The work done by the electric force as a testcharge (q0 = + 2.0 106 C) moves from A to B isW + 5 0 10 5 J ( ) Fi d U U UWAB = + 5.0 105 J. (a) Find U = UB UAbetween these points. (b) Determine V = VB VA.
EPF 0024 PHY II By Dr. John Ojur Dennis 12
SolutionSolution
(a) WU(a)
J100.5
5=
= ABWU
(b) = UV
V25J100.5
50
==
qV
V25C102.0
6 ==
EPF 0024 PHY II By Dr. John Ojur Dennis 13
Example 2p
Find U as a charge of (a) +2.20 106 C or(b) 1.10 106 C moves from a point A to B,(b) 1.10 10 C moves from a point A to B,given that V = VB VA = +24.0 V (positiveimplies B at higher potential compared to A).
EPF 0024 PHY II By Dr. John Ojur Dennis 14
SolutionSolution
(a) ( )( )V 0.24102.20
6==
CVqU
J105.28 5=
(b)( )( )V02410101
6==
CVqU
( )( )J1064.2
V 0.241010.1 5=
= C
EPF 0024 PHY II By Dr. John Ojur Dennis 15
Example 3 yExample 3x
The work done by the electricforce as a test charge(q =+2 0 106 C) moves from A
++++
A +q0(q0=+2.0 10 C) moves from Ato B is WAB = +5.0 105 J. (a)Determine U = UBUA. (b) Find
( )
F = q0E
V = VB VA. (c) What ismagnitude of E if distancebetween A and B is 0 25 m?
EB
between A and B is 0.25 m? _ ___
EPF 0024 PHY II By Dr. John Ojur Dennis 16
Solution
(a) J100.5 5=== ABAB WUUU
Solution
(a) J100.5 ABAB WUUU
(b)
== UVVV AB( )
V 25C102 0J100.5 6
50
== qAB
C102.0 6
(c) ( )( ) V/m100V 25 ==== AB VVVE(c) ( ) V/m100m 25.0 AB yyyE
Magnitude of E is 100 V/m and the -ve
EPF 0024 PHY II By Dr. John Ojur Dennis 17
Magnitude of E is 100 V/m and the ve indicates that E points in ve y direction
Example 4Example 4
A uniform electric field with a magnitude of6250 N/C points in the positive x direction.6250 N/C points in the positive x direction.Find the change in electric potential energywhen a + 12.5 C charge is moved 5.50 cmin (a) the positive x direction, (b) the negativex direction, and (c) in the positive y direction.
EPF 0024 PHY II By Dr. John Ojur Dennis 18
SolutionSolution
xqEVqUVE
= xqEU(a)
xqEVqUx
E === ,
( )( )( ) mJ 30.4m 0550.0N/C 6250C105.12 6 ==
xqEU(a)
( )( )( ) mJ30.4m0550.0N/C6250C105.12 6 ===
xqEU(c)
( )( )( )(b) ( ) 00 === qExqEUEPF 0024 PHY II By Dr. John Ojur Dennis 19
(b) ( ) 00 === qExqEU
Example 5Example 5
When an ion accelerates through a potentialdifference of + 2850 V the change in itsdifference of + 2850 V the change in itselectric potential energy is 1.37 1015 J.What is the sign and magnitude of the chargeon the ion?
EPF 0024 PHY II By Dr. John Ojur Dennis 20
SolutionSolution
15
= VqU
C10814V 2850
J10371
19
15
== .
VUq
C10814 19= .
It is a negative charge with magnitude of 4.81 1019 C
EPF 0024 PHY II By Dr. John Ojur Dennis 21
Todays topics include:Today s topics include:
2.4 Electric Potential of a Point Charge
2.5 Equipotential Lines and Surfaces
EPF 0024 PHY II By Dr. John Ojur Dennis 22
Objectives of this lecture:Objectives of this lecture:
To determine an expression for the potential dueto a point charge.p g
To explain the meaning of equipotential lines.p g q p
To solve problems related to these concepts.To solve problems related to these concepts.
EPF 0024 PHY II By Dr. John Ojur Dennis 23
2 4 Electric Potential of a Point Charge2.4 Electric Potential of a Point Charge
Fig 2.2 shows a point charge +q fixed at origin.g p g q gA +ve test charge +q0 is held at rest at A adistance rA from the origin.
The test charge experiencesl i f fa repulsive force of
magnitude given byrA
20
ArqqkF =
EPF 0024 PHY II By Dr. John Ojur Dennis 24
The work done by the electric field in moving theThe work done by the electric field in moving thetest charge +q0 from position rA to rB using integralcalculus is
1( )1
1. 20r
r
r
r
r
r
rAB
qqqqqqqq
drr
qkqdrrFdW
B
B
A
B
A
B
A
=== rF
(2.9)
F th b i th h i l t i
1 00000BAABr rqqk
rqqk
rqqk
rqqk
rqkq
A
=
+=
=
From the above expression the change in electricpotential energy of test charge is:
ABABBA
AB UUrqqk
rqqk
rqqk
rqqkWU ==
== 0000 (2.10)
EPF 0024 PHY II By Dr. John Ojur Dennis 25
Therefore U at any point for a point charge isgiven by (dropping the subscripts):
A d V i i b(2.11)
rkqqU 0=
And V is given by:AB kqkqUUUV === (2 12)
Finally, V at any point is given by (dropping theAB rrqq
V ===00
(2.12)
y y p g y ( pp gsubscripts:
kqV = (2.13)
EPF 0024 PHY II By Dr. John Ojur Dennis 26
r
The net potential at a point due to a group ofThe net potential at a point due to a group ofpoint charges can be calculated by summingpotentials. For n point charges, net potential isp p g , p
==++++= n in qkVVVVVV (2 14) == ==++++= i ii in rkVVVVVV 11321 (2.14)
EPF 0024 PHY II By Dr. John Ojur Dennis 27
Example 1Example 1
What minimum work is required by an externalforce to bring a charge q1 = +3.0 106 C from aforce to bring a charge q1 3.0 10 C from agreat distance away (take r = infinity) to a point0.5 m from a charge q2 = +20 106 C?
EPF 0024 PHY II By Dr. John Ojur Dennis 28
SolutionSolution
The work W required isThe work, W, required is
( )== AA UUUW ( ) 212121 =
=
AA
AA
rqqk
rqq
rqqk
( )( )( )m0 5
C1020C100.3/CNm1099.8 66
229 =
AA rrr
J 08.1 m0.5
=
EPF 0024 PHY II By Dr. John Ojur Dennis 29
Example 2Example 2
Calculate the electric potential at points A andp pB due to charges q1 and q2 as shown Fig.
yy
60 cmA B
30 cm40 cm
40 cm
q = +50 C q = 50 Cx26 cm 26 cm
EPF 0024 PHY II By Dr. John Ojur Dennis 30
q2 = +50 C q1 = 50 C
SolutionSolution
The potential at point A isThe potential at point A is
1212 qqkqkq
( ) C1005C1005 55A1
1
A2
2
A1
1
A2
2A1A2
+=+=+=
rq
rqk
rkq
rkqVVVA
( )V1057V10750V1051
m 6.0C100.5
m 3.0C100.5/CN.m1099.8
566
229
==
+=
V105.7V1075.0V105.1 ==
EPF 0024 PHY II By Dr. John Ojur Dennis 31
Solution (continued)Solution (continued)
At point B
B1
1
B2
2
B1
1
B2
2B1B2B
+=+=+=
rq
rqk
rkq
rkqVVV
( )m40
C100.5m40
C100.5/CN.m1099.8 55
229
B1B2B1B2
+=
rrrr
V 0 m4.0m 4.0
=
the potential is zero everywhere on the planeequidistant between the two charges. This planeis called an equipotential surface with V = 0
EPF 0024 PHY II By Dr. John Ojur Dennis 32
is called an equipotential surface with V 0.
Example 3
A charge q = 4.11 109 C is placed at the origin,d d h 2 i l d th iand a second charge 2q is placed on the x-axis
at x = 1.00 m. (a) Find V midway between the twocharges (b) V vanishes at some point P betweencharges. (b) V vanishes at some point P betweenthe charges; that is, for a value of x between 0and 1.00 m. Find this value of x.
+q 2q
+y
P
0
1 00 m
q
x
q+ x
EPF 0024 PHY II By Dr. John Ojur Dennis 33
1.00 m
SolutionSolution
(a) 21
= kqV(a)
( )( ) 21C10114/CN m10998 9229 =
=
BA rrkqV
( )( )V 9.73
m 50.0m 50.0C1011.4/CN.m1099.8
==
(b) Setting V = 0 we obtain( ) 03m 00.12 + xqkkqV ( )( ) ( )
103001
0m 00.1m 00.1
==+= xxxxV
EPF 0024 PHY II By Dr. John Ojur Dennis 34
m3
,03m 00.1 == xx
2.5 Equipotential Lines and surfaces2.5 Equipotential Lines and surfaces
The electric potential can be represented byequipotential lines or surfaces as shown by thegreen lines in Fig. 2.3.
EPF 0024 PHY II By Dr. John Ojur Dennis 35Fig. 2.3: Equipotential lines (green lines)
By definition, an equipotential line/surface isone in which all points on it are at the samepotential. That is to say, the potential difference(V) between any two points on it is zero(V) between any two points on it is zero.
E i t ti l li / f lEquipotential lines/surfaces are alwaysperpendicular to the electric field. Note that aconductor must be entirely at the sameconductor must be entirely at the samepotential in static case. The surface of aconductor is therefore an equipotential surface.
EPF 0024 PHY II By Dr. John Ojur Dennis 36
q p
Todays lecture include:Today s lecture include:
Capacitor
Dielectric
Storage of Electrical Energy
Electric Energy and Power in Current Electricity
EPF 0024 PHY II By Dr. John Ojur Dennis 37
Objectives of this lecture:Objectives of this lecture:
To provide expressions for capacitance of ap p pcapacitor.
To explain dielectric.
To analyze energy stored in a capacitor andelectric energy in current electricity.
To solve problems related to these concepts.EPF 0024 PHY II By Dr. John Ojur Dennis 38
p p
2 6 Capacitor2.6 Capacitor
A capacitor consists of a pair of metal plates of area At d b ll di t h i Fi 2 5 fseparated by a small distance as shown in Fig. 2.5 for a
parallel- plate capacitor.
Circuit symbol
EPF 0024 PHY II By Dr. John Ojur Dennis 39
Fig. 2.5: A Parallel-Plate Capacitor
2 6 1 Charging a Parallel-Plate Capacitor2.6.1 Charging a Parallel-Plate Capacitor
When p d (voltage) V isWhen p.d. (voltage) V isapplied to a capacitor, it ischarged (Fig 2.6). Charge Q
i d b h l t iacquired by each plate isdirectly proportional to V
Q
where C is proportionalityVQ CCVQ == or (2.15)
where C is proportionalityconstant called capacitanceof the capacitor. SI unit C/V
f d (F)
Fig. 2.6: Charging a Parallel-Plate Capacitor
EPF 0024 PHY II By Dr. John Ojur Dennis 40
or farad (F).
2.6.2 Calculating the Capacitance2.6.2 Calculating the Capacitance
For a parallel-plate capacitorFor a parallel plate capacitorwith area A and separation d(Fig. 2.7), the capacitance is
l ti l t A dalso proportional to A andinversely proportional to d:
A
22120
0
/N.mC1085.8 ==
dAC (2.15)
The proportionality constantis found to have the value 0,th iti it f f
0
Fig. 2.7: A parallel-
EPF 0024 PHY II By Dr. John Ojur Dennis 41
the permitivity of free space. plate capacitor
2.6.3 Calculating the E in a capacitor
Using equ. 2.6 and taking thelt t l t t b Vvoltage at +ve plate to be V
and at ve plate to equalzero (Fig. 2.8), we obtain anzero (Fig. 2.8), we obtain anexpression for E inside acapacitor thus:
xVE
= (2.16)
dV
dV
xxVV =
==
00
12
12
Fig 2 8: A parallel-
EPF 0024 PHY II By Dr. John Ojur Dennis 42
Fig. 2.8: A parallel-plate capacitor
Example 1Example 1
(a) Calculate the capacitance of a parallel-platecapacitor whose plates are 20 cm by 3 cm andare separated by a 1.0 mm air gap. (b) What isthe charge on each plate if the capacitor isconnected to a 12 V battery? (c) What is theconnected to a 12 V battery? (c) What is theelectric field between the plates?
EPF 0024 PHY II By Dr. John Ojur Dennis 43
SolutionSolution
AC(a)
m1006 232212
0
=
dAC (a)
pF53m100.1m100.6)/N.mC1085.8( 3
2212 =
=
( )( ) 1012( )( ) C104.6V12F1053 1012 === CVQ(b)V/m102.1
m100.1V12 4
3 === dVE(c)
EPF 0024 PHY II By Dr. John Ojur Dennis 44
2 7 Dielectric2.7 Dielectric
In most capacitors, an insulating sheet called aIn most capacitors, an insulating sheet called adielectric is placed between the plates. Thisincreases the capacitance of the capacitor.
Consider a parallel-plate capacitor whose platest d b i If th it iare separated by an air gap. If the capacitor is
isolated (i.e. not connected to a battery) andcarries a charge Q and has voltage V across itscarries a charge Q and has voltage V0 across itsterminals, the capacitance will be given byC0= Q/V0. Let the electric field be E0.
EPF 0024 PHY II By Dr. John Ojur Dennis 45
0 Q 0 0
A dielectric is placed in between the platesbecomes polarized as shown in Fig. 2.8.
EPF 0024 PHY II By Dr. John Ojur Dennis 46
Fig. 2.9: Dielectric effect on E inside a capacitor.
There will be net -ve charge on outer edge of theg gdielectric facing +ve plate, and net +ve chargefacing the -ve plate. An electric field is set upinside the dielectric opposite to original field Einside the dielectric opposite to original field E0and reduces it by a factor say. Net E inside thedielectric is now E = E0/ .0
Hence the voltage must decreased by a factorHence the voltage must decreased by a factor also, i.e. V = V0/ , since V = Ed.
EPF 0024 PHY II By Dr. John Ojur Dennis 47
The charge Q on plates remain constantThe charge Q on plates remain constant(remember capacitor is isolated). Therefore from
VVQC 0d VVQC 0
find we
and ==(2.17)
dA
CVQC 00
0
===Implies decrease in E by a factor will increase Cby the same factor .
EPF 0024 PHY II By Dr. John Ojur Dennis 48
Capacitance for parallel plate capacitor withdielectric between plates is therefore given by
AAC (2 18)
where = is called the permittivity of thedd
C == 0 (2.18)where = is called the permittivity of the
material.0
is known as dielectric constant, a factorindicating by how many fold capacitance isincreased Table 2 1 shows the dielectric
EPF 0024 PHY II By Dr. John Ojur Dennis 49
increased. Table 2.1 shows the dielectricconstants of some substances.
Table 2 1: at 20C for different substancesTable 2.1: at 20 C for different substances
Material Material Vacuum 1.0000Air (1 atm) 1.0006P ffi 2 2Paraffin 2.2Rubber, hard 2.8Vinyl (plastic) 2.8 4.5Paper 3 7Quartz 4.3Glass 4 7Porcelain 6 8Mica 7Ethyl alcohol 24
EPF 0024 PHY II By Dr. John Ojur Dennis 50
Ethyl alcohol 24Water 80
ExampleExample
A parallel-plate capacitor has plates of area0.0280 m2 and separation 0.550 mm. The spaceb t th l t i fill d ith di l t i fbetween the plates is filled with a dielectric ofdielectric constant . When the capacitor isconnected to a 12 0-V battery each of the platesconnected to a 12.0-V battery, each of the plateshas a charge of magnitude 3.62 108C. What isthe value of the dielectric constant ?
EPF 0024 PHY II By Dr. John Ojur Dennis 51
SolutionSolution
First we determine C thus:First we determine C thus:
( ) F10023C1062.3 98 === QCN t fi d
F1002.3V 0.12
===V
C
Next we find : 0= d
AC
( )( )( )( ) 70.602800/NC10858 m10550.0F1002.3 22212319
0
===
ACd
d
EPF 0024 PHY II By Dr. John Ojur Dennis 52
( )( )m0280.0/N.mC1085.8 222120 A
2 8 Storage of Electric Energy2.8 Storage of Electric Energy
A Charged capacitor stores electric energy equal to workd t h it Th lt V th it idone to charge it. The voltage V across the capacitor isdirectly proportional to the charge q already accumulatedand increases from 0 to V0 linearly .(Fig. 2.10).
V (V)V0
021 V
021 Q Q (C)0 Q0
EPF 0024 PHY II By Dr. John Ojur Dennis 53
Fig. 2.10: Voltage across a capacitor vs. charge accumulated
V (V)V0
The average voltage02
1 V
The average voltage(from Fig. 2.10)
0V +02
1 Q Q (C)0 Q0
0210
20 VVVav =+= (2.19)
The average energy stored in a capacitor
002
1000 2
VQVQVQU avav =
== (2.20)
EPF 0024 PHY II By Dr. John Ojur Dennis 54
Dropping the subscript 0 in eq.2.20 and usingpp g p q g
f
and 21 CQVQVU ==
we obtain alternative expressions for energystored in a capacitor as
2
or(2.21)C
QQVU2
21
21 ==
or
221
21 CVQVU ==
EPF 0024 PHY II By Dr. John Ojur Dennis 55
Energy stored in a capacitor is stored in theelectric field between the plates. Using V=Edand C=o (A/d)
A
Ad is volume Therefore we may define the( ) AdEEd
dACVU 202
1202
1221 =
== (2.21)
Ad is volume. Therefore we may define theenergy density u as energy per unit volume:
21 EUu (2 22)and for a Capacitor with dielectric
021 E
Adu == (2.22)
EPF 0024 PHY II By Dr. John Ojur Dennis 56
202
1221 EEu == (2.23)
Example 1Example 1
What is the potential difference between theplates of a 3.3-F capacitor that stores
ffi i t t t 75 W li ht b lbsufficient energy to operate a 75-W light bulbfor one minute.
EPF 0024 PHY II By Dr. John Ojur Dennis 57
SolutionSolution
21 2= CVU
22 ===
CPt
CUV
PtU
( )( )( ) V 52F3 3
s 60 W752 ==CC
F3.3
EPF 0024 PHY II By Dr. John Ojur Dennis 58
Example 2Stopped here
Example 2
A parallel-plate capacitor has plates with area ofA parallel plate capacitor has plates with area of1.2 cm2 and separation 0.88 mm. The spacebetween the plates is filled with dielectric ofdielectric constant 2.0. (a) What is the potentialdifference between the plates when the charge
l t i 4 7 C (b) Will t ton plates is 4.7C (b) Will your answer to part(a) increase, decrease, or stay the same if thedielectric constant is increased? Explain (c)dielectric constant is increased? Explain. (c)Calculate the potential difference for the casewhere the dielectric constant is 4.0.
EPF 0024 PHY II By Dr. John Ojur Dennis 59
SolutionSolution
QdQ(a)
( )( ) V1091m1088.0C107.4 6360
==
==
AQd
CQV
(a)
( )( )( )( ) V109.1m 10.21N.mC108.852.0 242212 == (b) ill d b i i i l(b) V will decrease because it is inversely
proportional to .
(c)( )( )( )( ) V107.9m1021/N mC108 854 0 m1088.0C107.4 4242212
36
==
V
EPF 0024 PHY II By Dr. John Ojur Dennis 60
( )( )m10.21/N.mC108.854.0
Example 3Example 3
Find the electric energy density between theplates of a 225-F parallel-plate capacitor.plates of a 225 F parallel plate capacitor.The potential difference between the plates is315 V, and the plate separation is 0.200 mm.
EPF 0024 PHY II By Dr. John Ojur Dennis 61
SolutionSolution
202
1= Eu
( )2
20
2 =
dV
( )( )( )23
2NmC12
m10200.02
V 3151085.8 2
2
=
3J/m 11.0 =
EPF 0024 PHY II By Dr. John Ojur Dennis 62
2.9 Energy and Power in Current Electricitygy y
Electric Energy can be easily transformed intoElectric Energy can be easily transformed intoother forms of energy.
E.g. Electric motors transform electric energyinto mechanical energy, Heaters transforminto mechanical energy, Heaters transformelectric energy into heat energy, while lightbulbs transform electric energy into light energy.
EPF 0024 PHY II By Dr. John Ojur Dennis 63
For a small amount of charge Q moved acrossa p. d. V, U changes by the amount
( )VQU = (2 24)Power P is the time rate of change of energy,
and therefore
( )Q (2.24)and therefore
( ) IVtVQ
tUP =
== (2.25)
SI unit of power is the watt (W).tt
EPF 0024 PHY II By Dr. John Ojur Dennis 64
Eq. 2.25 is power transformed by a deviceEq. 2.25 is power transformed by a devicewhere I is the current passing through it and Vis the potential difference across it
and also gives the power delivered by ag p ysource such as a cell or battery.
EPF 0024 PHY II By Dr. John Ojur Dennis 65
2 9 1 Alternative expressions for power
2.9.1 Alternative expressions for power
For conductors, using V = IR, alternativeexpressions for power are obtained as follows:
( ) RIIRIIVP 2=== (2.26)
orVVVIVP
2
(2 27)R
VR
IVP === (2.27)
EPF 0024 PHY II By Dr. John Ojur Dennis 66
2 9 2 Fire Hazard from Electrical Appliances2.9.2 Fire Hazard from Electrical Appliances
Electrical appliances have small resistance ifElectrical appliances have small resistance. ifthe current through electric wires is large, thewire heats up and produce thermal energy at ap p gyrate which equals to I2R and may start a fire.
To prevent overload, a fuse or circuit breaker isused. These are switches that open the circuitwhen current exceeds some particular value(Fig 2.10).
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Fig. 2.10: (a) fuse (b) circuit breakerFig. 2.10: (a) fuse (b) circuit breaker
(a) Fuses
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(b) Circuit breaker
Fig. 2.11: Connection of appliances to fuseg pp
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ExampleExample
Calculate the resistance of 40 W automobilehead-light designed for 12 V.
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SolutionSolution
( )V12 22V ( ) 6.3
W40V12 22 ===
PVR
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2 9 3 Paying Electrical Bills
2.9.3 Paying Electrical Bills
Electric bills are paid based on total electricenergy U used. From U = P t we find analternative unit for energy that isalternative unit for energy, that is,1 joule (J) = 1 watt second (Ws).
Larger units such a kilowatt-hour (kWh) maybe used for energy 1kWh = 1000 W 3600 sbe used for energy. 1kWh 1000 W 3600 s= 3.6 106 Ws = 3.6 106 J.
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Example 1Example 1
An electric heater draws 15.0 A on a 240 Vline How much energy does it use and howline. How much energy does it use and howmuch does it cost per month (30 days) if itoperates 3 hours per day and the rate is 10.5p p ycents per kWh. Assume the current flowssteadily in one direction.
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SolutionSolution
= IVP
( )( ) kWh 324 Wh10324d 30 h/d 3V 240A 15 3 ===== IVtPtU
34.02 $ $/kWh 0.105 kWh 324Cost ==
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Example 2Example 2
A 75-W light bulb operates on a p. d. of 95 V.Find (a) the current in the bulb and (b) theresistance of the bulb. (c) If this bulb is replacedwith one whose resistance is half the valuef d i t (b) i it ti tfound in part (b), is its power rating greater orless than 75-W? By what factor?
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SolutionSolution
75 W 0.79 A95 V
PIV
= = =(a)
2 2(95 V) 120VR = = = (b) 120 75 W
RP
= = = (b)
(c) greater by a factor of 2
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Example 3Example 3
Find the power dissipated in a 25- electrich t t d t 120 V tl theater connected to a 120-V outlet.
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SolutionSolution
2
=R
VP
( ) W58025V 120
2
==R
25
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Example 4Example 4
The current in a 120-V reading lamp is 2.3 A.If the cost of electrical energy is $ 0.075 perIf the cost of electrical energy is $ 0.075 perkilowatt-hour, how much does it cost tooperate the light for an hour?
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SolutionSolution
( )( )( )hr 1V 120A 3.2 === IVtPtU
kWh 276.0 Wh276 ==
( )( )021.0$
kWh$ 075.0kWh 276.0Cost==
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