Electrostatic for Murni

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    Chapter 11:Electrostatics

    The study of electriccharges at rest, the

    forces between themand the electric fields

    associated with them.

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    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

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    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.

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    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.

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    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

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    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

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    Mathematically,

    chargespointobetween twdistance:r

    2

    k

    r

    Qq

    F

    -229 CmN1009constant(Coulomb)ticelectrosta:k x.

    2r

    QqF

    where

    force(Coulomb)ticelectrostaofmagnitude:F

    chargeofmagnitude:,qQ

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    Since

    04

    1k

    , hence the Coulombs law can be written as

    2

    04

    1

    r

    QqF

    where

    air)or(vacuumspacefreeoftypermittivi:0

    ).( 212120 mNC10x858

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    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

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    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:

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    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)

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    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

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    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)

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    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

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    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

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    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.

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    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

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    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

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    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

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    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

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    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.

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    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.

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    g. Two equal negative charges, +q and -q(exercise).

    h. Two unequal negative charges, -2q and +q(exercise).

    +q -q

    +q-2q

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    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

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    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 :

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    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

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    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

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    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 :

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    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 ))((

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    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)

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    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.

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    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

    +

    +

    +

    +

    -

    -

    -

    -

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    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

    +

    +

    +

    +

    +

    +

    -

    -

    -

    -

    -

    -

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    CHAPTER 11

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    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

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    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

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    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

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    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.

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    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.

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    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 :

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    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 :

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    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

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    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 :

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    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

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    Potential Difference

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    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 :

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    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

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    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)

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    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

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    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

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    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)

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    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

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    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 :

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    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