electrostatic for murni
TRANSCRIPT
-
8/3/2019 Electrostatic for Murni
1/55
1
Chapter 11:Electrostatics
The study of electriccharges at rest, the
forces between themand the electric fields
associated with them.
-
8/3/2019 Electrostatic for Murni
2/55
2
CHAPTER 11ELECTROSTATICS (4 HOURS)
LESSON 1
OBJECTIVE:
a) Identify types of charges
b) State Coulomb's Law
c) Apply Coulombs Law for a system of point charges
2
04 r
QqF
2r
kQqF
-
8/3/2019 Electrostatic for Murni
3/55
3
Electrostatics - is the study of electrical charges in staticcondition.
-a matter consists of positivelycharged nucleus and negatively charged
electrons at outer region of the matter.
-
8/3/2019 Electrostatic for Murni
4/55
4
11.1.1 Identify type of charges
The electric charge has the following important properties :
There are two kinds of charges in nature positive and negativecharge.
opposite sign attract one another attractive force.
same sign repel one another repulsive force.
The total charge in an isolated system is constant(conserved)Principle of conservation of charges
Charge is quantized.
-
8/3/2019 Electrostatic for Murni
5/55
5
The force between two charges is inversely proportional tothe distance between two charges.
The number of charges is conserved- Charge is not created, only exchanged
- Objects become charged because negative charge istransferred from one object to another
11.1.1 Identify type of charges
-
8/3/2019 Electrostatic for Murni
6/55
6
11.1.2 Coulombs Law
Statesthe magnitude of the electrostatic (Coulomb/electric)force between two point charges is proportional to theproduct of the charges and inversely proportional to the
square of the distance between them.
+ +
r
2qQF
F
-
8/3/2019 Electrostatic for Murni
7/55
7
Mathematically,
chargespointobetween twdistance:r
2
k
r
Qq
F
-229 CmN1009constant(Coulomb)ticelectrosta:k x.
2r
QqF
where
force(Coulomb)ticelectrostaofmagnitude:F
chargeofmagnitude:,qQ
-
8/3/2019 Electrostatic for Murni
8/55
8
Since
04
1k
, hence the Coulombs law can be written as
2
04
1
r
QqF
where
air)or(vacuumspacefreeoftypermittivi:0
).( 212120 mNC10x858
-
8/3/2019 Electrostatic for Murni
9/55
9
If q1 and q2 are charges of opposite sign, the force (F) actingon each charge is attractive as shown in figure below.
This mean that Fis directed towards the neighbouringcharge and will result in both charges moving towards eachother.
If q1 and q2 are both positive or both negative charges, the
force (F) acting on each charge is repulsive. This mean that Fis directed away from the neighbouring
charge and will result in a separation of the two charges ifthey are free to move.
+
r-2q1q F
F
-
8/3/2019 Electrostatic for Murni
10/55
10
Example 1 :
Two point charges, q1=-20 nC and q2=90 nC, are separated by a
distance of 4.0 cm as shown in figure below.
Find the magnitude and direction of
a. the electric force that q1 exerts on q2.
b. the electric force that q2 exerts on q1.
(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)
Solution: q1=2.0 10-8C, q2=9.0 10
-8C, r=4.0 10-2m
21F12 chargeonchargebyforce:
-
cm0.4
+ 2q1q
-
cm0.4
+ 2q1q 12F
21F
where
12F21 chargeonchargebyforce:
-
8/3/2019 Electrostatic for Murni
11/55
11
a. By applying the Coulombs law equation :
b. By using the Coulombs law equation :
Conclusion :
The magnitude of both forces is the same but opposite in direction
obey the Newtons third law.
The characteristic of electric force exert on both charges isattractive force.
22
889
12
)104(
)100.9)(100.2)(100.9(
F
2112 FF
2
2112
k
r
qqF
N100.1 212
F Direction : to the left (q1)
2
12
21
k
r
qq
F
N100.1 221
F Direction : to the right (q2)
-
8/3/2019 Electrostatic for Murni
12/55
12
Example 2 :
Two rain drop A and B falling side by side 10mm
apart carry charges of +4.0 pC and 5.0 pCrespectively. What is the force which one rain dropact on the other?
(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)
-
mm0.10
+ 2q1q
-
8/3/2019 Electrostatic for Murni
13/55
13
SOLUTION:
By applying the Coulombs law equation :
By using the Coulombs law equation :
Conclusion :
This is an attractive force since the two rain drop areoppositely charged.
23
12129
12
)1010(
)100.5)(100.4)(100.9(
F
2
2112
k
r
qqF
N108.1 912
F Direction : to the left (q1)
2
1221
k
r
qqF
N108.1 921
F Direction : to the right (q2)
-
8/3/2019 Electrostatic for Murni
14/55
14
Example 2 : (exercise)Two point charges are placed on the x-axis as follows :Charge q1 = +4.00 nC is located at x = 0.200 m, charge q2 = +5.00 nC is atx = -0.300 m. Find the magnitude and direction of the total electric force
exerted by these two charges on a negative point charge q3 = -6.00 nC thatis placed at the origin.(Young & freedman,pg.829,no.21.20)(Given 0=8.85 x 10-12 C2 N-1 m-2)Ans. : 2.4 N to the right
-
8/3/2019 Electrostatic for Murni
15/55
15
CHAPTER 11
ELECTROSTATICS (4 HOURS)
LESSON 2
OBJECTIVE:
a) Explain electric Field
b) Define electric field strength,
c) Sketch the electric field lines of isolated point charge,
two charges and parallel plate of uniform charge.
oq
FE
-
8/3/2019 Electrostatic for Murni
16/55
16
11.2 Electric Field
Definition is defined as a region of space around isolated chargewhere an electric force is experienced if a positivetest charge placed in the region.
Electric field around charges can be represented by drawing a seriesof lines. These lines are called electric field lines (lines of force).
The direction of electric field is tangent to the electric field line at eachpoint.
-
8/3/2019 Electrostatic for Murni
17/55
17
Figures below show the electric field patterns around the charge.
a. Single positive charge
(the lines point radially inward
toward the charge)
b. Single negative charge
(the lines point radially outwardfrom the charge)
+q -q
Field direction
-
8/3/2019 Electrostatic for Murni
18/55
18
c. Two equal point charges of opposite sign, +q andq
+q -q
(the lines are curved and they are directedfrom the positive charge to the negativecharge.
Field direction
-
8/3/2019 Electrostatic for Murni
19/55
19
d. Two equal positive charges, +q and +q
(pointXis neutral point )
is defined as a point (region) where the totalelectric force is zero.
It lies along the vertical dash line.
+q +qX
Field direction
-
8/3/2019 Electrostatic for Murni
20/55
20
e. Two opposite unequal charges, +2q andq
(note that twice as many lines leave +2q as there
are lines entering
q,
number of lines isproportional to magnitude of charge.)
+2q-q
Field direction
-
8/3/2019 Electrostatic for Murni
21/55
21
f. Two opposite charged parallel metal plates
The electric field lines are perpendicular to the surface of
the metal plates.
The lines go directly from positive plate to the negativeplate.
-
8/3/2019 Electrostatic for Murni
22/55
22
The properties of electric field lines:
the field points in the direction tangent to the field line atany point
closer the lines, the stronger the field.
Electric field lines start on positive charges and end onnegative charges, and the number starting or ending isproportional to the magnitude of the charge.
The field lines never crossbecause the electric field donthave two value at the same point.
-
8/3/2019 Electrostatic for Murni
23/55
-
8/3/2019 Electrostatic for Murni
24/55
24
g. Two equal negative charges, +q and -q(exercise).
h. Two unequal negative charges, -2q and +q(exercise).
+q -q
+q-2q
-
8/3/2019 Electrostatic for Murni
25/55
25
11.2.1Electric Field Strength (intensity), The electric field strength at a point,
Definition is defined as the electric (electrostatic) force per unitpositive charge that acts at that point in the samedirection as the force.
Mathematically,
It is a vector quantity.
The units of electric field strength is N C-1 or V m-1.
E
0q
FE forceelectrictheofmagnitude:F
where
chargetestofmagnitude:0q
strengthfieldelectrictheofmagnitude:E
E
-
8/3/2019 Electrostatic for Murni
26/55
26
11.2.1Electric Field Strength (intensity),
Since
E
2
0
r
kqqF
, then the equation above can be written as
0
2
0
q
rkqq
E2r
kqE or 2
0r4
qE
where
chargepointisolatedofmagnitude:q
chargepointisolatedandpointebetween thdistance:r
Note :
-
8/3/2019 Electrostatic for Murni
27/55
27
Note :
The direction of the electric field strength,Edepends on the signof isolated point charge.
The direction of the electric force, Fdepends on the sign of
isolated point charge and test charge. For example A positive isolated point charge.
a. positive test charge
b. negative test charge
q)( veq0 EF
r
q)( veq0 E
F
r
A negative isolated point charge
-
8/3/2019 Electrostatic for Murni
28/55
28
A negative isolated point charge.
a. positive test charge
b. negative test charge
In the calculation of magnitudeE, substitute the magnitude ofthe charge only.
q )( veq0 E
F
r
q )( veq0 E
F
r
-
8/3/2019 Electrostatic for Murni
29/55
29
Example 3 :
A small object carrying a charge of +20 C
experiences a force of 6.0 105 N at angle 300 whenplaced at a point in an electric field. What are themagnitude and direction of the electric field at thatpoint?
Solution:qEF
q
FE
C1020
N100.66
5
E
1NC3
E
Example 4 :
-
8/3/2019 Electrostatic for Murni
30/55
30
Example 4 :
Two point charges, q1=1 C and q2=-4 C, are placed 2 cm and 3 cmfrom the point A respectively as shown in figure below.
Find
a. the magnitude and direction of the electric field intensity at point A.
b. the total electric force exerted on q0=-4 C if it is placed at point A.
(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)
Solution: q1=1 C, q2=4 C, q0=4 C, r1=2x10-2 m, r2=3x10-2 m
a. By applying the equation of electric field strength, the magnitude of
Eat point A.Due to q1 :
+ - 2q1q
cm2 cm3
A
+ - 2q1q
cm2cm3
A 1AE 2AE
22
9
2
1
11A
10x2
110x09
r
kqE
)(
))(.(
113
1A CN10x252E
. Direction : to the right (q2)
Due to q : 9 410x09kq ))((
-
8/3/2019 Electrostatic for Murni
31/55
31
Due to q2 :
therefore the electric field strength at point A due to the charges isgiven by
b. From the definition of the electric field strength,
thus the total electric force exerted on q0 is given by
222
2
22A
10x3
410x09
r
kqE
)(
))(.(
113
2A CN10x4E Direction : to the right (q2)
2A1AA EEE
0
AA
q
FE
1313
A 10x410x252E .
113
ACN10x256E .
Direction : to the right (q2)
A0A EqF
N10x52F 14A .
).)(( 13A 10x2564F
Direction : to the left (q1)
-
8/3/2019 Electrostatic for Murni
32/55
32
CHAPTER 11
ELECTROSTATICS (4 HOURS)
LESSON 3OBJECTIVE:
a) Explain quantitatively with the aid of a diagram the
motion of a charge in a electric field.
-
8/3/2019 Electrostatic for Murni
33/55
33
Direction of a moving charge which is parallel to
E
E
Charges are released in an electric field
The positive charge moves in the direction ofthe field
+
+
+
+
-
-
-
-
-
8/3/2019 Electrostatic for Murni
34/55
34
Direction of a moving charge which is parallel to
E
E
Charges are released in an electric field
The negative charge moves in the opposite direction
+
+
+
+
+
+
-
-
-
-
-
-
-
8/3/2019 Electrostatic for Murni
35/55
CHAPTER 11
-
8/3/2019 Electrostatic for Murni
36/55
36
CHAPTER 11
ELECTROSTATICS (4 HOURS)
LESSON 4
OBJECTIVE:a) Define electric potential.
b) Use for a point charge and a system of charges.
c) Define potential different between two points.
d) Use to calculate the potential difference
between point A and point B
r
QV
o4
BAAB VVV
0q
WV
-
8/3/2019 Electrostatic for Murni
37/55
37
Electric potential, Vof a point in the electric field
Definition is defined as the work done in bringing positive testcharge from infinity to that point in the electric field.
or
0q
WV
then the equation above can be written as
donework:W
chargetest:0q
where
r
kqqW 0
11.3 Electric Potential, V
-
8/3/2019 Electrostatic for Murni
38/55
38
Since
0
0
q
rkqq
V
orr
q
4
1V
0
r
kqV
chargepoint:q
chargepointthepoint withebetween thdistance:rwhere
spacefreeoftypermittivi:0 ).(21212
0 mNC10x858
11.3 Electric Potential, V
-
8/3/2019 Electrostatic for Murni
39/55
39
Electric potential is a scalar quantity.
The S.I. unit for electric potential is the Volt (V) or
J C-1.
The total electric potential at a point in space isequal to the algebraic sum of the constituentpotentials at that point.
-
8/3/2019 Electrostatic for Murni
40/55
40
Note :
The theoretical zero of electric potential of a charge is atinfinity.
The electric potential energy of a positively charged particle
increases when it moves to a point of higher potential.
The electric potential energy of a negatively chargedparticle increases when it moves to a point of lowerpotential.
-
8/3/2019 Electrostatic for Murni
41/55
41
Since charge q can be positive or negative, the electricpotential can also be positive or negative.
work done is negativework done by the electric force(system).
work done is positivework done by the external forceor on the system.
In the calculation of V, the sign of the charge must be
substituted in the equation of V.
Example 5 :
-
8/3/2019 Electrostatic for Murni
42/55
42
Figure below shows a point A at distance 10 m from the positive pointcharge, q=5C.
Calculate the electric potential at point A and describe the meaning ofthe answer.
(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)
Solution:q=5 C, r=10 m
By applying the equation of the electric potential at a point,
)10(
)5)(100.9( 9r
kqVA
19 @105.4 CJVVA
Meaning : 4.5 109 joule of work is done in bringing 1 C positivecharge from infinity to the point A.
+qA
m10
Example 6 :
-
8/3/2019 Electrostatic for Murni
43/55
43
Two point charges, q1=+0.3 C and q2=-0.4 C are separated by adistance of 6 m as shown in figure below.
Calculatea. the electric field strength andb. the electric potentialat point A ( 3 m from the charge q1).(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)
Solution:q1=+0.3 C, q2=-0.4 C
a. By applying the equation of electric field strength, the magnitude of
Eat point A.Due to q1 :
+ - 2q1qA
m6
+ - 2q1qA
m3r1 m3r2
1AE
2AE
2
9
2
1
11
)3(
)3.0)(100.9(
r
kqEA
18
1 103 CNEA Direction : to the right (q2)
Due to q2 :9
2 )4.0)(100.9( kqE
-
8/3/2019 Electrostatic for Murni
44/55
44
q2
therefore the electric field strength at point A due to the charges isgiven by
b. By applying the equation of electric potential, the value of Vat pointA is
22
2
22
)3(
))((
r
qEA
18
2 CN104AE Direction : to the right (q2)
2A1AA EEE
2A1AA VVV
88 104103 AE
18CN107 AE
Direction : to the right (q2)
2
2
1
1
2
2
1
1A
r
q
r
qk
r
kq
r
kqV
3
4.0
3
3.0100.9 9AV
V103 8AV
Example 7 :
-
8/3/2019 Electrostatic for Murni
45/55
45
Two point charges, q1=+12 nC and q2=-12 nC are separated by adistance of 8 cm as shown in figure below.
Determine the electric potential at point P( 6 cm from the charge q2).(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)
Solution:q1=+1210-9C, q2=-12 10
-9C
1q + - 2q
P
m8 c
m6c
1q+ -
2q
P
m1082
m106 22rm1010
21
r
-
8/3/2019 Electrostatic for Murni
46/55
Potential Difference
-
8/3/2019 Electrostatic for Murni
47/55
47
Potential Difference
Potential difference between two points in an electric field,
Definition is defined as the work done in bringing a positive testcharge from a point to another point in the electric field.
From the figure 3.8a, the potential difference between point A and B, VABis given by
0
BAAB
q
WV BAAB
VVV and
0
BABAqWVV
or
A.pointtoBpointfromchargetestpositivebringingindonework:BAWwhere
Apointatpotentialelectric:AV
Bpointatpotentialelectric:BVchargetest:0q
Note :
-
8/3/2019 Electrostatic for Murni
48/55
48
If the positive test charge moving from point A to point B, thus thepotential difference between this points is given by
therefore
0q
W
VVVAB
ABBA
B.pointA topointfrom
chargetestpositivebringingindonework:ABW
where
ApointandBpointbetweendifferencepotential:BAV
BAAB VV
-
8/3/2019 Electrostatic for Murni
49/55
49
Example 9:
Two point charges q1=+2.40 nC and q2=-6.50 nC are 0.100 mapart. Point A is midway between them, point B is 0.080 m
from q1 and 0.060 m from q2 as shown in figure below.
(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)
1
q+ -
2
q
B
A
m0600 .m0800 .
m0500. m0500 .
Find
a. the electric potential at point A,
b. the electric potential at point B,
c. the work done by the electric field
on a charge of 2.5 nC that travels
from point B to point A.
(Young &
freedman,pg.900,no.23.21)
-
8/3/2019 Electrostatic for Murni
50/55
50
Solution:
q1=+2.4010-9 C, q2=-6.50 10
-9 C,
r1A=r2A=0.050 m, r1B=0.080 m , r2B=0.060 m
a. By applying the equation of electric potential, the value of Vat point
A is
A2A1A VVV
V738VA
A2
2
A1
1A
r
kq
r
kqV
b. By applying the equation of electric potential, the
-
8/3/2019 Electrostatic for Murni
51/55
51
y pp y g q p ,
value of Vat pointB is
c. Given q0=2.5010-9C
The work done in bringing charge, q0 from point B topoint A is
given by
B2B1B VVV
V750VB
B2
2
B1
1B
rkq
rkqV
AB0BAVqW
J103 8 WBA
)( BA0BA VVqW
-
8/3/2019 Electrostatic for Murni
52/55
52
Example 10 :
A test charge q0=+2.3x10-4 C is 5 cm from a point charge q. A work doneof +4 J is required to overcome the electrostatic force to bring the testcharge q0 to a distance
8 cm from charge q.
Calculate :
a. the potential difference between point 8 cm and 5 cm from the pointcharge, q.
b. the value of charge q.
c. the magnitude of the electric field strength for charge q0 at point 5 cm
from the charge q(given Coulombs constant, k = 9.0109 N m2 C-2)
-
8/3/2019 Electrostatic for Murni
53/55
53
Solution: q0=+2.3010-4 C
a. Given WAB= +4J,
From the figure above, rA= 510-2 m, rB= 8 10-2 mBy applying the equation of potential difference, the value of VBA is
0
ABBA
q
WV
V1074.1 4 VBA
qBA
m105 2 m108 2
eF
F
-
8/3/2019 Electrostatic for Murni
54/55
54
b. The electric potential at point A due to pointcharge, q :
)105(
q)109(k2
9
A
Ar
qV
q108.1 11AV
The electric potential at point B due to point charge, q :
-
8/3/2019 Electrostatic for Murni
55/55
The potential difference between point A and B is
c. By using the equation of electric field strength, thus
VVV BAAB 1074.14BAAB VVV
C1058.2 7 q
q10x125.1q10x8.110x74.1 11114
2
k
A
Ar
qE
and
15
CN1029.9
EA
q10125.1)108(
q)109(k 112
9
B
Br
qV